Physics Problems Q&A - Recent questions and answers in Physics Problems
http://physics.qandaexchange.com/?qa=qa/physics-problems
Powered by Question2AnswerElectric field intensity at the centre of a solid hemisphere with azimuthal volume density
http://physics.qandaexchange.com/?qa=3360/electric-intensity-centre-hemisphere-azimuthal-volume-density
<p><a rel="nofollow" href="https://imgur.com/a/n0iohng">https://imgur.com/a/n0iohng</a></p>
<p>I am using Guru's Electromagneic Theory Fundamentals: <a rel="nofollow" href="https://www.amazon.ca/Electromagnetic-Field-Theory-Fundamentals-Singh/dp/0521116023">https://www.amazon.ca/Electromagnetic-Field-Theory-Fundamentals-Singh/dp/0521116023</a></p>
<p>For finding the electric field intensity, I am using the equation:</p>
<p><a rel="nofollow" href="https://imgur.com/a/RHS6SfY">https://imgur.com/a/RHS6SfY</a></p>
<p>I am unsure about what the value of r prime should be for a solid. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3360/electric-intensity-centre-hemisphere-azimuthal-volume-densitySun, 20 Jan 2019 03:09:55 +0000Rotating ball inside rotating cylinder.
http://physics.qandaexchange.com/?qa=3359/rotating-ball-inside-rotating-cylinder
<p>How do I proof: <img src="https://cdn.pbrd.co/images/HWY3y5g.png" alt=""> </p>
<p>Textbook solution :<br>
<img src="https://cdn.pbrd.co/images/HXsICHVJ.png" alt=""></p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3359/rotating-ball-inside-rotating-cylinderFri, 18 Jan 2019 05:18:03 +0000Frequency modes of the rectangular shell
http://physics.qandaexchange.com/?qa=3357/frequency-modes-of-the-rectangular-shell
<p>This is task i received from my professor:</p>
<blockquote><p>The shapes of three natural modes having the frequencies $\omega_1, \omega_2, \omega_3$ of the rectangular shell are presented in the figure. The exciting pressure $p(t)$ applied uniformly all over the one side of the shell has the form $p(t) = Pe^{jωt}$. <br>
Make a sketch of the normal displacement of the gravity point of the shell against frequency, if the excitation frequency varies within bounds $0.5\omega_1< \omega <2\omega_1$ and static displacement of that point equals to $u_0$.</p>
</blockquote>
<p>Links to photo of frequency nodes (sorry for low quality)-><a rel="nofollow" href="https://ibb.co/b1nh3j9">1</a> .</p>
<p>Can somebody help me and tell me how i should get started?</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3357/frequency-modes-of-the-rectangular-shellMon, 14 Jan 2019 21:42:44 +0000Answered: tension in wires from which a block is suspended
http://physics.qandaexchange.com/?qa=3349/tension-in-wires-from-which-a-block-is-suspended&show=3354#a3354
<p>In order to compare $T$ and $W$, one needs to express one in terms of the other. As explained in the diagram itself, since the block is in translational equilibrium, we have:</p>
<p>$$\vec{T_1}+\vec{T_2}=\vec{W_b}$$</p>
<p>Projecting this relation onto the $Oy$ axis, the components of the tensions are both $T\sin\theta$ (because the angle between $T$ and $Oy$ is $\frac{\pi}{2}-\theta$ and the angle between $T$ and $Ox$ is $\theta$). Therefore:</p>
<p>$$2T\sin\theta=W\implies T=\dfrac{W}{2\sin\theta}$$</p>
<p>The information <em>the strings are almost horizontal</em> is equivalent to $\theta\cong 0$. For very small angles, the sine function is as well extremely small, and almost equal to the angle $\theta$ itself. Take a look at the following examples (I've included them here solely to make sure that you understand why $\sin\theta\to 0$ as $\theta\to 0$)</p>
<ul>
<li>$\sin(0.01)\cong 0.00999983333333\cong 0.01$</li>
<li>$\sin(0.0001) \cong 0.00009999999983\cong 0.0001$</li>
<li>$\sin(0)=0$</li>
</ul>
<p>So the denominator $2\sin\theta\cong0$ and therefore $T\to\infty$. As $W=mg$, clearly $W$ is a finite value (making the reasonable assumption that the mass is not inifite), and therefore <em>clearly</em> $T>W$. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3349/tension-in-wires-from-which-a-block-is-suspended&show=3354#a3354Fri, 11 Jan 2019 19:52:11 +0000How to solve this throwing ball off cliff question
http://physics.qandaexchange.com/?qa=3350/how-to-solve-this-throwing-ball-off-cliff-question
<p><img src="https://cdn.pbrd.co/images/HVdO2s0.png" alt=""></p>
<p>For b) I was curious why i couldn't use $s = ut + \frac12at^2$ as my equation as i had all the necessary known values. Instead to get the correct answer i had to use $v=u + at$.</p>
<p>Why is this so? Thank you for your time.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3350/how-to-solve-this-throwing-ball-off-cliff-questionSun, 06 Jan 2019 15:45:23 +0000Time period of small oscillation- disk and particle system
http://physics.qandaexchange.com/?qa=3346/time-period-of-small-oscillation-disk-and-particle-system
<p><img src="https://cdn.pbrd.co/images/HUWMQA2.jpg" alt=""></p>
<p>Attempt: </p>
<p>$y_{COM} = \dfrac{R}{3} = \dfrac {25}3$</p>
<p>$I = \dfrac 12 (2)R^2 + 1R^2 = 2\times 25 = 50 $</p>
<p>Now, T is given by: $T =2\pi \sqrt{\dfrac{I}{mgl}}$ where l is the distance of the centre of mass from the centre of rotation</p>
<p>So $T =2\pi \sqrt{\dfrac{50}{3 \times 10 \times \dfrac{25}{3}}} = 2\pi\sqrt{\dfrac{1}{5}}$</p>
<p>No option :( </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3346/time-period-of-small-oscillation-disk-and-particle-systemFri, 04 Jan 2019 22:33:35 +0000Wave optics- number of minimas.
