Physics Problems Q&A - Recent questions and answers in Physics Problems
http://physics.qandaexchange.com/?qa=qa/physics-problems
Powered by Question2AnswerAnswered: SHM with elastic collision.
http://physics.qandaexchange.com/?qa=3260/shm-with-elastic-collision&show=3265#a3265
<p>The free oscillation (without the wall) can be described by $\phi=\beta\cos\omega t$ where angle $\phi$ made with the vertical is +ve to the right, and $\omega=\sqrt{\frac{g}{\ell}}$.</p>
<p>The time $t_1$ taken from $\phi=+\beta$ to $\phi=-\alpha$ is given by $-\alpha=\beta\cos\omega t_1$. So $$\frac{\alpha}{\beta}=-\cos\omega t_1=\sin(\omega t_1-\frac{\pi}{2})$$ $$\omega t_1=\frac{\pi}{2}+\sin^{-1}\frac{\alpha}{\beta}$$</p>
<p>The period of oscillation is $$T=2t_1=\frac{2}{\omega}(\frac{\pi}{2}+\sin^{-1}\frac{\alpha}{\beta})=2\sqrt{\frac{\ell}{g}}(\frac{\pi}{2}+\sin^{-1}\frac{\alpha}{\beta})$$</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3260/shm-with-elastic-collision&show=3265#a3265Fri, 16 Nov 2018 16:23:52 +0000Questions from a jumping kangaroo
http://physics.qandaexchange.com/?qa=3263/questions-from-a-jumping-kangaroo
<p><img src="http://i68.tinypic.com/2lc8wnn.jpg" alt=""></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3263/questions-from-a-jumping-kangarooFri, 16 Nov 2018 15:38:20 +0000Diffusion with reflecting boundary
http://physics.qandaexchange.com/?qa=3254/diffusion-with-reflecting-boundary
<p><img src="https://cdn.pbrd.co/images/HNaMhdt.png" alt=""></p>
<p>What I have tried:</p>
<p><img src="https://cdn.pbrd.co/images/HNaMSPP.jpg" alt=""></p>
<p><a rel="nofollow" href="https://imgur.com/a/ESWy1k5">https://imgur.com/a/ESWy1k5</a> (sorry but I did not manage to use imgur as you instructed).</p>
<p>Note that my question is that how is it possible that I get $t = \infty$. It does not make sense for me (physically speaking). Where did I get wrong?</p>
<p>Please if there is information you require let me know. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3254/diffusion-with-reflecting-boundaryWed, 14 Nov 2018 20:07:35 +0000Answered: Angle made by the plane of hemisphere with inclined plane
http://physics.qandaexchange.com/?qa=3241/angle-made-by-the-plane-of-hemisphere-with-inclined-plane&show=3244#a3244
<p><img src="https://i.imgur.com/8DvcKSs.png[/img]" alt=""><br>
Centroid C is at the midpoint of the axis OA of the hemisphere, ie distance OC=R/2. In equilibrium position C must lie vertically above point of contact P.</p>
<p>Apply Sine Rule in triangle OPC : $$\frac{R/2}{\sin\theta}=\frac{R}{\sin\beta}$$ $$\sin\beta=2\sin\theta=2\sin30^{\circ}=1$$ $$\beta=180^{\circ}-(\theta+\alpha)=90^{\circ}$$ $$\alpha=90^{\circ}-\theta=60^{\circ}$$</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3241/angle-made-by-the-plane-of-hemisphere-with-inclined-plane&show=3244#a3244Mon, 05 Nov 2018 14:15:06 +0000Radial distribution of particle separation in a liquid at small distances
http://physics.qandaexchange.com/?qa=3233/radial-distribution-particle-separation-liquid-distances
<blockquote><p>Draw schematically the radial distribution function $g(r)$ for a Lennard-Jones fluid at low and high particle densities. Discuss in both cases the behavior at small $r$. At high densities $g(r)$ has some damped oscillations. Explain their origin and what would happen at long distances.</p>
</blockquote>
<p><strong>What I know:</strong></p>
<p>The radial distribution function $g(r)$ describes how density varies as a function of distance from a reference particle (<em>Wikipedia</em>).</p>
<p>Specifically in Lennard-Jones case, we have (Pasteboard still does not work for me):</p>
<p><a rel="nofollow" href="https://imgur.com/a/pvXgax4">https://imgur.com/a/pvXgax4</a></p>
<p>Here we can observe that if repulsion outweighs attraction, the curve will die out on the x axis. On the contrary, if it is the other way around, the curve will grow exponentially. For the sake of clarity, I would remark y = 0 as the Ideal Gas behaviour. </p>
<p>I have read that $g(r)$ vanishes at short distances, because the probability of finding two particles close to each other vanishes due to the repulsive part of the potential. At high densities $g(r)$ can show some damped oscillations: <a rel="nofollow" href="https://imgur.com/a/i32D39I">https://imgur.com/a/i32D39I</a>. </p>
<p>These oscillations express the preference of the particles to be found at specific distances from a reference particle at the origin. For instance in the LJ case, a first layer of particles will be localized closed to the minimum of LJ's potential, which is the origin of the first peak in $g(r)$.</p>
<p>This layer prevents other particles from getting close to it, which is what causes the first<br>
minimum in $g(r)$.</p>
<p><strong>What I do not know.</strong></p>
<p>1)The explanation: 'These oscillations express the preference of the particles to be found at specific distances from a reference particle at the origin'. <strong>How can particles have 'preference' to be at different distances? I think this is not a good physical explanation. I have been thinking about this and I would say that as there is a high density of particles at small distances, the repulsion force is exerted on these, which triggers such oscillations. Finally there is a point where they are so close to each other that the repulsion force provides them with an initial kinetic energy which will be equal to a final electrostatic potential energy of zero (selecting our zero of electrostatic potential energy at y=0) , and that is the moment when the curve dies out on the x axis. Do you agree with this explanation?</strong></p>
<p>2)Are we dealing with underdamped oscillations at high densities? Could you give an insight into these kind of oscillations (if what I have just said is not enough or incorrect)?</p>
<p>3)When we are dealing with long distances, would $g(r)$ tend to be one? What would we observe, the Ideal Gas behaviour? Why? Is it because there is no interaction between the particles?</p>
<p>As always, I am interested in explaining such phenomena from a physical point of view. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3233/radial-distribution-particle-separation-liquid-distancesThu, 01 Nov 2018 16:00:05 +0000Liquid in a capacitor as dielectric (IE Irodov 3.144)
http://physics.qandaexchange.com/?qa=3225/liquid-in-a-capacitor-as-dielectric-ie-irodov-3-144
<blockquote><p>A parallel plate capacitor is located horizontally so that one of its plates is submerged into the liquid while the other is over its surface. The permittivity of the liquid is equal to $\epsilon$, its density is equal to $\rho$. To what height will the level of the liquid in the capacitor rise after its plates get a charge of surface charge density $\sigma$?</p>
</blockquote>
<p>This is a type of controversial problem, as it contains a different answer by the different author, two different answers provided for this are </p>
<p> $$1) \ h=\dfrac{(\epsilon-1)\sigma^2}{2\epsilon_o\epsilon\rho g}$$ $$2) \ h=\dfrac{(\epsilon^2-1)\sigma^2}{2\epsilon_o\epsilon^2\rho g}$$</p>
<p><a rel="nofollow" href="https://ibb.co/fYmek0">Solution for 1</a> and <a rel="nofollow" href="https://youtu.be/oyfo-TJEG28?t=31">solution for 2</a></p>
<p>Please help!</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3225/liquid-in-a-capacitor-as-dielectric-ie-irodov-3-144Tue, 30 Oct 2018 01:37:42 +0000Answered: Diffusion with an even number of reflecting boundaries
http://physics.qandaexchange.com/?qa=3211/diffusion-with-an-even-number-of-reflecting-boundaries&show=3220#a3220
<p><strong>(a)</strong></p>
<p>Your prediction is correct. With reflecting boundaries the limiting concentration tends to a uniform distribution. With absorbing boundaries the limiting concentration tends to zero.</p>
<p>The type of boundary makes a difference. The starting concentration is the same, the end concentration is different, so there must be an increasing difference in between the two cases. One similarity is that in both cases the final distribution is <strong>uniform</strong> : with reflecting boundaries it is a uniform non-zero value, with absorbing boundaries it is a uniform zero value.</p>
<p>The question asks you to explain the behaviour at the boundaries. The concentration at the reflecting boundary rises steadily to the final limiting value. I interpreted this as asking about $\frac{\partial c}{\partial t}$. However, perhaps it is asking about $\frac{\partial c}{\partial x}$, prompting you to state that $\frac{\partial c}{\partial x}=0$ as you suggested.</p>
<p><strong>(b)</strong></p>
<p>The limiting concentration is $N/a$, which should be marked on your sketch. It is $N/a$ because concentration in 1D is number of particles per unit length : there are $N$ particles spread uniformly over a length of $a$.</p>
<p>Yes that is a good point : $\frac{\partial c}{\partial x}=0$ at reflecting boundaries. This is not true for higher derivatives. </p>
<p><strong>Why $\frac{\partial c}{\partial x}=0$ at a reflecting boundary</strong> </p>
<p>Consider first the distribution without the RH reflecting boundary. Close to any point the distribution is approximately linear, eg $c=A(\frac{a}{2}-x)+B$ near the RH boundary position $x=+\frac{a}{2}$. The reflection of this concentration in the RH boundary is $c'=A(x-\frac{a}{2})+B$. When the reflecting boundary is replaced, the concentration to the left of the RH boundary is the sum of the unreflected distribution and its reflection, ie $c'+c=2B$, which is a constant.</p>
<p>It is not obvious to me how this condition could explain the limiting distribution. I do not think it does explain it. However, it does fulfill the requirement for a reflecting boundary.</p>
<p><strong>Why the limiting distribution is uniform</strong></p>
<p>Remove the boundaries. Divide the x axis into an infinite number of intervals or boxes of length $a$ to the left and right of the box at the origin between $x= \pm a/2$. </p>
<p>As the initial distribution evolves it spreads into the adjacent boxes, becoming broader and flatter as well as smaller in height. The highest concentration is always in the central box, and concentration falls monotonically in both directions outside this box. After a long time the distribution is far broader than the box width $a$. In each box the concentration is approximately constant, with a small linear decrease or increase to right and left of centre respectively. </p>
<p>When the LH and RH reflecting boundaries of the central box are replaced, the distribution in the central box becomes the sum of the distributions in all adjacent boxes. Distributions in odd numbered boxes are reversed before they are added. Because there are approximately equal numbers of boxes with decreasing/increasing concentrations, and approximately equal numbers of odd and even boxes, the linear increasing/decreasing contributions tend to cancel out more exactly as the number of boxes increases. The constant contributions all add up to another constant function, ie a uniform distribution, which is equal in area to the area under the spreading Gaussian function.</p>
<p>This is an <strong>averaging process</strong>. The distribution is continuously expanded and chopped into an increasing number of slices, and these slices are added together. Every part of the distribution gets mixed with every other part. The differences, such as peaks and troughs, are averaged out.</p>
<p><strong>(c)</strong></p>
<p>I do not see why the author thinks that this problem is the same as drift-diffusion in a square potential well. It is not the same. There is no drift here. The peaks in concentration remain at their initial positions, they do not move.</p>
<p><strong>Why drift does not affect the final distribution</strong></p>
<p>If the initial concentration drifts as well as diffuses then there is no difference in the final distribution, which is again uniform. The reason is that drift does not prevent the distribution from spreading out. It continues to diffuse at the same rate. Also, the peak concentration is confined to the central box; the concentration decays to left and right of the centre just as it did without drift. </p>
<p>The same averaging process as above takes place : the distribution is chopped up into a larger and larger number of slices, which are added together - outer, middle and inner portions of the distribution get mixed together and lose their individual features. </p>
<p><strong>The Boltzmann Equation</strong></p>
<p>The Boltzmann Equation does not seem to be of any use in this problem. It relates particle numbers $n$ and energy levels. In an infinite square well the particle KE does not change because the particles cannot gain enough energy to escape. The kinetic energy of each particle is also fixed : either by definition, or because all collisions which the particle make are elastic. </p>
<p>(In the Random Walk model of diffusion, <strong>ideal</strong> particles change direction at random, without any cause, and keep the same speed. However, real diffusing particles only change direction because of causes such as collisions. These are either collisions with invisible particles, such as smoke particles colliding with much smaller air molecules, or collisions with each other. This leads to a distribution of speeds, such as the <strong>Maxwell-Boltzmann Distribution</strong>. However, for most purposes each diffusing particle is assumed to have the same <strong>root-mean-square speed</strong>.) </p>
<p>Another scenario is continuous boundaries, in which a particle which exits across the left boundary re-enters immediately across the right boundary. The approach to equilibrium is not quite the same as with reflecting boundaries, but the limiting distribution is identical.</p>
<p><strong>(d)</strong> </p>
<p>I do not understand what this is asking for. You already have an expression for $g_n(x,t)$ from earlier. What else is there to be done here?</p>
<p><strong>Why an infinite number of Gaussians is required to model the boundary conditions</strong></p>
<p>The explanation you found seems to be as follows :</p>
<p>Start with a diffusing Gaussian distribution at $x=0$. If you place an identical 2nd Gaussian distribution at $x=+a$ then the RH boundary at the midpoint $x=+a/2$ will always satisfy the boundary condition $\frac{\partial c}{\partial x}=0$ because of symmetry. However, the bc does not hold at $x=-a/2$. </p>
<p>If you add a 3rd Gaussian at $x=-a$ then the LH boundary at $x=-a/2$ is not symmetrical, and neither now is the RH boundary, so the bc does not hold at either boundary. </p>
<p>However, adding more pairs of Gaussians either side of the centre makes the boundaries at $x=\pm a/2$ look more symmetrical. The number of Gaussians either side differs by only one. The final 'odd' Gaussian is so far away that only the tail of this Gaussian overlaps the boundary, so the effect it makes on the bc gets smaller and smaller as it gets further away from the centre.</p>
<p>In the limit, with an infinite comb of Gaussians, both boundaries at $x=\pm a/2$ are symmetrically placed because each has an infinite number of Gaussians on either side of it, every one with its mirror image. There can be at most one extra Gaussian on the opposite side of the centre, but this is so far away that is has an insignificant effect on the bc $\frac{\partial c}{\partial x}=0$, which will again hold.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3211/diffusion-with-an-even-number-of-reflecting-boundaries&show=3220#a3220Mon, 29 Oct 2018 02:46:20 +0000Induced charge dentisity at boundry surface.
http://physics.qandaexchange.com/?qa=3216/induced-charge-dentisity-at-boundry-surface
<blockquote><p>The space between the plates of a parallel plates capacitor is filled consecutively with two dielectrics layers $1$ and $2$ having the thickness $d_1$ and $d_2$ respectively and permittivities $\epsilon_1$ and $\epsilon_2$. The area of each plate is equal to $S$ find:<br>
1. The capacitance of the capacitor <br>
2. The density $\sigma'$ of the bound charges on the boundary plane if the voltage across the capacitor equals $V$ and the electric field is directed from layer $1$ to layer $2$.</p>
</blockquote>
<p>I managed to find $C_{eq}$ easily, for the second part I don't understand that word " <em>boundary plane</em>", I only know $\sigma'=\sigma\bigg(1-\dfrac{1}{\epsilon}\bigg)$ holds when there would have been just one plate, how to work on two such consecutive plates. Please help.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3216/induced-charge-dentisity-at-boundry-surfaceSun, 28 Oct 2018 05:02:22 +0000Answered: Charges induced on conducting plates.
http://physics.qandaexchange.com/?qa=3210/charges-induced-on-conducting-plates&show=3212#a3212
<p><strong>Revised Answer</strong></p>
<p>(I have again misinterpreted your question, which is simpler than I had anticipated.)</p>
<p>This question is Problem 3.44 in Griffiths' "Introduction to Electrodynamics". A solution is given as Exercise 2 in <a rel="nofollow" href="https://physicspages.com/pdf/Griffiths%20EM/Griffiths%20Problems%2003.43-4.pdf">Physics Pages</a> using <strong>Green's Reciprocity Theorem</strong>, but it is not easy to understand. </p>
<p>In the simplified problem suggested by your author, we have 3 parallel conducting plates A, B, C. The middle plate B carries total charge $+Q$. The other two plates are either neutral or grounded.</p>
<p>Suppose the area of each face of each conductor is $A$ and the surface charges on the left and right faces of plate B are $+\sigma_1$ and $+\sigma_2$ where $\sigma_1+\sigma_2=\frac{Q}{A}$. The surface charges on the adjacent faces of the outer plates A, C must be $-\sigma_1, -\sigma_2$ respectively. This is because there is no electric field inside any of the conductors, so all electric field lines starting on one face of B must end on the adjacent face of A or C.</p>
<p><img src="https://cdn.pbrd.co/images/HKsipIJ.png" alt=""></p>
<p>The electric field due to a plane face with surface charge density $\sigma$ is $E=\frac{\sigma}{2\epsilon_0}$. So the field between B and A is $E_1=\frac{\sigma_1}{\epsilon}$ and that between B and C is $E_2=\frac{\sigma_2}{\epsilon}$.</p>
<p><strong>Assuming plates A and C are either grounded or at the same potential,</strong> then the potential<br>
differences with plate B are equal : $V_{BA}=V_{BC}$ which means that $$E_1 x=E_2 (\ell-x)$$ where $\ell$ is the distance between A and C. </p>
<p>Using the above expressions for surface charge we get $$\sigma_1 x=\sigma_2 (\ell-x)$$ $$Q_1 x=Q_2 (\ell-x)$$ Now $Q_1+Q_2=Q$. Therefore the charges induced on A and C are $$Q_1=-\frac{\ell-x}{\ell}Q$$ $$Q_2=-\frac{x}{\ell}Q$$</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3210/charges-induced-on-conducting-plates&show=3212#a3212Sat, 27 Oct 2018 15:32:09 +0000Answered: Relation between flux through lateral surface of a cylinder and flat parts.
http://physics.qandaexchange.com/?qa=3192/relation-between-flux-through-lateral-surface-cylinder-parts&show=3194#a3194
<p>Your 1st equation can be interpreted as saying that the flux through the section is equal to $\frac{1}{\epsilon_0}$ times the charge $Q$ contained within a <strong>cone</strong> whose base is the section and whose vertex is the centre of the sphere.</p>
<p>This result can be obtained directly using Gauss' Law. The electric field inside a uniformly charged sphere is radial. So the flux across the slanting curved face of the cone is zero, because this face is radial. The only flux out of the cone is across its base. The electric field across this base varies in magnitude and direction. Nevertheless, the total flux across it equals the charge $Q$ enclosed by the cone divided by $\epsilon_0$.</p>
<p>Your 2nd equation is derived using Gauss' Law and gives the total flux through the surface of the cylinder.</p>
<p>The similarity between the formulas for $\phi_1$ and $\phi_2$ is entirely due to geometry. Gauss' Law says that total flux through a surface of <strong>any shape</strong> containing uniform charge density $\rho$ is $$\phi=\frac{\rho V}{\epsilon_0}$$ The only difference for each shape is the volume $V$.</p>
<p>The volumes of cone and cylinder depend in the same way on base area $\pi a^2=\pi (R^2−r_0^2)$ and height $r_0$. They are both of the form $k r_0 \pi a^2$. For a cone $k=\frac13$ while for a cylinder $k=1$.</p>
<p>You can see by looking at extremes that your conjecture (that the flux through the flat ends of a cylinder is $\frac13$ of the total flux) must be false. For a short fat cylinder ($r_0 \ll a$) almost all of the flux will be through the flat ends. For a long thin cylinder ($r_0 \gg a$) almost none of the flux will be through the flat ends.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3192/relation-between-flux-through-lateral-surface-cylinder-parts&show=3194#a3194Mon, 22 Oct 2018 15:56:31 +0000Answered: Oscillations of a rotating mass on a spring
http://physics.qandaexchange.com/?qa=3191/oscillations-of-a-rotating-mass-on-a-spring&show=3193#a3193
<p>The equilibrium length of the spring $r_0$ is not its natural length. $r_0$ is the radius at which the mass orbits the pivot. It increases with $\Omega$.</p>
<p>The question is asking about <strong>oscillations in the radial direction</strong>. </p>
<p>Suppose we have chosen a value for $\Omega$ and the mass is currently at radius $r$. In the rotating frame of reference there is a centrifugal outward force $mr\Omega^2$ and an elastic inward force $k(r-a)$ where $a$ is the <strong>natural length of the spring</strong>. When these two forces are exactly balanced and the mass is rotating at a constant radius about the pivot, this is the equilibrium position. so $$mr_0\Omega^2=k(r_0-a)$$ which you can solve to find $r_0$.</p>
<p>However, we could displace the mass a small distance from its equilibrium position. The forces would then no longer be balanced. If the equilibrium position is stable the mass will return to it with a non-zero velocity, overshoot, and oscillate about the equilibrium position. The frequency $\omega$ of this oscillation is what you are being asked to find.</p>
<p>To find $\omega$ you need to write the equation of radial motion of the mass in the usual form for simple harmonic motion $$\ddot x +\omega^2 x=0$$ To obtain the equation of motion suppose that $r$ is increased by a small amount $x$. Then applying $F=m\ddot x$ we have $$m(r_0+x)\Omega^2-k(r_0+x-a)=m\ddot x$$ This can be simplified by substituting for $r_0$ as found above.</p>
<p><strong>Comment</strong></p>
<p>Energy is not conserved because the motor is doing work to keep the mass, spring and axle rotating at constant angular velocity. For the same reason angular momentum is not conserved either. A constant value of $\Omega$ could otherwise be achieved using a flywheel, or making the mass of the axle very much larger than $m$. In this case energy and momentum are conserved because the system, which is isolated, now consists of the axle, mass and spring instead of only the mass and spring.</p>
<p>If the motor is switched off before the mass is displaced then both energy and angular momentum are conserved. In this case the oscillation frequency would be different, because $\Omega$ would vary as $r$ changes, unless the mass of the axle or flywheel are much larger than $m$.</p>
<p>A related question on this site is <a rel="nofollow" href="http://physics.qandaexchange.com/?qa=3059/spinning-connected-springs-system-feynman-exercises-14-20">Spinning connected springs system</a>.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3191/oscillations-of-a-rotating-mass-on-a-spring&show=3193#a3193Mon, 22 Oct 2018 14:47:35 +0000Deducing the mass of the Sirius system - Feynman exercises 3.17
http://physics.qandaexchange.com/?qa=3185/deducing-the-mass-of-the-sirius-system-feynman-exercises-17
<p><img src="https://cdn.pbrd.co/images/HJlk9yQ.jpg" alt=""></p>
<p>I do not even know how to try to deduce the mass M of the Sirius system in terms of that of the sun, as I do not not how to interpret the given data. </p>
<p>Could you give me a hint? </p>
<p><strong>EDITED AT THIS POINT</strong></p>
<p><strong>1) Estimating the distance to the Sirius binary system.</strong></p>
<p>I calculated the hypotenuse of the triangle $S E_1 SS$ (Sun - position of the Earth located at the top of a vertical circular orbit - Sirius System):</p>
<p>$$ \theta = 0.378^{\circ} arc \times \frac{1}{3600} \times \frac{2 \pi}{360} = 1.83 \times 10^{-6}arc sec $$</p>
<p>$$D = \frac{D'}{sin(\theta)} = 8.16\times 10^{16} m$$</p>
<p>Where:</p>
<p>D = the distance from the Earth to the Sirius binary system.</p>
<p>D' = Distance from the Sirius system to the Earth ($15 \times 10^{10} m$).</p>
<p>Here's an illustration of what I have done:</p>
<p><img src="https://cdn.pbrd.co/images/HJtrflM.png" alt=""></p>
<p><strong>2) Measuring the semi-major axis of the ellipse and estimating the period of the orbit from the figure.</strong></p>
<p>To compute the semi-major axis I have assumed the major axis is under the scale [0'',12'']. </p>
<p>As Sammy Gerbil said :'To measure the semi-major axis from the figure, magnify the diagram to fill your screen, place the edge of a sheet of paper along the diagonal line marked on the ellipse, mark off the ends of this line on the paper, then transfer the paper edge to the scale and read off the length in arc seconds. I get 13.8" for the major axis length, so the semi-major axis is a=6.9". The calculation is then:'</p>
<p>$$a=8.16\times 10^{16}m\times \frac{6.9}{3600}\times \frac{2\pi}{360} = 2.73 \times 10^{12}m$$</p>
<p>Now let's estimate the period of the orbit.</p>
<p>Is this the right thought? </p>
<p>$$\frac{dA}{dt} = SM \times SM \times \frac{d \theta}{2dt}$$</p>
<p>It will not be accurate as 'the two sides of most of the triangles are different'.</p>
<p>$A_1$:</p>
<p><img src="https://cdn.pbrd.co/images/HK5k6ab.jpg" alt=""></p>
<p>On process...</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3185/deducing-the-mass-of-the-sirius-system-feynman-exercises-17Sat, 20 Oct 2018 15:50:53 +0000Find the angle between the $x$-axis and a vector
http://physics.qandaexchange.com/?qa=3105/find-the-angle-between-the-%24x%24-axis-and-a-vector
<blockquote><p>The $x$ component of vector $A$ is 25.0 m and the $y$ component is 40.0 m. <br>
(a) What is the magnitude of $A$? <br>
(b) What is the angle between the direction of $A$ and the positive direction of x?</p>
</blockquote>
<p>For (b) I tried using the formula $\tan \theta = \frac{a_y}{a_x} = \frac{40}{-25} = -1.6$, thus $\arctan(-1.6)=58$ degrees which does not match the answer key: $122$ degrees.</p>
<p>Any help is appreciated.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3105/find-the-angle-between-the-%24x%24-axis-and-a-vectorSat, 20 Oct 2018 11:59:19 +0000Answered: Onset of bouncing for rolling hoop with off-centre mass (Irodov ex 1.265)
http://physics.qandaexchange.com/?qa=3173/onset-bouncing-for-rolling-hoop-with-off-centre-mass-irodov&show=3174#a3174
<p><strong>Revised Answer</strong></p>
<p>There were some errors in my previous answer.</p>
<hr>
<p>It is not quite true that bouncing can happen <strong>only</strong> when particle A reaches the top of the hoop. I think what you mean is that bouncing <strong>occurs for the lowest value of $v_0$</strong> when particle A is at the top of the hoop. At faster speeds it can occur before A reaches the top.</p>
<p>Your calculation so far is correct. Bouncing starts when the upward centrifugal force caused by the rotation of particle A around its average position exceeds the combined weight of the hoop and particle A. When the velocity of the centre C of the hoop is $v$ the velocity of A relative to C is also $v$. Therefore bouncing starts when $$m\frac{v^2}{R} \gt 2mg$$ $$v^2 \gt 2gR$$ Bouncing can occur before A reaches its highest position if the <strong>vertical upward component</strong> of the centrifugal force exceeds $2mg$.</p>
<p>As you realise, we have to use conservation of energy to find $v$, the seed of the hoop when A reaches its highest position. </p>
<p>When A is at its lowest point it is stationary and has no KE. The COM of the hoop is moving with velocity $v_0$ so the hoop has translational KE $\frac12 mv_0^2$. The hoop also has rotational KE. In the COM frame every point on the hoop is moving at speed $v_0$ so the hoop has rotational KE of $\frac12 mv_0^2$. The total energy when A is at its lowest point is therefore $mv_0^2$.</p>
<p>When A reaches its highest point the centre of the hoop has some unknown speed $v$ and particle A has speed $2v$. The translational KE of the hoop and particle A are $\frac12 mv^2$ and $2mv^2$ respectively. The hoop also has rotational KE of $\frac12 mv^2$ as calculated above. The total KE in this position is $3mv^2$. The gravitational PE of A has increased by $2mgR$.</p>
<p>Total mechanical energy is the same in both positions. Applying the inequality above we get $$mv_0^2=3mv^2+2mgR$$ $$v_0^2-2gR = 3v^2 \gt 3(2gR) = 6gR$$ $$v_0^2 \gt 8gR$$</p>