http://physics.qandaexchange.com/?qa=3345/wave-optics-number-of-minimas
<p><img src="https://cdn.pbrd.co/images/HUWujTQ.jpg" alt=""></p>
<p>Attempt: </p>
<p>Let m be the integer associated with $420 nm$<br>
and n be the integer associated with $540 nm$</p>
<p>$d \sin \theta = k \lambda $ $k\in Z$<br>
Clearly, </p>
<p>$m_{max} = 180$ </p>
<p>$n_{max} = 140 $</p>
<p>$(2m+1) \lambda_1 = (2n+1)\lambda_2$ (condition for dark fringes to overlap)</p>
<p>$\implies \dfrac{2m+1}{2n+1} = \dfrac 97$</p>
<p>$\implies m = \dfrac{1+7n +2n }{7}$</p>
<p>Hence, we obtain, for m to be an integer: $2n+1 = 7k$</p>
<p>$\implies n = \dfrac{7k-1}{2}$ where $k \in Z$</p>
<p>Now note that k must be odd since odd-1 = even </p>
<p>Thus, using $n \le 140$, $k_{max} = 39$</p>
<p>Now, we have to consider only odd values of k which are $1,3,...39$ = 20 numbers </p>
<p>Thus, we have 20 minimas on the upper side and 20 on the lower,<br>
Total minimas = $20+20 = 40$</p>
<p>But answer is $D$</p>
<p><strong>Question 1:</strong> What is wrong with my method? </p>
<p><strong>Question 2:</strong> Considering that this is a JEE Mains problem, what is the fastest 2 minute way to do it? </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3345/wave-optics-number-of-minimasFri, 04 Jan 2019 22:01:11 +0000Answered: oblique collision of a rigid rod with a plane surface
http://physics.qandaexchange.com/?qa=3336/oblique-collision-of-a-rigid-rod-with-a-plane-surface&show=3344#a3344
<p>The normal contact impulse $J$ between the rod and plane has two effects : it changes the velocity of the CM of the rod, and it makes the rod rotate about its CM. </p>
<p>The impulsive force is related to the change in momentum of the CM of the rod, and the impulsive torque to the change in angular momentum of the rod, as follows : $$J=m(v_0+v_1)$$ $$J\frac{L}{2}\cos\theta=I\omega$$ Here $v_0, v_1$ are the vertical velocities of the CM before and after the collision, and $I=\frac{1}{12}mL^2$ is the moment of inertia of the rod about its CM. </p>
<p>The rod is not rotating initially. After the collision, the velocity of the end A which collided with the ground is $v_2'=\frac{L}{2}\omega$ relative to the CM of the rod, directed perpendicular to the rod, ie at an angle of $\theta=60^{\circ}$ to the vertical. The CM of the rod is moving upwards with speed $v_1$ so the vertical component of the velocity of A relative to the ground after collision is $$v_2=v_2'\cos\theta+v_1$$ </p>
<p>The elasticity of the collision is known ($e=1$) so the <strong>Law of Restitution</strong> can be applied to the relative velocities of approach and separation at the point of contact : $$v_2=ev_0$$</p>
<p>From the above equations you can eliminate $v_2, v_2', \omega$ to find the rebound speed $v_1$ of the CM of the rod and thereby the height to which the CM rises after collision.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3336/oblique-collision-of-a-rigid-rod-with-a-plane-surface&show=3344#a3344Wed, 02 Jan 2019 22:07:06 +0000Answered: Diffusion with reflecting boundary
http://physics.qandaexchange.com/?qa=3254/diffusion-with-reflecting-boundary&show=3343#a3343
<p>You have not gone wrong. You simply have not gone far enough.</p>
<p>You have obtained a quartic equation $$z^4-2z+1=0$$ This has 4 solutions, not only the obvious one $z=1$. The solution $z=1$ requires $t=+\infty$ so this is not the solution you require. You need to find some more solutions. Then check which one describes the situation you are trying to find.</p>
<p>In fact there is one other real solution and 2 imaginary solutions. Probably you should solve numerically (this is a physics question, not maths). </p>
<p>One simple method is to make a first guess $z<em>0$ then reiterate $$z</em>{n+1}=\frac12 (z_n^4+1)$$ With an initial guess of $z_0=0.5$ I get the following iterations :<br>
0.5<br>
0.53125<br>
0.5398259163<br>
0.5424604827<br>
0.543295467<br>
0.543562654<br>
0.5436484118<br>
0.543675964<br>
0.5436848186<br>
0.5436876646</p>
<p>These converge to 0.5437, though not quickly. </p>