<p>This answer agrees with solutions on <a rel="nofollow" href="https://www.youtube.com/watch?v=8_BMJAPuTWI">YouTube</a> and <a rel="nofollow" href="https://brilliant.org/problems/hold-your-ground/">Brilliant.org</a>.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3173/onset-bouncing-for-rolling-hoop-with-off-centre-mass-irodov&show=3174#a3174Fri, 19 Oct 2018 22:49:49 +0000Normal reaction for particle at rest on a sphere
http://physics.qandaexchange.com/?qa=3171/normal-reaction-for-particle-at-rest-on-a-sphere
<p><img src="https://cdn.pbrd.co/images/HIVCKHf.png" alt=""></p>
<p>The answer is C. </p>
<p>It is not difficult to understand why force $F$ decreases as the small metal ball is raised to the top of the sphere. The normal reaction $R$ from the hemisphere bears a greater share of the weight of the metal ball. </p>
<p>But why is the normal reaction $R$ the same for all positions of the ball on the sphere? Is there an intuitive explanation?</p>
<p>See discussion in <a rel="nofollow" href="https://chat.stackexchange.com/transcript/message/47213469#47213469">Problem Solving Strategies chatroom</a> on Physics Stack Exchange.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3171/normal-reaction-for-particle-at-rest-on-a-sphereWed, 17 Oct 2018 22:19:45 +0000Answered: The Random Walk
http://physics.qandaexchange.com/?qa=3164/the-random-walk&show=3170#a3170
<p>Yes the binomial probability approach can be applied to this problem.</p>
<p>The walker takes a series of steps. At each step he has only 2 choices : move left or move right. Which he chooses is determined completely at random and each has a definite probability. Note that it is not necessary for the probabilities to be equal, but they must add up to $1$ - ie these must be the only 2 outcomes which are possible, and they must be mutually exclusive. </p>
<p>This is exactly equivalent to tossing a coin $N$ times. Instead of $H, T$ you have $R, L$. The position of the walker on the line of integers is equivalent to the number of heads minus the number of tails.</p>
<p>This correspondence is pointed out after equation 6.10 in the notes which you reproduced.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3164/the-random-walk&show=3170#a3170Wed, 17 Oct 2018 14:31:45 +0000Answered: The length of the spring
http://physics.qandaexchange.com/?qa=3166/the-length-of-the-spring&show=3167#a3167
<p>It is not obvious how to deal with the glue. I shall discuss that later.</p>
<p>The main mistake you made is to assume that equal and opposite forces $F$ are applied at both ends of the spring. In the question there appears to be only one force $F$ applied to the RHS of the spring. Therefore the spring will accelerate. The tension in it will not be uniform, and the extension will not be the same as when there are equal forces at both ends. </p>
<p>How you have dealt with the glue is reasonable. You have assumed that the coils of the spring do not expand at all until the glue has broken. However, because the tension is not constant along the spring (as explained above), only those parts of the spring in which the tension is greater than $100N$ will extend. </p>
<p>Effectively the glue and spring are <strong>in parallel</strong> because the extension must be the same before the glue (or spring) breaks. The tension in the glue and spring need not be the same : the sum of tensions equals the applied force. If the glue and spring were <strong>in series</strong> then the tension in both would be the same but the extension could be different. Moreover, when the glue breaks then the spring would break also - the coils would come apart.</p>
<p>You have assumed that the glue has infinite <strong>stiffness</strong> (aka spring constant), so that it does not extend before it breaks. The question does not tell us what its stiffness is. It need not be infinite, but this is probably the only assumption we can make. If the stiffness is not infinite then we will get a different answer.</p>
<hr>
<p><strong>Comment :</strong> You might wonder why the question specifies that the force $F$ is <strong>gradually increased</strong>. This is because a force applied suddenly at one end causes the spring to oscillate as it accelerates. The maximum extension is then twice what it would be if a constant force is applied. See <a rel="nofollow" href="http://physics.qandaexchange.com/?qa=562/total-energy-of-the-block-and-spring&show=562#q562">Total energy of the block and spring</a>.</p>
<hr>
<p><strong>Solution :</strong></p>
<p>The tension in the accelerated spring varies linearly by mass from $F=200N$ at the RHS to $0N$ at the LHS. The tension in the middle is $100N$. This is where the glue comes unstuck. So the left half remains at its initial length of $0.5m$; only the right half extends. </p>
<p>The right half is now a spring with half its original length so the spring constant is $2k=1000N/m$. The tension at the left end of this spring is $100N$ and that at the right end is $200N$ so the average tension is $150N$. Its extension is $\frac{150N}{1000N/m}=0.15m$.</p>
<p>The new length of the whole spring is $1.15m$.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3166/the-length-of-the-spring&show=3167#a3167Mon, 15 Oct 2018 22:04:19 +0000Answered: Maximum height after collision in a bowl
http://physics.qandaexchange.com/?qa=3146/maximum-height-after-collision-in-a-bowl&show=3162#a3162
<p>The two blocks reach the bottom of the bowl at the same time with the same speed $u=U=\sqrt{2gh}$, as discussed in the comments. These are the initial speeds in the collision.</p>
<p>Momentum and kinetic energy are both conserved during the elastic collision. Instead of applying conservation of KE it is simpler to apply the <strong>Law of Restitution</strong> with a coefficient $e=1$. This implies that the relative velocity of separation equals the relative velocity of approach.</p>
<p>If the final speeds are $v, V$ for small and large blocks respectively, taking the +ve direction as left for the smaller block and right for the larger, then conservation of linear momentum and the law of restitution give $$(M-m)u=mv-MV$$ $$2u=v+V$$ From these we get $$v=\frac{3M-m}{M+m}u$$ $$V=\frac{3m-M}{m+M}u$$ Note that if $M \gt 3m$ then the larger block does not rebound ($V\lt 0$); it continues moving in its initial direction, chasing the smaller block. The smaller block always rebounds because $3M \gt m$ if $M \gt m$ hence $v \gt 0$ always.</p>
<p>The heights reached after the collision by the smaller and larger blocks are respectively $$h'=\frac{v^2}{2g}=(\frac{3M-m}{m+M})^2h$$ $$H'=\frac{V^2}{2g}=(\frac{3m-M}{M+m})^2h$$</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3146/maximum-height-after-collision-in-a-bowl&show=3162#a3162Tue, 09 Oct 2018 14:20:27 +0000Answered: Average of the internal energy of a system
http://physics.qandaexchange.com/?qa=3122/average-of-the-internal-energy-of-a-system&show=3159#a3159
<p>Setting the mathematics aside, the solution suggested by Count Iblis in a comment to your question in Mathematics SE, is convincing and physically intuitive. It uses the <strong>Equipartition of Energy Theorem.</strong> This states that when a system is in thermal equilibrium at temperature $T$, each degree of freedom has an average kinetic energy of $\frac12 kT$. </p>
<p>In the present case there are 5 degrees of freedom per diatomic molecule : 3 orthogonal directions of translational motion, and 2 orthogonal axes of rotation. Because the atoms within the diatomic molecule are point particles there is no rotation about the axis joining them.</p>
<p>So the result is very simply obtained : the average energy per molecule is $5\times \frac12 kT=\frac52 kT$.</p>
<hr>
<p>If the bond between the atoms acted as a spring, so that the molecule could vibrate as an harmonic oscillator, there would be another $\frac12 kT$ from the kinetic energy and also $\frac12 kT$ from the potential energy, making $\frac72 kT$ in total. </p>
<p>As you have suggested there ought also to be some average potential energy associated with vertical motion in the gravitational field. Because this is linear in the displacement variable $z$ ($U=mgz$), rather than quadratic like the potential for the harmonic oscillator ($U=\frac12 kz^2$), the average energy is $kT$ instead of $\frac12 kT$. In general, for a potential function proportional to $z^p$ the average energy would be $\frac{1}{p}kT$.</p>
<p>However, the gas is confined to a laboratory-sized container which prevents molecules from exploring the full range of the potential energies from $0$ to $kT$. At room temperature $kT$ is very much larger than the difference in gravitational PE for typical diatomic gas molecules such as nitrogen and oxygen. An extremely tall container would be required. For laboratory-sized containers and atmospheric gases the capacity of the gravitational field for storing energy can be neglected. But if the particles were much larger, such as sediment or colloid particles suspended in a liquid, the gravitational PE would become significant even for small containers.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3122/average-of-the-internal-energy-of-a-system&show=3159#a3159Tue, 09 Oct 2018 00:11:27 +0000Answered: Find angular acceleration of disc and tension in the string
http://physics.qandaexchange.com/?qa=3155/find-angular-acceleration-of-disc-and-tension-in-the-string&show=3157#a3157
<p>This is similar to problems asking for the tension $T$ in the string attached to a falling mass $m$, the other end of which is attached to a flywheel or to a block which slides on a horizontal surface. Beginners are tempted to say either that $T=mg$ because that is the weight of the load, or that $T=0$ because the load is assumed to be in free fall. The truth lies between both extremes. The point at which the string is attached to the flywheel or the sliding block is also accelerating, at rate $a \lt g$. This changes the force in the string to $T=m(g-a)$. </p>
<p>Likewise here, if the disk were fixed in place, or very much more massive than the small particles - in both cases the disk does not (significantly) rotate. Then the points at which the strings are attached would not accelerate. The tension in them would then be the centripetal force required to keep the small particles moving instantaneously in a circle of radius $2R$, which is $T=ma_c$ where $a_c=\frac{v_0^2}{2R}$ is the centripetal acceleration, as you calculated. </p>
<p>However, the disk is relatively light so it can quite easily rotate, which means that the points of attachment accelerate vertically in the direction of the strings, reducing the tension in the strings below $T_c$.</p>
<p>What you need to do is draw separate FBDs for the disk and one mass, using the same unknown tension $T$ in each string. For the disk the 2 tensions $T$ form a couple which causes angular acceleration $\alpha$ about its centre. This results in a linear acceleration $a=R\alpha$ of the point at which the string is attached. The tension in the string is then $T=m(a_c-a)$. </p>
<p>You obtain 2 simultaneous equations relating $T$ and $\alpha$, enabling you to find both. </p>
<p>Your mistake is the same as the students who assume the tension in the string attached to a falling mass is $T=mg$ and then go on to calculate the acceleration of the flywheel or sliding block based on this value. </p>
<p>Note that although the particles may have obtained their velocity from an impulse force, this is not necessary, and it does not result in the tension in the string being an impulse force. The tension continues for a finite time. </p>
<p>If the impulse given to the particles had a component along the string, there would be an impulse tension as well as a continuing tension. But because the impulse is perpendicular to the string there is no impulse tension. An alternative, equivalent way of setting the system into motion is to accelerate the particles while holding the disk in place, then release the disk when the strings reach the vertical position. An impulse is not necessary to satisfy the given initial conditions.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3155/find-angular-acceleration-of-disc-and-tension-in-the-string&show=3157#a3157Mon, 08 Oct 2018 21:38:57 +0000Answered: Thevenin's theorem on Capacitor.
http://physics.qandaexchange.com/?qa=3144/thevenins-theorem-on-capacitor&show=3145#a3145
<p>The mistake you have made is to replace the EMF labelled $2E$ with an open circuit. If you replace it with a short-circuit then the vertical resistance $R$ and the internal resistance $R$ are both shorted out - there is no resistance between F and G.</p>
<p>It is a good idea to re-draw the circuit after shorting the EMFs :<br>
<img src="https://cdn.pbrd.co/images/HHnbpb1.png" alt=""><br>
The EMF labelled $E, R$ cannot be shorted because it appears to have an internal resistance of $R$. There is no indication that the EMF labelled $2E$ has any internal resistance. When this is shorted there is no resistance between the points F and G. Points F, G, H are then at the same potential, so the 3 resistors $R$ (horizontal), $2R$ (diagonal) and $2R$ (vertical) are in parallel.</p>
<p>The resistance between A and B is therefore $\frac12 R$.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3144/thevenins-theorem-on-capacitor&show=3145#a3145Sun, 07 Oct 2018 16:54:05 +0000Answered: Finding path for which line integral of force is zero.
http://physics.qandaexchange.com/?qa=3123/finding-path-for-which-line-integral-of-force-is-zero&show=3135#a3135
<p>The main difficulty with such questions is applying correct logic. The options <strong>could be true</strong> for some cases, but we must decide if they are <strong>necessarily true</strong> for all cases.</p>
<p>The line integral $W=\int_O^C F.ds$ is the work done on the particle by the external force. By the Work-Energy Theorem this equals the change in kinetic energy of the particle. We are told that there is no change in the KE of the particle for any of the 3 paths OAC, OBC, OPC. So for each of these paths the work done is $W=0$. </p>
<p>Therefore <strong>option 1 is correct</strong> : at least 3 such paths exist along which the line integral of force is zero.</p>
<p>A conservative force is one for which the work done between two points such as O and C is independent of the path taken. This is true for the 3 paths identified, because the work done is zero for each. But we cannot assume that it is true for all other paths between O and C. Even if it is true for all paths between O and C we do not know if it is true for all paths between any other pairs of points. So <strong>option 2 is not necessarily true</strong>.</p>
<p>Conversely, it is possible that the work done could be zero for all paths between all pairs of point. We don't have enough evidence to decide this issue. So <strong>option 3 is not necessarily true either</strong>.</p>
<p>If we were told that C coincides with O then we would know that option 4 would be false, because we would know there are at least 3 closed paths along which the line integral is zero. But we don't know if C coincides with O, so we don't know whether the line integral is zero for any closed path. We cannot be sure whether option 4 is true or false. All we can conclude is that <strong>option 4 is not necessarily true</strong>.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3123/finding-path-for-which-line-integral-of-force-is-zero&show=3135#a3135Sat, 06 Oct 2018 17:01:18 +0000Maximum energy stored in a spring-block system
http://physics.qandaexchange.com/?qa=3124/maximum-energy-stored-in-a-spring-block-system
<blockquote><p>A block of mass $m$ is attached with an ideal spring of spring constant $k$ is kept on a smooth horizontal surface. Now the free end the spring is pulled with a constant velocity $u$ horizontally. The maximum energy stored in the spring and block system during subsequent motion is?</p>
</blockquote>
<p>I think,<br>
Spring will keep on extending till it gain velocity $u$ to stop free end's expansion, then spring is elongated it will apply force on mass $m$ till it again gain its natural length, by now maximum work has been done, so if I calculate $v_m$ then $E=\frac{1}{2}mv^2$, so how to calculate $v_m$?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3124/maximum-energy-stored-in-a-spring-block-systemSat, 06 Oct 2018 05:01:17 +0000Answered: The ratio of temperatures of two sides of a disk near a concave mirror
http://physics.qandaexchange.com/?qa=3117/the-ratio-of-temperatures-two-sides-disk-near-concave-mirror&show=3121#a3121
<p>Your reasoning about the geometry is <strong>not quite correct</strong>. The radiation which is focussed onto the right face of the disk does not come from a circle of radius $2r$ but from an <strong>annulus</strong> of inner radius $r$ and outer radius $2r$. The disk itself obscures the inner area of radius $r$ of the incident light from reaching the mirror, so the area of radiation reflected onto the right face is only $\pi(2r)^2-\pi r^2=3\pi r^2$. </p>
<p>You have assumed that $\Delta T \propto A$ for each side of the disk, where $A$ is its effective area which collects radiant power. But you have not given any justification for this assumption. I think that you might have to consider each face as a perfect absorber-emitter (ie a black body) and use <strong>Stephan's Law</strong>, which relates power and temperature. </p>
<p>Perhaps you also need to bring the ambient temperature of the surroundings into your calculation, or demonstrate why it is not relevant.</p>
<p>There may be a complication : some radiation emitted by the right face of the disk is reflected back onto the same side of the disk. This does not happen for the left side of the disk, and adds to the total radiation power received by the right side. However, perhaps we are expected to assume that the mirror is highly reflecting only in a narrow band of wavelengths centred on the wavelength of the incident radiation. The disk radiates as a black body at all wavelengths, so only a very small fraction of this radiation would be reflected by the mirror. So this effect could be assumed to be negligible.</p>
<hr>
<p>The radiant power $P$ emitted by a surface is related to its temperature $T$ by <strong>Stephan's Law</strong> : $$P=\sigma \epsilon AT^4$$ where $\sigma$ is Stephan's constant, $\epsilon$ is the emissivity of the surface ($\epsilon=1$ for a perfect black body emitter) and $A$ is the area. When power is absorbed from the surroundings at temperature $T$ we use the same equation except that <strong>absorptivity</strong> $\alpha$ replaces $\epsilon$. At thermal equilibrium, as occurs here, $\alpha=\epsilon$. </p>
<p>At equilibrium temperature the emitted power $P$ is also the total power received by the surface - in this case the sum of power from the source of parallel light of intensity $I$ and from the surroundings, which are at temperature $T_0$.</p>
<p>For the two sides (left 1 and right 2) we have $$P_1=\sigma \epsilon AT_1^4=IA+\sigma\epsilon AT_0^4$$ $$P_2=\sigma \epsilon AT_2^4=3IA+\sigma \epsilon AT_0^4$$ because (as shown above) the right side receives $3\times$ as much radiation as the left side.</p>
<p>Multiplying the 1st equation by 3 and subtracting the 2nd we get $$3T_1^4-T_2^4=2T_0^4$$ $$3(\frac{T_1}{T_0})^4-(\frac{T_2}{T_0})^4=2$$</p>
<p>In order to proceed we must assume that the faces of the disk are not much hotter than the surroundings - ie that $\delta T_1=T_1-T_0$ and $ \delta T_2= T_2-T_0$ are both very much smaller than $T_0$. This assumption is perhaps implied by the notation $\delta T$.</p>
<p>Then neglecting terms higher than the 1st power in $\frac{\delta T}{T_0}$ we get $$2=3(\frac{T_1}{T_0})^4-(\frac{T_2}{T_0})^4\approx 3(1+4\frac{\delta T_1}{T_0})-(1+4\frac{\delta T_2}{T_0})$$ $$\frac{\delta T_1}{\delta T_2}\approx \frac13$$</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3117/the-ratio-of-temperatures-two-sides-disk-near-concave-mirror&show=3121#a3121Tue, 02 Oct 2018 16:40:59 +0000Answered: Solving the diffusion equation with an absorbing boundary
http://physics.qandaexchange.com/?qa=3113/solving-the-diffusion-equation-with-an-absorbing-boundary&show=3120#a3120
<p>In this situation the Method of Images would apply as follows :</p>
<p>At $t=0$ you have a dense concentration of particles at a particular point $+x_0$. Over time they diffuse outwards with a Gaussian probability density profile. However, this expanding distribution is disrupted by the absorbing boundary at $x=0$. The result is that the initially symmetrical Gaussian distribution becomes increasingly asymmetrical as it spreads out, as in the diagram below.</p>
<p><img src="https://cdn.pbrd.co/images/HGuAUaz.png" alt=""></p>
<p>In order to take account of the absorbing boundary, the Method of Images places an equal dense concentration of <strong>anti-particles</strong> at the image point $-x_0$. These anti-particles have the same properties as the ordinary particles released at $+x_0$, so they diffuse outwards with the same speed hence the same Gaussian distribution. However, when they come into contact with ordinary particles they annihilate, just like electrons and positrons.</p>
<p>The two distributions are exactly anti-symmetrical about the boundary. There are always equal numbers of each type of particle at the boundary, so the sum at the boundary is always zero.</p>
<p>So if you have a solution $G (x,t)$ centred on $+x_0$ without the absorbing boundary, then to account for the absorbing boundary you should subtract the same solution $G (x,t)$ centred on $-x_0$. </p>
<p>As with the electrostatic and optical method of images, this trick only works on the <em>object</em> side of the boundary. It cannot tell you the distribution of ordinary particles on the <em>image</em> side, where there could be other sources and other absorbing boundaries. In the absence of such information, you must assume that you result applies only for $x \ge 0$ : for all $x \lt 0$ you must enforce the condition $G(x,t)=0$. </p>
<p><img src="https://cdn.pbrd.co/images/HGuF1iJ.png" alt=""></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3113/solving-the-diffusion-equation-with-an-absorbing-boundary&show=3120#a3120Mon, 01 Oct 2018 21:54:25 +0000Answered: Is centripetal force not always true for Planets?
http://physics.qandaexchange.com/?qa=3114/is-centripetal-force-not-always-true-for-planets&show=3116#a3116
<p>When $r_1=r_2=r$ the orbit is a circle of radius $r$. The gravitational force on the planet is then the centripetal force because the point to which the gravitational force on the planet is directed is the same as the centre of the circle in which it is moving.</p>
<p>However if $r_1 \ne r_2$ then the orbit is an ellipse not a circle. The local radius of curvature $r$ of the ellipse at each of the extremes $P_1, P_2$ (where $r_1, r_2$ are measured) is not $r_1, r_2$ respectively. See diagram below. </p>
<p>The gravitational forces at $P_1, P_2$ are directed towards the Sun $S$ and <strong>not</strong> towards the local centre of curvature $C$. To see this examine what happens when the planet has moved a short distance from $P_1$ to $P_1'$. Then the gravitational force is directed towards $S$ not $C$ - these are 2 different directions. This is true however close $P_1'$ is to $P_1$.</p>
<p>Therefore the gravitational force at $P_1, P_2$ is not equal to the centripetal force at these points.</p>
<p><img src="https://cdn.pbrd.co/images/HGiCu1x.png" alt=""> </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3114/is-centripetal-force-not-always-true-for-planets&show=3116#a3116Sun, 30 Sep 2018 15:12:47 +0000Angle of pendulum in accelerated incline
http://physics.qandaexchange.com/?qa=3107/angle-of-pendulum-in-accelerated-incline
<p>Consider the following system consisting of a box sliding down a plane. The coefficient of friction between the plane and the box is $\mu$. A pendulum is attached to the top of the box as shown.</p>
<p><img src="http://i65.tinypic.com/1z37azr.png" alt=""></p>
<p>The acceleration of the box+pendulum is $a=g(\sin\theta-\mu\cos\theta)$ I believe.<br>
In a non-inertial frame attached to the box, the free-body diagram for the pendulum is</p>
<p><img src="http://i63.tinypic.com/2eyyx3s.png" alt=""></p>
<p>My goal is to find the angle $\phi$ from the equilibrium of these 3 forces. I have to pick x and y axes to decompose these forces. If I pick the x axis along the fictitious force and the y perpendicular to it I get (I think) </p>
<p>$$m_P a+T\sin\phi=m_P g\sin\theta$$ </p>
<p>and </p>
<p>$$T\cos\phi=m_{P}g\cos\theta$$</p>
<p> which I can then solve for $\phi$ :</p>
<p>$$\tan\phi=\frac{g\sin\theta-a}{g\cos\theta}=\mu$$</p>
<p>Is this right?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3107/angle-of-pendulum-in-accelerated-inclineFri, 28 Sep 2018 04:00:23 +0000Answered: Minimum distance from the mirror should the boy be to see his full image
http://physics.qandaexchange.com/?qa=3099/minimum-distance-from-the-mirror-should-the-boy-see-full-image&show=3100#a3100
<p>Your solution is correct. The official answer is wrong.</p>
<p>The question should ask for the <strong>maximum</strong> distance, not minimum. The closer you are to the mirror the more of your image you can see at any angle of tilt. That error might alert you to the possibility that the official answer is also wrong.</p>
<p>If you find that you don't have the "right" answer, and you cannot spot your mistake, the next thing to do is check the reasonableness of both answers. What does each answer predict when you input values which you know the answer to?</p>
<p>The obvious trial value here is $\theta=0$. Then you can always see your feet, at any distance from the mirror. Your answer predicts $d=\infty$, which is what you expect. The official formula predicts $d=\frac12 h$, which is not what you expect. A sketch shows that you can still see your feet closer or further than $d=\frac12 h$.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3099/minimum-distance-from-the-mirror-should-the-boy-see-full-image&show=3100#a3100Tue, 25 Sep 2018 12:23:37 +0000Answered: Lengths of vertical cables and maximum tension - Feynman exercises 2.35
http://physics.qandaexchange.com/?qa=3088/lengths-vertical-cables-maximum-tension-feynman-exercises&show=3089#a3089
<p>I do not understand your calculations. (You are working partly in kgwt and partly in Newtons, which is confusing. Stick to one unit and use throughout.)</p>
<p>I would start from the outside and work inwards :</p>
<p>There are 3 distinct joints (nodes) to examine. At each of them there are 3 forces in balance.</p>
<p>The tension in the 4 longitudinal cable ends can be found because the vertical components of this tension supports the whole weight of the bridge ($4.80\times 10^4 kg$). </p>
<p>The 12 vertical cables each carry the same weight, which in total is again the full weight of the bridge. Balancing forces at joint B allows you to find the force and angle of cable BA.</p>
<p>Repeat the same process for joint A to find the tension and angle of cable AC where C is the innermost joint/node.</p>
<p>The lengths of the vertical cables can be found from the angles calculated above.</p>
<p>Balancing forces horizontally at each joint shows that the horizontal component of tension is the same in each section of the cable : $$T_1\cos\theta_1=T_2\cos\theta_2=T_3\cos\theta_3=...$$ This shows that the maximum tension occurs where $\cos\theta$ has the smallest value, which is where the angle is maximum - ie at the ends. The maximum tension is in the outermost cable : you already calculated it above. </p>
<hr>
<p><strong>Calculation</strong></p>
<p>Balance forces vertically and horizontally in turn at the joints A, B, C, D shown in the following diagram, starting from the outer end D :</p>
<p><img src="https://cdn.pbrd.co/images/HFQSg3o.png" alt=""></p>
<p>At the 4 end-points D of the two horizontal cables, the vertical component of the tension $T_1$ in the 1st section of the cable is balanced by a reaction force equal to $\frac14 W$ where $W=4.80\times 10^4 kgwt=4.70\times 10^5 N$ is the full weight of the bridge. $$T_1\sin45^{\circ}=\frac{1}{\sqrt2}T_1=\frac14 W$$ $$T_1=\frac{\sqrt2}{4}W=1.70\times 10^4 kgwt=1.66\times 10^5N$$ </p>
<p>At joint B the balance of vertical and horizontal forces gives $$T_1\sin\theta_1=T_2\sin\theta_2+\frac{1}{12}W$$ $$T_1\cos\theta_1=T_2\cos\theta_2$$ from which we get $$\frac{T_2\sin\theta_2}{T_2\cos\theta_2}=\tan\theta_2=1-\frac{W}{12T_1\sin\theta_1}=1-\frac{W}{3W}=\frac23$$ $$\theta_2=0.588 rad=33.7^{\circ}$$ $$\cos\theta_2=\frac{3}{\sqrt{13}}, \sin\theta_2=\frac{2}{\sqrt{13}}$$ $$T_2=\frac{T_1\cos\theta_1}{\cos\theta_2}=\frac{\sqrt{13}}{12}W=5.10\times 10^4 kgwt=5.00\times 10^5 N$$</p>
<p>At joint A we have similar forces so we can immediately write down that $$\tan\theta_3=1-\frac{W}{12T_2\sin\theta_2}=1-\frac{W}{2W}=\frac12$$ $$\theta_3=0.464rad=26.6^{\circ}$$ $$\cos\theta_3=\frac{2}{\sqrt5}, \sin\theta_3=\frac{1}{\sqrt5}$$ $$T_3=\frac{T_1\cos\theta_1}{\cos\theta_3}=\frac{\frac14}{\frac{2}{\sqrt5}}W=\frac{\sqrt5}{8}W=1.34\times 10^4 kgwt=1.31\times 10^5 N$$</p>
<p>We do not need to go any further. I leave it to you to calculate the lengths of the vertical cables.</p>
<p><strong>Note :</strong> There may be something wrong with my calculation because the largest decrease is from $\theta_3=26.6^{\circ}$ to $\theta_4=0^{\circ}$. The largest decrease should be at the outside and the smallest at the inside. I am too tired to check this right now.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3088/lengths-vertical-cables-maximum-tension-feynman-exercises&show=3089#a3089Sun, 23 Sep 2018 15:59:03 +0000Statics - Minimum angle before sliding
http://physics.qandaexchange.com/?qa=3084/statics-minimum-angle-before-sliding
<p><img src="https://imgur.com/a/mixJS6f" alt=""></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3084/statics-minimum-angle-before-slidingThu, 20 Sep 2018 08:01:27 +0000Answered: Spinning 4 connected springs system - Feynman exercises 14.20
http://physics.qandaexchange.com/?qa=3059/spinning-connected-springs-system-feynman-exercises-14-20&show=3064#a3064
<p>(a) Yes, your mistake is with the elastic force, which you have written as $k\Delta L$. This is the tension $T$ in each of the springs, but it is not the force pulling each mass towards the centre.</p>
<p>There are 2 tension forces $T$ acting on each mass $m$. The component of each in the direction of the centre is $T\cos45^{\circ}=\frac{1}{\sqrt2}T$. Therefore the total elastic force on each mass towards the centre is $\frac{2}{\sqrt2}T=\sqrt2 T$.</p>
<p>Writing $F=ma$ with centripetal acceleration $a=\omega^2x$ we have $$\sqrt2 T=m\omega^2 x$$ $$\sqrt2 k\Delta L=m\omega^2(\frac{L+\Delta L}{\sqrt2})$$ $$2k\Delta L=m\omega^2(L+\Delta L)$$ $$\Delta L=\frac{m\omega^2L}{2k-m\omega^2}$$</p>
<p>(b) The above equation tells us that there is no equilibrium position for $2k \lt m\omega^2$, because $\Delta L$ would then be -ve, so the spring force would be outward - ie there would be no centripetal (inward) force. But it does not tell us if any equilibrium position is stable. </p>
<p>As you found out, to do that we have to examine how the resultant force on the mass changes as we increase/decrease the radius $x$ close to the equilibrium position $x_0$. If the resultant force becomes -ve (inward) for $x \gt x_0$ and +ve (outward) for $x \lt x_0$ then the equilibrium is stable. </p>
<p>Mathematically, we need to examine $\frac{dF}{dx}$ at the equilibrium position $x_0$. If it is $\lt 0$ - ie resultant force decreasing as we move outwards - then the equilibrium position is stable.</p>
<p>In the rotating frame of reference the resultant force $F(x)$ on each mass $m$ is the sum of the outward (+ve) centrifugal force and the elastic central force, which can be inward (-ve) or outward (+ve). Therefore : $$F(x)=m\omega^2 x-2k(x-\frac{L}{\sqrt2})$$ in which I have written the extension $\Delta L$ in terms of $x=\frac{L+\Delta L}{\sqrt2}$, and $F(x)$ is measured in the outward direction, the same as $x$. </p>
<p>Now the question has been somewhat ambiguous about $\omega$, so there are 2 cases to consider. We are not told whether (i) $\omega$ is fixed - eg regulated by an external torque (a <a rel="nofollow" href="https://en.wikipedia.org/wiki/Governor_(device)"><strong>governor</strong></a>) which keeps $\omega$ constant regardless of whether the masses move inwards or outwards, or (ii) after $\omega$ is given a fixed value at an equilibrium position the regulator is disconnected - then it is the <strong>angular momentum</strong> $m\omega x^2$ which is constant.</p>
<p><strong>(i) angular velocity $\omega$ is constant</strong><br>
$$\frac{dF}{dx}=m\omega^2-2k$$ We already know that we must have $2k \gt m\omega^2$ for all equilibrium positions, so in this case $\frac{dF}{dx} \lt 0$ for all equilibrium positions, hence <strong>all equilibrium positions are stable</strong>.</p>
<p><strong>(ii) angular momentum $m\omega x^2$ is constant</strong></p>
<p>The angular velocity and radius at the equilibrium position are $\omega_0, x_0$. If the mass moves to a different radius $x$ the angular velocity $\omega$ changes but the angular momentum remains constant : $m\omega x^2 = m \omega_0 x_0^2$. Then $m\omega^2 x=\frac{m\omega_0^2 x_0^4}{x^3}$. Therefore : $$F(x)=\frac{m\omega_0^2 x_0^4}{x^3}-2k(x-\frac{L}{\sqrt2})$$ $$\frac{dF}{dx}=-\frac{3m\omega_0^2 x_0^4}{x^4}-2k$$ which is clearly $\lt 0$ for all values of $x$. So in this case also we find that <strong>all equilibrium positions are stable</strong>.</p>
<p>The only condition for stable equilibrium is $2k \gt m\omega^2$.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3059/spinning-connected-springs-system-feynman-exercises-14-20&show=3064#a3064Sat, 15 Sep 2018 23:23:22 +0000Answered: Find the range of a projectile on an inclined plane
http://physics.qandaexchange.com/?qa=3055/find-the-range-of-a-projectile-on-an-inclined-plane&show=3056#a3056
<p>Use axes $Ox, Oy$ which are respectively parallel and perpendicular to the inclined plane. The components of the launch velocity along these axes are $v_x=v_0 \cos(\beta-\alpha)$ and $v_y=v_0\sin(\beta-\alpha)$, while the components of gravity are $-g_x=-g\sin\alpha$ and $-g_y=-g\cos\alpha$. </p>
<p>The co-ordinates of the arrow at time $t$ after launch are $$x=v_x t-\frac12 g_x t^2$$ $$y=v_y t-\frac12 g_y t^2$$ When the arrow lands then $y=0$ and $t=T$ the time of flight and $x=R$ the range. So $$v_y=\frac12 g_y T$$ $$T=\frac{2v_y}{g_y}=\frac{2v_0}{g}\frac{\sin(\beta-\alpha)}{\cos\alpha}$$ and $$R=v_xT-\frac12 g_xT^2=\frac{2v_0^2}{g}\frac{\cos(\beta-\alpha)\sin(\beta-\alpha)}{\cos\alpha}-\frac{2v_0^2}{g}\frac{\sin\alpha \sin^2(\beta-\alpha)}{\cos^2\alpha}$$ $$=\frac{2v_0^2\sin(\beta-\alpha)}{g\cos^2\alpha}[\cos(\beta-\alpha)\cos\alpha-\sin(\beta-\alpha)\sin\alpha]$$ $$=\frac{2v_0^2}{g}\frac{\sin(\beta-\alpha)\cos\beta}{\cos^2\alpha} $$</p>
<p>To find the angle of launch for maximum range note that $$2\cos A\sin B = \sin (A+B) - \sin (A-B)$$ therefore $$2\cos\beta\sin(\beta-\alpha)=\sin(2\beta-\alpha)-\sin\alpha$$ The 2nd term is fixed because $\alpha$ is fixed, but $\beta$ can vary. The maximum value this expression can have occurs when the 1st term is $+1$, ie when $$2\beta-\alpha=\frac12\pi$$ $$2(\beta-\alpha)= \frac12 \pi-\alpha$$ That is, the maximum range is achieved when the direction of launch bisects the angle between the inclined plane and the gravity vertical. The maximum range is $$R_{max}=\frac{v_0^2}{g}\frac{(1-\sin\alpha)}{\cos^2\alpha}=\frac{v_0^2}{g}\frac{(1-\sin\alpha)}{(1-\sin\alpha)(1+\sin\alpha)}=\frac{v_0^2}{g(1+\sin\alpha)}$$ </p>
<p><em>Source : <a rel="nofollow" href="https://aapt.scitation.org/doi/10.1119/1.13680">Letter to the Editor of American Journal of Physics by A P French, 1984</a></em></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3055/find-the-range-of-a-projectile-on-an-inclined-plane&show=3056#a3056Thu, 13 Sep 2018 21:54:21 +0000Answered: Find velocity of approach of the centres
http://physics.qandaexchange.com/?qa=3053/find-velocity-of-approach-of-the-centres&show=3054#a3054
<p><strong>Your Solution</strong></p>
<p>You have made a good start on the problem, but you have not made use of the information that the outer radius of the spool is $n$ times the inner radius. </p>
<p>Your assumption that $\frac{dl}{dt}=u\cos\theta$ is not quite correct. This is the component of velocity of the top end of the rope in the current direction of the rope. However, the bottom end of the rope also has a component of velocity in the same direction, which is $v\sin\theta$. So you should have $$\frac{dl}{dt}=u\cos\theta-v\sin\theta$$ Also note that $\frac{dx}{dt}=-v$ because $x$ is decreasing. Then your equation reduces to $u=u$.</p>
<p>However, you are almost there. $\frac{dl}{dt}$ tells you how fast the string is unwinding from the spool. This is $\frac{v}{n}$. Then $$\frac{v}{n}=u\cos\theta-v\sin\theta$$ $$v=\frac{\cos\theta}{\sin\theta+\frac{1}{n}}u$$ which is the same as my solution below, but simpler!</p>
<hr>
<p><strong>Approximate (Longer) Solution</strong></p>
<p>Suppose that when the string is vertical it has length $a$. This is its maximum length. When the spool has rolled a distance $x$ to the right of the vertical a length $\frac{x}{n}$ of the string has wound onto the inner drum; the length of the string in this position is $a-\frac{x}{n}$ (see diagram below). The height of the top of the string above the centre of the spool is now $y$ where $$y^2=(a-\frac{x}{n})^2 - x^2$$ </p>
<p><strong>Note :</strong> I am assuming that the inner radius $r$ of the spool is very much smaller than the maximum length $a$ of the string. Then as the spool rolls on a horizontal surface the point at which the string is attached moves approximately along a horizontal straight line. Without this assumption, the point of attachment moves from the right side of the inner drum when the string is vertical to the top of the inner drum when the string is horizontal. If $r$ is comparable with $a$ this motion makes a significant difference to the length of the string and complicates the calculation (see <strong>Exact Solution</strong> below). </p>
<p><img src="https://cdn.pbrd.co/images/HDBSTGG.png" alt=""><br>
Differentiating the above expression wrt time we get $$yu=(a-\frac{x}{n})\frac{v}{n}+xv$$ $$a-\frac{x}{n}=n(py-x)$$ where $u=\dot y, v=-\dot x$ are the vertical and horizontal speeds of the ends of the string and $p=\frac{u}{v}$. Substituting into the 1st equation to eliminate the unknown initial length $a$ of the string, we get $$y^2=n^2(y-px)^2-x^2$$ Divide by $y^2$ and write $\frac{x}{y}=t=\tan\theta$. Then : $$1=n^2(p-t)^2-t^2$$ $$\frac{u}{v}=p=t + \frac{1}{n}\sqrt{1+t^2}$$ I have taken the +ve root because when $t=0$ then $v$ must be +ve. In fact $v=nu$ in this position, as expected. Therefore $$v=\frac{\cos\theta}{\sin\theta+\frac{1}{n}}u$$</p>
<p><strong>Exact Solution</strong></p>
<p>If the radius $r$ of the inner drum is significant compared with the length of the string then after the right end $U$ of the spool has moved distance $x$ to the right, the point of contact between string and spool rises from $O$ which is level with the centre $S$ of the spool when $\theta=0$, to $T$ which is above the level of the centre $S$ and also a little closer to $O$. See following diagram.<br>
<img src="https://cdn.pbrd.co/images/HE0gFvU.png" alt=""></p>
<p>As before, an amount $\frac{x}{n}$ of the string has been wound onto the inner drum at point $U$. However, the contact point has moved round to $T$, so an additional arc length $UT=r(\theta-\phi)$ has been wound onto the spool.</p>
<p>Angle $QTS$ is a right angle, therefore $$QS^2=OQ^2+OS^2=QT^2+ST^2$$ $$y^2+(x-r)^2=(a-\frac{x}{n}-r(\theta-\phi))^2+r^2$$ This is very much more complicated than the <strong>Approximate Solution</strong> above. It will result in a transcendental equation, which can only be solved numerically. This is because we shall again have to make trigonomentric substitutions to eliminate $x, y, \phi$ and will be left with eg $\theta, \tan\theta$ in the same equation.</p>
<p>I shall not attempt to solve this.</p>
<hr>
<p><strong>Comment</strong></p>
<p>I have assumed that the problem is purely one of geometrical constraint and kinematics. I have assumed that the string never becomes slack. I think this assumption is correct. But I had wondered, Is it possible that the spool could accelerate so much that the string does becomes slack at some point? It is not obvious why this does not happen. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3053/find-velocity-of-approach-of-the-centres&show=3054#a3054Thu, 13 Sep 2018 01:29:59 +0000Answered: Find maximum charge on ball and maximum current through inductor .
http://physics.qandaexchange.com/?qa=3046/find-maximum-charge-ball-maximum-current-through-inductor&show=3052#a3052
<p>Your method for solving the problem is correct, but you have made some mistakes.</p>
<p>First, you have assumed that the current $i$ through the inductor towards Earth equals $-\frac{dq}{dt}$ where $q$ is the charge on the sphere. This is not quite correct because charge is also arriving on the sphere from the electron beam at the constant rate $i_0$. So you should have $$\frac{dq}{dt}=i_0 -i$$</p>
<p>Then we can differentiate and substitute into your 1st equation : $$\frac{q}{C}=L\frac{di}{dt}=-L\frac{d(i_0-i)}{dt}$$ $$\frac{1}{C}\frac{dq}{dt}=\frac{1}{C}(i_0-i)=-L\frac{d^2(i_0-i)}{dt^2}$$ $$\frac{d^2(i_0-i)}{dt^2}+\frac{1}{LC}(i_0-i)=0$$ $$(i_0-i)=A\sin(\omega t+\phi)$$ where $\omega^2 =\frac{1}{LC}$. I have used $\sin$ but you can instead use $\cos$ then the value of $\phi$ is different.</p>
<p>$A, \phi$ are constants which are to be determined from the initial conditions. When $t=0$ there is no charge on the sphere and no current flowing through the inductor ($i=0$) and also no voltage across the inductor ($V_L=L\frac{di}{dt}=0$) therefore $$i_0=A\sin\phi$$ $$-\frac{di}{dt}=0=\omega A\cos\phi$$ $$\implies \phi=\frac{\pi}{2}, A=i_0$$ $$i=i_0(1-\cos\omega t)$$ We can see from this that the maximum current through the inductor is $2i_0$, which occurs when $\cos\omega t=-1$.</p>
<p>So what is $i_0$? This is where you make another mistake or two. The rate at which charge falls onto the sphere is $$i_0=\text{cross-section area x velocity of electrons x density of beam x electron charge }=-\pi R^2 une$$ (<em>Comment :</em> This is not realistic because as -ve charge accumulates on the sphere other electrons in the beam will be repelled from it, so $i_0$ will not be a constant, as assumed above, but will decrease and will be a function of $q$ and therefore of time $t$. However, I think we are expected to ignore this effect, which would make the problem vastly more complicated.)</p>
<p>What about maximum charge? We integrate the equation for $i(t)$ using the initial condition $q=0$ when $t=0$ : $$i(t)=i_0-\frac{dq}{dt}=i_0(1-\cos\omega t)$$ $$\frac{dq}{dt}=i_0\cos\omega t$$ $$q=\frac{i_0}{\omega}\sin\omega t$$ Maximum charge occurs when $\sin\omega t=\pm 1$ and is $$Q=\pm \frac{i_0}{\omega}=\pm \sqrt{LC}\pi R^2 une$$ </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3046/find-maximum-charge-ball-maximum-current-through-inductor&show=3052#a3052Sat, 08 Sep 2018 16:09:33 +0000A 5.0kg object slides 2.0m down a smooth plane. Find its potential energy and component of weight parallel to the plane
http://physics.qandaexchange.com/?qa=3036/object-slides-potential-energy-component-weight-parallel
<p>A 5.0kg block of ice is at rest at the top of a smooth inclined plane. The block is released and slides 2.0m down the plane. Assuming there is no friction between the block and the surface, calculate:<br>
1. The gravitational potential energy at the top of the plane,<br>
2. The component of the weight parallel to the plane,<br>
3. The acceleration of the block,<br>
4. The velocity of the block at the bottom of the plane,<br>
5. The kinetic energy at the bottom of the plane.</p>
<p>Since there's no friction, the change in potential energy will be equal to the kinetic energy at the end. Am I right in thinking that the work done on the block is also equal to the kinetic energy gained by the block?</p>
<p>If so, that gives: <br>
$$W = E_k = \Delta E_p$$<br>
$$F \cdot s \cdot \cos\theta = \frac{m \cdot v^2}{2} = m \cdot g \cdot \Delta h$$ <br>
Substituting in values gives: <br>
$$98.1 \cos \theta = 2.5v^2 = 49.05 \Delta h$$<br>
Without knowing the angle the slope is on, I'm not sure where to go from here.<br><br>
I've been assuming that the block travels at an angle, 2 metres down the slope. Perhaps it goes down 2 metres vertically? <br>
$$49.05 \cdot s \cdot \cos \theta = 2.5v^2 = 98.1$$<br>
This would help with parts 1, 4, and 5; but I wouldn't know how to answer parts 2 and 3.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3036/object-slides-potential-energy-component-weight-parallelFri, 07 Sep 2018 12:24:58 +0000Answered: Problem on Capacitors.
http://physics.qandaexchange.com/?qa=2629/problem-on-capacitors&show=3037#a3037
<p>net capacitance of the system is series combination of "elemental capacitors" each of thickness $dx$</p>
<p>$\dfrac{1}{C_{net}}=\dfrac{1}{A\epsilon_{0}}\displaystyle \int_{0}^{d}\dfrac{dx}{e^{ax}}\implies C_{net}=\dfrac{aA\epsilon_{0}}{1-e^{-ad}}$</p>
<p>$V=\dfrac{Q}{C_{net}}=Q\left[\dfrac{1-e^{-ad}}{aA\epsilon\_{0}}\right]\tag{1}$</p>
<p>by inserting inserting dielectric D remains same </p>
<p>we have ,</p>
<p>$\vec {D}=electric \ displacement=\epsilon_{0}K E \ { {\hat x}}$</p>
<p>again by definition of D i.e, it equal to the amount of <strong>free charge</strong> on one plate divided by the area of the plate $\left(\dfrac{Q}{A}\right)$</p>
<p>$D=\left(\dfrac{Q}{A}\right)=\epsilon_{0}K E \ { }\implies E=E(x)=\dfrac{Q}{A\epsilon_{0}{K}}$</p>
<p>in above, put$ K=e^{ax}$ </p>
<p>$E(x)=\dfrac{Q}{A\epsilon_{0}e^{ax}}$</p>
<p>if you calculate $-\displaystyle \int _{0}^{d} E(x) dx $ using above expression of E(x) you will get same voltage as above <strong>eq.(1)</strong> </p>
<p>i think <strong>answer in your book is wrong</strong> </p>
<p>and also, charge $Q(x)$ will not vary as function of $x$because both $C_{net}$ and $V$ are fixed constant quantities</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=2629/problem-on-capacitors&show=3037#a3037Fri, 07 Sep 2018 12:24:52 +0000Answered: Minimum width of belt and minimum speed of the disc
http://physics.qandaexchange.com/?qa=3040/minimum-width-of-belt-and-minimum-speed-of-the-disc&show=3044#a3044
<p>(a) Your calculation is very good but you have not reached the end of it.</p>
<p>The distance of $2.5m$ which the disk travels across the belt is measured <strong>diagonally</strong>. It is the hypotenuse of a $3:4:5$ triangle. The perpendicular distance across the belt is $2.0m$. So that is the minimum width.</p>
<p>(b) This is the same problem as in <a rel="nofollow" href="http://physics.qandaexchange.com/?qa=205/a-block-is-pushed-onto-a-conveyor-belt&show=205#q205">A block is pushed onto a conveyor belt</a>. </p>
<p><img src="https://cdn.pbrd.co/images/HBMuJqm.png" alt=""></p>
<p>Initially the velocity of the disk over the ground is $OA=4m/s$ along the $Oy$ axis. When the disk has come to rest relative to the conveyor belt its velocity over the ground is $OB=3m/s$ along the $Ox$ axis. At intermediate times the velocity of the disk over the ground is represented by points on the line $AB$. </p>
<p>At point $C$ the magnitude $OC$ of the velocity over the ground is minimum, where $OC\perp AB$. $OAB$ and $COB$ are similar triangles. Therefore $OC=\frac45 \times 3=2.4m/s$.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3040/minimum-width-of-belt-and-minimum-speed-of-the-disc&show=3044#a3044Fri, 31 Aug 2018 22:41:14 +0000$\Psi$ in function of 'position'
http://physics.qandaexchange.com/?qa=3028/%24-psi%24-in-function-of-position
<p>The general exercise:<br>
<img src="https://cdn.pbrd.co/images/HApljRF.jpg" alt=""></p>
<p><img src="https://cdn.pbrd.co/images/HAQZH0G.jpg" alt=""></p>
<p>Let's focus on c). </p>
<p>The value of $\sigma$:</p>
<p>$$\sigma =\sqrt{ < x^2 > -< x >^2 }= \frac{1}{\lambda \sqrt{2}}$$</p>
<p>We are asked to plug in $\sigma$:</p>
<p>$$| \Psi (\sigma ) | = \lambda e^{\frac{-2}{\sqrt{2}}}$$</p>
<p>Where $\lambda$ is a positive constant.</p>
<p>I calculated < x > and its value is 0. Check this link to obtain further details:</p>
<p><a rel="nofollow" href="https://math.stackexchange.com/questions/2893056/misleading-expected-value">https://math.stackexchange.com/questions/2893056/misleading-expected-value</a></p>
<p><img src="https://cdn.pbrd.co/images/HBKGTII.png" alt=""></p>
<p><strong>About the probability</strong></p>
<p>I have difficulties solving this probability as this time I got:</p>
<p>$P_i = \int_{-\sigma}^{\sigma} \lambda e^{-2 \lambda |x|} dx =$ </p>
<p>$= \lambda \int_{-\sigma}^{0} e^{-2 \lambda |x|} dx$ </p>
<p>$+ \lambda \int_{0}^{+\sigma} e^{-2 \lambda |x|} dx = 0$</p>
<p>Where did I get wrong? I have been trying to get this probability but none outcome seems to make sense.</p>
<p>$$P_o = 1 - P_i $$</p>
<p>Where $P_i$ and $P_o$ mean inside and outside respectively</p>
<p>Thank you</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3028/%24-psi%24-in-function-of-positionSat, 25 Aug 2018 21:27:13 +0000Answered: The amplitude of the resulting oscillation is $\frac{xqQ}{8\pi ^2 \epsilon _0 a^3 k}$ .Find the value of 'x'?
http://physics.qandaexchange.com/?qa=3022/amplitude-resulting-oscillation-frac-xqq-epsilon-find-value&show=3023#a3023
<p>Your calculation of $T$ is correct - and possibly simpler than integration, which is the usual method for finding the electrostatic repulsion force $F$. Perhaps your difficulty is that the result which you get for the amplitude does not match the answer which you have been given for $x$? </p>
<p>The wire is initially in equilibrium between two forces : the repulsive force $F$ between the two charges $+Q$ and $+q$ which pushes the wire to the right, and the 2 tensions $T$ in each spring which pull the wire to the left : $F=2T$. The initial extension of the springs is the amplitude of oscillations : $T=kA$. </p>
<p>You are not asked to find the <strong>period</strong> of oscillations, so it is not necessary to write the equation of motion and deduce the angular frequency. To find the period we would need to know the mass of the wire, which is not stated. The amplitude will be the same regardless of the mass of the wire.</p>
<p>You have found $T$. Finding the amplitude $A$ is then straightforward. If the result $x=2a$ is not what you expected, there may be an error in the question or in the given answer.</p>
<p>$qQ/4\pi \epsilon_0 a^2$ has the unit of force ($N$), and $k$ has unit of force per length ($N/m$). So in order for $xqQ/8\pi^2\epsilon_0 a^3 k$ to have the unit of length (amplitude) then $x$ must also have the unit of length.</p>
<p><strong>Note :</strong> Do not confuse $x$ with the extension of the springs. In this question $x$ is an unknown factor in an equation. The initial extension of the springs is the amplitude $A$.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3022/amplitude-resulting-oscillation-frac-xqq-epsilon-find-value&show=3023#a3023Fri, 24 Aug 2018 17:14:45 +0000Answered: Normalization of a wave function
http://physics.qandaexchange.com/?qa=3018/normalization-of-a-wave-function&show=3019#a3019
<p>Your mistake is that you have not taken enough care about the definition of the function $|x|$. </p>
<p>For $-\infty \lt x \le 0$ the function $y=|x|$ is a reflection of the region $0 \le x \lt +\infty$. Likewise the function $y=e^{-2\lambda |x|}$ in the domain $-\infty \lt x \le 0$ is a reflection of the same function in the domain $0 \le x \lt +\infty$. </p>
<p>Therefore the integral over the domain $-\infty \lt x \lt +\infty$ is twice that over the domain $0 \le x \lt \infty$ : $$\int_{-\infty}^{+\infty} e^{-2\lambda |x|} dx=2\int_0^{+\infty} e^{-2\lambda x} dx=-\frac{1}{\lambda}[e^{-2\lambda x}]_0^{+\infty} =\frac{1}{\lambda}$$</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3018/normalization-of-a-wave-function&show=3019#a3019Thu, 23 Aug 2018 17:19:08 +0000Answered: Finding the matricial form of Brinkmann's metric
http://physics.qandaexchange.com/?qa=2994/finding-the-matricial-form-of-brinkmanns-metric&show=3014#a3014
<p>The coordinate variables are: <strong>v, x, y, u</strong></p>
<p>As you stated the metric is: </p>
<p>$$ds^2 = g_{i j} dx^i dx^j$$</p>
<p>The matrix has to symmetric and the value of $\delta_{i j}$ depends on <strong>i</strong> and <strong>j</strong> as follows:</p>
<p>If i = j, $\delta_{i j}$ = 1</p>
<p>If $i \neq j$, $\delta_{i j}$ = 0</p>
<p>Now let's find out the value of the matrix's diagonal:</p>
<p>1) $g_{1 1}$</p>
<p>$$ds^2 = g_{1 1} dv^2$$</p>
<p>As this factor does not appear in the given Brinkmann's metric, $g_{1 1} = 0$. Note our first variable is v.</p>
<p>2) $g_{2 2}$</p>
<p>$$ds^2 = g_{2 2} dx^2$$</p>
<p>This factor does appear in the given Brinkmann's metric and $\delta_{i j}$ provides us with the answer, $g_{2 2} = -1$</p>
<p>3) $g_{3 3}$</p>
<p>For the same reason that $g_{2 2} = -1$, $g_{3 3} = -1$</p>
<p>4) $g_{4 4}$</p>
<p>$$ds^2 = g_{4 4} du^2$$</p>
<p>Thus: </p>
<p>$$g_{4 4} = -K_{i j}x^ix^j$$</p>
<p>You can obtain the rest of the matrix proceeding like shown. To illustrate an example different from the diagonal:</p>
<p>5) $g_{1 4}$</p>
<p>$$ds^2 = g_{1 4} dvdu$$</p>
<p>Because of the matrix symmetry we know:</p>
<p>$$g_{1 4} = g_{4 1}$$</p>
<p>As the coefficient of $du dv$ is 1 we know:</p>
<p>$$g_{1 4} + g_{4 1} = 1$$</p>
<p>Therefore:</p>
<p>$$g_{1 4} = g_{4 1} = \frac{1}{2}$$</p>
<p>The entire matrix:<br>
<img src="https://cdn.pbrd.co/images/HzWgPIH.jpg" alt=""></p>
<p>PD: I suggest if you want to learn more about metrics, you check the lectures of Leonard Susskind on youtube, they are great. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=2994/finding-the-matricial-form-of-brinkmanns-metric&show=3014#a3014Sun, 19 Aug 2018 19:53:59 +0000Answered: Obtaining standard deviation using different equations
http://physics.qandaexchange.com/?qa=3002/obtaining-standard-deviation-using-different-equations&show=3011#a3011
<blockquote><p>I think my mistake has to be at b), calculating the average value of $(\Delta j)^2$.</p>
</blockquote>
<p>Yes, this part of your calculation does seem to be incorrect. In calculating the average you have to weight each value of $(\Delta j)^2$ by its frequency $N(j)$. You do not appear to have done that.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3002/obtaining-standard-deviation-using-different-equations&show=3011#a3011Sun, 19 Aug 2018 17:57:07 +0000Answered: Oblique collision of three disks
http://physics.qandaexchange.com/?qa=3005/oblique-collision-of-three-disks&show=3006#a3006
<p>The solution which you and Avnish Kabaj found (see diagram below) is almost certainly the one which the question-setter or examiner is looking for. However, I think the problem is more difficult than was intended, as you might see from considering what happens if the stationary disks were different sizes. It is not clear to me whether a satisfactory solution to this problem exists.<br>
<img src="https://cdn.pbrd.co/images/HAztt9L.png" alt=""><br>
<strong><strong><strong><strong><strong><strong><strong><strong><strong><strong><strong><strong><strong><strong><strong><strong>****</strong></strong></strong></strong></strong></strong></strong></strong></strong></strong></strong></strong></strong></strong></strong></strong></p>
<p>There are 2 related difficulties with this problem.</p>
<p>The 1st is that there is a simultaneous collision (ie simultaneous contact) between 3 rigid bodies. Such collisions are <strong>indeterminate</strong> : changing initial conditions infinitesimally such that one pair of disks collide first results in different outcomes depending on which pair collide first. See <a rel="nofollow" href="https://physics.stackexchange.com/q/296767">Multiple colliding balls</a> and links therein. This is (probably) unrealistic. </p>
<p>All real bodies are deformable to some extent. Extending the object model to include laws of deformation (eg Hooke's Law) enables the outcome of collisions with multiple simultaneous contacts to be determined uniquely. However, if we wish to keep the rigid body model then the options are either (1) ensure that no more than 2 objects are in contact at any one time (which is always possible if you look on a small enough time scale because rigid body collisions are by definition instantaneous ), or (2) adopt an <em>ad hoc</em> or empirical rule to define how the initial momentum will be distributed among the 3 or more bodies which are in contact. </p>
<p>The 2nd (related) difficulty is that the Coefficient of Restitution may not be definable for such collisions, because there may be a different ratio of relative velocity of separation and approach for each pair of objects. For example, if the stationary disks have different sizes they will leave at different angles with different velocities, resulting in a different COR for each of the smaller disks which depends on the angle of incidence for each disk. The COR should depend only on the material, which is the same for both disks; it should not depend on angle of incidence. This difficulty is hidden here because the smaller disks have been chosen to have the same radius. </p>
<p>The alternative definition of COR, based on ratio of total kinetic energy after and before collision, is less ambiguous but does not necessarily give the same value, whereas it always does in two-body collisions.</p>
<p><strong>In short, this problem may not be sufficiently well defined to have a single solution.</strong> </p>
<p>In my solution below I handle the collision by transforming it into sequential two-body collisions, by splitting the large disk into two equal parts, which collide with the smaller disks then collide with each other and coalesce. However, this method also fails to give a consistent value of COR for the two definitions, so it is not correct. </p>
<hr>
<p>Yes you are correct : coefficient of restitution must be measured along the common normal. However, I think this is complicated by the above problem of having two simultaneous collisions, as I shall explain below.</p>
<p>Impulse is transferred along the common normal, which is the line joining the centres of the larger disk A and smaller disks B. This line lies at an angle of $\theta$ to the horizontal, where $\sin\theta=1/3$. See LH diagram below.</p>
<p>The smaller disks B are initially at rest, so their final velocities are along the common normal. Let this final velocity of disks B be $v$ and the initial velocity of the larger disk A be $U$. Disk A has the same mass is $m$ as disks B. Conservation of linear momentum in the horizontal direction gives $$mU=2mv\cos\theta$$ $$v=\frac{U}{2\cos\theta}$$.<br>
<img src="https://cdn.pbrd.co/images/HzTYmbm.png" alt=""></p>
<p>The solution which you and Avnish Kabal proposed in the <a rel="nofollow" href="https://chat.stackexchange.com/transcript/message/46289104#46289104">PSE Problem-Solving Strategies Chatroom</a> resolves the vector velocity of A along and perpendicular to two different non-orthogonal directions at the same time. You cannot do this : the vector sum of the 4 components is greater than the original vector, which is clearly wrong. Your calculation assumes that the 2 collisions between disk A and each disk B are independent of each other by allowing disk A to approach each with velocity $U\cos\theta$ at the same time, but also inter-dependent by allowing momentum to be exchanged, otherwise momentum would not be conserved in each collision separately. </p>
<p>An alternative method of handling the simultaneous collision of the disk A with both disks B is to view A as two smaller disks C each of mass $\frac12m$ travelling together. Each disk C approaches disk B horizontally with speed $U$ the same as A, collides with it and rebounds so that linear momentum is conserved separately. See RHS of diagram above.</p>
<p>The disks C cannot be stationary after this collision with B because linear momentum would not be conserved. Instead the disks C must rebound vertically from B, then collide with each other and coalesce. This ensures that after they coalesce they have mass $m$ and are at rest again with no momentum, so they are equivalent to the larger disk A.</p>
<p>The vertical momentum with which each disk C rebounds from disk B must equal the final vertical momentum of disk B : $$\frac12mV=mv\sin\theta$$ $$V=2 v\sin\theta$$ where $V$ is the vertical velocity of the disks C.</p>
<p>Now we have enough information to calculate the Coefficient of Restitution, $e$ between the 2 disks C and B, which we assume is the same as between the 3 disks A and B. </p>
<p>The relative speed of approach between disks C and B along the common normal is $U\cos\theta$. The relative speed of separation along the same normal is $$v+V\sin\theta=v(1+2\sin^2\theta)=\frac{U(1+\frac12\sin^2\theta)}{2\cos\theta}$$ Therefore: $$e_V=\frac{U(1+2\sin^2\theta)}{2U\cos^2\theta}=\frac{1+\frac29}{2\times \frac89}=\frac{11}{16}$$ </p>
<p>The total kinetic energies before and after collision are $$K_1=\frac12 mU^2, K_2=2\times \frac12 mv^2$$ An alternative way of defining the COR is related to the fraction of kinetic energy retained after collision : $$e_K=\sqrt{\frac{K_2}{K_1} }=\sqrt2 \frac{v}{U}=\sqrt2\frac{1}{2 \cos\theta}=\frac{3}{4}$$ </p>
<p>This answer does not match any of the given options but is close to (b), which is $\frac{9}{16}$. The reason is clearly the way in which I have handled the simultaneous collision. It does not give a consistent method of calculating COR. KE is lost in the collision between the imaginary disks C when they coalesce. However, perhaps no method is consistent :</p>
<p>The solution which both you and Avnish Kabaj have obtained is that $$v=\frac{U}{2\cos\theta}=\frac{3}{4\sqrt2}U$$ which results in $$e_V=\frac{v}{U\cos\theta}=\frac{U}{2U\cos^2\theta}=\frac{9}{16}$$ which applies to each collision separately. The definition of $e$ using the ratio of kinetic energies is the same as above, viz. $e_K=\frac34$ which applies to both collisions collectively. However, note that $e_K=\sqrt{e_V}$. If we had obtained instead that $e_K=e_V^2$ then we might explain this by saying that the two contacts were sequential - ie a fraction $e_V$ of KE is lost in each contact.</p>
<p>If COR is defined consistently then these two methods should give the same result - ie $e_V=e_K$. Neither my method nor yours gives the a consistent result for COR. </p>
<p><strong>Conclusion :</strong> it may not be possible to define the COR of a multi-body collision.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3005/oblique-collision-of-three-disks&show=3006#a3006Sun, 19 Aug 2018 05:08:46 +0000Answered: Rising balloons
http://physics.qandaexchange.com/?qa=3000/rising-balloons&show=3003#a3003
<p>(a) Each segment is in static equilibrium under the forces acting on it. The only forces with a horizontal component are the tensions in the segments of rope. So the horizontal tension forces pulling on the left and right sides of each segment are equal. For example $T_1\cos\theta_1=T_3\cos\theta_3$ and $T_2\cos\theta_2=T_4\cos\theta_4$ etc. This argument applies for all segments, so it means that the horizontal component of tension in each segment is the same : $T_i \cos\theta_i=T_H=\text{ a constant}$.</p>
<p>(c) $\tan \theta_{N+1}= -\tan \theta_0$ because $\theta_{N+1}= -\theta_0$. </p>
<p>This occurs because the slope of the rope segments changes from +ve to -ve as you move from left to right.</p>
<p>$\theta$ is +ve when measured anticlockwise from the $+x$ direction. After reaching the middle of the rope $\theta$ becomes -ve because the angle is then measured clockwise from the $+x$ direction.</p>
<p>(d) I do not see where you are asked to prove or assume that $\tan \theta_{i-1}=\tan \theta_0$. This is not correct. It implies that $\theta_i=\theta_0$ which is obviously wrong.</p>
<p>$$ \sin \theta=\tan \theta \cos\theta$$ therefore $$F=T_0 \sin\theta_0 -T_1\sin\theta_1=T_0\cos\theta_0 \tan \theta_0 -T_1\cos\theta_1 \tan\theta_1=T_H(\tan \theta_0 -\tan \theta_1)$$ $$F=T_H(\tan \theta_1-\tan \theta_2)$$ $$F=T_H(\tan\theta_2-\tan\theta_3)$$ and so on. In general $$F=T_H (\tan \theta_{i-1} -\tan \theta_i )$$ </p>
<p>Adding these i equations together we get $$iF=T_H( \tan_0-\tan \theta_i)$$ $$\tan \theta_i=\tan \theta_0 -\frac{iF}{T_H}=\frac{NF}{2T_H} -\frac{2iF}{2T_H}=\frac{(N-2i)F}{2T_H}$$</p>
<p>The equations for $x_i, y_i$ are derived from : $$y_i =l\sin \theta_0 +l\sin\theta_1 +l\sin\theta_2 +\text{...}+l \sin \theta_{i-1}$$ $$=l\sum_0^{i-1} \sin\theta_j$$ and similar for $x_i$.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3000/rising-balloons&show=3003#a3003Sat, 18 Aug 2018 22:28:33 +0000Answered: What is electromotive force induced between its ends
http://physics.qandaexchange.com/?qa=2998/what-is-electromotive-force-induced-between-its-ends&show=3001#a3001
<p>You should expect the EMF induced in the quarter circle to be constant because of the symmetry of the magnetic field of the dipole : whatever angle the quarter circle turns through, the magnetic field distribution in the plane of the quarter circle always looks the same.</p>
<p>Imagine that the quarter circle is located on the surface of a sphere of radius $r$. As it rotates about the vertical axis of the sphere, it sweeps out a region $S$ on the surface of the sphere. After one complete revolution $S$ is the surface of a hemisphere.</p>
<p>The magnetic field due to the magnetic dipole has radial, tangential and azimuthal components. But we are only interested in the radial component $B(r)$ which crosses the surface of the sphere. The flux of $B(r)$ across the upper hemisphere is the flux through the area swept out by the quarter ring as it makes one complete revolution. This is the flux $\phi$ which you need to use in Faraday's Law $E=-\frac{d\phi}{dt}$.</p>
<p>The radial component of the magnetic field of the dipole is $$B(r)=\frac{\mu_0 m}{4\pi r^3} (2\cos\theta)$$ where $m$ is the magnitude of the dipole moment $\mathbf{m}$ and $\theta$ is the angle between the radius and the dipole axis. (See eg <a rel="nofollow" href="https://en.wikipedia.org/wiki/Magnetic_dipole#External_magnetic_field_produced_by_a_magnetic_dipole_moment">wiki: Magnetic Dipole External Field</a>.)</p>
<p>The area of an elementary hoop of the sphere of radius $R=r\sin\theta$ and width $rd\theta$ at polar angle $\theta$ is $$dA=2\pi R. rd\theta=2\pi r^2 \sin\theta d\theta$$ So the flux through the upper hemisphere is $$\Phi=\int_0^{\pi/2} B(r)dA =\frac{\mu_0 m}{2r}<br>
\int_0^{\pi/2} \sin2\theta d\theta=\frac{\mu_0 m}{2r}$$</p>
<p>The rate at which this flux is cut by the quarter circle equals the EMF induced in the ring : $$E=\frac{d\phi}{dt}=\frac{\omega}{2\pi}\Phi=\frac{\mu_0 m\omega}{4\pi r}$$ because $$\phi=\frac{t}{T}\Phi=\frac{\omega}{2\pi}\Phi t$$ where $T$ is the time for the quarter circle to make one rotation.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=2998/what-is-electromotive-force-induced-between-its-ends&show=3001#a3001Sat, 18 Aug 2018 17:34:48 +0000Answered: Average velocity Of the particle in a large time interval.
http://physics.qandaexchange.com/?qa=2986/average-velocity-of-the-particle-in-a-large-time-interval&show=2997#a2997
<p>Your method of solution is correct. Your mistakes (I think) are in the calculations of $T_1, T_2$ and $x$, all of which should depend on angle $\theta$. Your formula for $T_2$ depends on $\theta$ but that for $T_1$ does not. Likewise your formula for $x$ appears to be $2R_1-2R_2\sin\theta$. The 2nd term depends on $\theta$ but the 1st term does not.</p>
<hr>
<p>The motion in each region is a circular arc of radius $r$. The centripetal force for this motion is provided by the magnetic force on the particle : $$\frac{mv^2}{r}=Bqv$$ $$r=\frac{mv}{qB}$$ The distance moved along the $x$ axis is the chord of the arc, which is $2r\sin\theta$ where $\theta$ is the angle between the arc and the $x$ axis. </p>
<p>After tracing an arc in both regions, the motion repeats. During each cycle the resultant distance moved in the $+x$ direction is $$x=2(r_1-r_2)\sin\theta=\frac{2mv\sin\theta}{q}(\frac{1}{B_1}-\frac{1}{B_2})=\frac{2mv\sin\theta (B_2-B_1)}{qB_1B_2}$$</p>
<p>The angle subtended by each arc is $\theta_1=2\theta$ and $\theta_2=2(\pi-\theta)$ respectively. The times for the particle to traverse each arc are $\frac{r_1\theta_1}{v}, \frac{r_2\theta_2}{v}$ respectively. The total time for each cycle of the motion is therefore $$t=\frac{1}{v}(r_1\theta_1+r_2\theta_2)=\frac{1}{v}(2\theta\frac{mv}{qB_1}+2(\pi-\theta)\frac{mv}{qB_2})=\frac{2m((\pi-\theta) B_1+\theta B_2)}{qB_1B_2}$$</p>
<p>The average velocity of the particle in the $+x$ direction is $$V=\frac{x}{t}=\frac{(B_2-B_1)v\sin\theta}{(\pi-\theta)B_1+\theta B_2}=\frac{(B_2-B_1)v\sin\theta}{\pi B_1+\theta (B_2-B_1)}$$</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=2986/average-velocity-of-the-particle-in-a-large-time-interval&show=2997#a2997Mon, 13 Aug 2018 18:05:06 +0000Answered: Mutual indutance of two long parallel wires
http://physics.qandaexchange.com/?qa=2990/mutual-indutance-of-two-long-parallel-wires&show=2993#a2993
<p>It is not clear what you are asking. Mutual inductance is not mentioned in the problem statement. Perhaps you are making the problem more complicated than it is by asking about mutual inductance.</p>
<p>Current-carrying wires cannot exist on their own without some circuit. I think the key to understanding what is going on in such questions is to ask "Where is/are the circuit/s?" </p>
<p>If identical parallel wires AB and DC carry the same current in opposite directions (so that both currents increase/decrease together) then they can be considered to be part of the same circuit, eg the loop ABCD of circuit I in the upper diagram below. If so you are really asking for the <strong>self-inductance</strong> of this circuit. </p>
<p>On the other hand if the two parallel wires AB and A'B' are part of separate circuits, such as the loops ABCD and A'B'C'D' in circuits I and II below, then you are indeed asking for the <strong>mutual inductance</strong> of these two circuits. This depends on the relative orientation of the circuits.</p>
<p><img src="https://cdn.pbrd.co/images/HyFKdGN.png" alt=""></p>
<p><strong>Self Inductance of Circuit I</strong></p>
<p>The sides AB, CD are so long that the magnetic fields due to the short sides AD, BD and corners effects are assumed to be negligible.</p>
<p>Assume that the current is distributed uniformly over the cross-section of the wire in the circuit. Using Ampere's Law the magnetic field around each wire has the form $$B(r)=\frac{r^2}{a^2}\frac{\mu_0 I}{2\pi r} \text{ for } r \le a$$ $$B(r)=\frac{\mu_0 I}{2\pi r} \text{ for } r \gt a$$ where $a$ is the radius of the wire and $r$ is the distance from each wire. </p>
<p>The magnetic flux through a unit length of circuit in the direction AB is $$\Phi=2\int_0^d B(r)dr=\frac{\mu_0 I}{\pi}(\int_0^a \frac{r}{a^2}dr+\int_a^d \frac{1}{r}dr)=\frac{\mu_0 I}{\pi}[\frac12+\ln\frac{d}{a}]$$ The self inductance per unit length $L$ of the circuit is defined by $$\Phi = LI$$ therefore $$L=\frac{\mu_0}{\pi}[\frac12+\ln\frac{d}{a}]$$</p>
<p>For comparison see <a rel="nofollow" href="https://physics.stackexchange.com/q/184728">Inductance of two parallel wires</a>.</p>
<p><strong>Mutual Inductance between Circuits I & II</strong></p>
<p>Assume that the 2 circuits lie in the same plane.</p>
<p>The flux through unit length of circuit I due to circuit II is </p>
<p>$$\Phi_{12}=\Phi_{A'B'}-\Phi_{C'D'}=\int_x ^{d+x} B(r)dr+\int_{2d+x} ^{d+x} B(r)dr=\frac{\mu_0 I}{2\pi}\ln\frac{(d+x) ^2}{(2d+x)x}$$</p>
<p>The mutual inductance per unit length $M_{12}$ is defined by </p>
<p>$$\Phi_{12}=M_{12} I_2$$ </p>
<p>therefore </p>
<p>$$M_{12}=\frac{\mu_0}{2\pi}\ln\frac{(d+x)^2}{(2d+x)x}$$</p>
<p>This means that if the current in circuit II is changing then an emf is induced in circuit I of magnitude </p>
<p>$$E=\frac{d\Phi_{12}}{dt}=M_{12}\frac{dI_2}{dt}$$</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=2990/mutual-indutance-of-two-long-parallel-wires&show=2993#a2993Sat, 11 Aug 2018 12:08:11 +0000Both molar specific heat of an ideal monatomic gas at constant volume and pressure
http://physics.qandaexchange.com/?qa=2984/both-molar-specific-ideal-monatomic-constant-volume-pressure
<blockquote><p>2 mols of oxygen are heated up from a temperature of 20 C and a pressure 1 atm to a temperature of 100 C. Compute:<br>
a) Having constant volume, how much heat must be provided to the gas throughout the process?<br>
b) Having constant pressure, how much work is done throughout the process?<br>
c) Compute in a) and b) cases the variation in the internal energy.</p>
</blockquote>
<p>My try:</p>
<p>The molar specific heat of an ideal gas at constant volume is:</p>
<p>$$C_v = \frac{3}{2}R$$</p>
<p>The molar specific heat of an ideal gas at constant pressure is:</p>
<p>$$C_p = \frac{5}{2}R$$</p>
<p>Then at a) it is just about using $Q = nC\Delta T$. My issue here is that the given outcome at a) uses n = 1:</p>
<p>$$Q = nC_v\Delta T = 997.68J$$ </p>
<p>I obtained 1995.36J (I used n=2).</p>
<p>To compute Q at constant pressure:</p>
<p>$$Q = nC_p\Delta T = 1662.80J$$ </p>
<p>I obtained 3325.6J (I used n=2).</p>
<p>I do not understand why it uses n=1 as it is specified that the gas is formed by 2 mols of oxygen. </p>
<p>b) When heat is supplied at constant pressure the gas disseminates and exerts a positive work (such as on a piston).</p>
<p>However the answer is given as a negative number, which means work is done by the system and not over it. Thus the given answer:</p>
<p>$$W = -nR\Delta T = -1330.24J$$ </p>
<p>Here I got the same but positive.</p>
<p>c) Using the first principle of thermodynamics:</p>
<p>$$ \Delta E_i = Q + W$$</p>
<p>When volume is constant:</p>
<p>$$ \Delta E_i = Q_v + W = -332.56J$$</p>
<p>When pressure is constant:</p>
<p>$$ \Delta E_i = Q_p + W = 332.56J$$</p>
<p>What I got before seeing the answers:</p>
<p>When volume is constant:</p>
<p>$$ \Delta E_i = Q_v + W = 3325.60J$$</p>
<p>When pressure is constant:</p>
<p>$$ \Delta E_i = Q_p + W = 4655.84J$$</p>
<p>As you can see all comes down to the definition: $Q = nC_v\Delta T$ I have made some research and in Tipler and $C_v$, $C_p$ and Q are defined: </p>
<p>$$C_v = n\frac{3}{2}R$$</p>
<p>$$C_p = n\frac{5}{2}R$$</p>
<p>$$Q = C\Delta T$$</p>
<p>Eventually it is the same, since I multiplied by 2 (number of mols) in the definition: $Q = nC\Delta T$ and I did not do it my first $C_v$ and $C_p$ definitions. Therefore I think my outcomes are right but what do you think?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=2984/both-molar-specific-ideal-monatomic-constant-volume-pressureThu, 09 Aug 2018 08:59:16 +0000Mean free path of ideal gas
http://physics.qandaexchange.com/?qa=2974/mean-free-path-of-ideal-gas
<p>There's a box with a wall in mid dividing it in two sections, each filled with same ideal gas in both sections at 150 K in one section and at 300 K in another. How am I supposed to calculate ratio of mean free paths (MFP)in 2 sections?</p>
<p>My attempt: MFP ~ Volume / Number of particles</p>
<pre><code> => MFP ~ Temperature / Pressure
</code></pre>
<p>Now, Assuming pressure to be same on both sections, ratio must be half. But that is incorrect. Why?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=2974/mean-free-path-of-ideal-gasTue, 07 Aug 2018 13:38:41 +0000Answered: Finding current through resistor as a function of time in the presence of 2 inductors
http://physics.qandaexchange.com/?qa=2940/finding-current-through-resistor-function-presence-inductors&show=2968#a2968
<blockquote><p>Now, when switch is shifted, there's flux change through both the<br>
inductors. Both will develop polarity with positive part on the same<br>
side as the positive side of the battery</p>
</blockquote>
<p>The problem is somewhat pathological in that the voltage across the inductors cannot be ordinary functions of time.</p>
<p>Why? The voltage across an inductor is proportional to the time derivative of the current through which means that in order for the inductor voltage to be defined for all time $t$, the inductor current must be continuous.</p>
<p>But, in this problem, the current through the inductors <em>cannot</em> be continuous at the time the switch is 'shifted' (thrown) to position 2.</p>
<p>Just before the switch is thrown, the resistor (and $L$ inductor) current is $i_R = \frac{\mathcal{E}}{R}$ while the current through the $2L$ inductor is zero.</p>
<p>Now, just after the switch is thrown, the current through the two inductors must, by KCL, be equal and so, one (or both) inductor current(s) must be discontinuous which implies that the inductor voltages do not exist at that time (the inductor voltages have a delta 'function' at the switching time).</p>
<p>One might try placing a large resistance $r \gg R$ in parallel with the $L$ inductor which will permit a continuous current and thus a well defined voltage across each inductor.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=2940/finding-current-through-resistor-function-presence-inductors&show=2968#a2968Sat, 04 Aug 2018 23:15:33 +0000