<p>Alternatively apply the <a rel="nofollow" href="http://www.sosmath.com/calculus/diff/der07/der07.html">Newton-Raphson Method</a>. Using $$f(z)=z^4-2z+1=0$$$$f'(z)=4z^3-2$$ $$z_{n+1}=z_n-\frac{f(z_n)}{f'(z_n)}$$ again with $z_0=0.5$ we get<br>
0.5<br>
0.5416666667<br>
0.5436837222<br>
0.5436890127<br>
0.5436890127</p>
<p>which converges much more rapidly.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3254/diffusion-with-reflecting-boundary&show=3343#a3343Wed, 02 Jan 2019 16:42:47 +0000Answered: floating cubical block
http://physics.qandaexchange.com/?qa=3319/floating-cubical-block&show=3342#a3342
<p>I think you have misinterpreted the question. </p>
<p>The copper block is not hollow, it is solid copper. There is nothing inside it except copper. Mercury and water are not poured into the copper block. They are poured into a vessel in which the copper block is able to float.</p>
<p>Initially the copper block is floating inside a vessel (eg a beaker) which contains a layer of liquid mercury at the bottom. Some water is added on top of the mercury, surrounding the copper block, until the water is level with the top of the copper block. Now the copper block is floating in a layer of mercury and a layer of water. The question is asking for the depth of the layer of water. </p>
<p>The words "the copper in the block just gets submerged" in your question should read "the copper block just gets submerged".</p>
<hr>
<p>Mercury is a liquid. Water is a liquid. The LHS of your equation gives the pressure in the mercury level with the base of the copper block (neglecting air pressure). The RHS is the pressure due to copper block, which you can think of as another liquid.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3319/floating-cubical-block&show=3342#a3342Wed, 02 Jan 2019 15:26:29 +0000Answered: Cylinder lying on conveyor belt
http://physics.qandaexchange.com/?qa=3323/cylinder-lying-on-conveyor-belt&show=3340#a3340
<p>In essence this question has been answered already in <a rel="nofollow" href="http://physics.qandaexchange.com/?qa=470/rotation-in-truck">rotation of wheels and axle in accelerating truck</a>. There I show that the linear acceleration $a_0$ (relative to the ground) and the rotational acceleration $\alpha$ of a solid disk or cylinder which rolls without slipping on a lorry (or conveyor belt) which has translational acceleration $a$ are given by $$a_0=\frac13 a$$$$R\alpha=2a_0=\frac23 a$$ where $R$ is the radius of the cylinder. </p>
<p>The friction force is $$f=\frac13 Ma \le \mu Mg$$ so to avoid slipping the acceleration of the belt must be limited to $a \le 3\mu g$. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3323/cylinder-lying-on-conveyor-belt&show=3340#a3340Wed, 02 Jan 2019 08:31:03 +0000Answered: Final charge on the capacitor and Charge on capacitor as a function time
http://physics.qandaexchange.com/?qa=3311/final-charge-the-capacitor-charge-capacitor-function-time&show=3318#a3318
<p>The diagram is faulty. Switch S1 needs to be in the same branch as the capacitor to prevent it from discharging before S1 is closed. But we can guess what they mean.</p>
<p>Using KVL is not much more difficult than what you have done already. It should give you the correct answer - for example, as follows.</p>
<p>Mark 3 loop currents $i_1, i_2, i_3$ as shown in the diagram below. Then applying KVL around each loop we get $$\frac{Q}{C}=3Ri_1-Ri_2$$ $$-\frac{Q}{C}=4Ri_2-Ri_1-2Ri_3$$ $$E=4Ri_3-2Ri_2$$ We are interested in the current through the capacitor which is $i=i_1-i_2$. Substituting and eliminating $i_1, i_2, i_3$ we obtain <br>
$$i=-\frac{dQ}{dt}=\frac{1}{2CR}(Q-\frac14 CE)$$<br>
$$Q-\frac14 CE=(Q_0-\frac14 CE)e^{-t/2RC}$$ </p>
<p>After a long time ($t \to \infty$) the factor $e^{-t/2RC} \to 0$. Then the charge on the capacitor becomes $Q_{\infty}=\frac14CE$.</p>
<p><img src="https://i.imgur.com/y964baO.png" alt=""></p>
<p>A quicker solution is to find the equivalent resistance $R'$ across the capacitor after shorting the cell. The time constant of the decay (or increase) of charge on the capacitor is then $C R'$. If the initial charge is $Q_0$ and the final charge is $Q_1$ then the time-variation of the charge is given by </p>
<p>$$ Q(t) - Q_1 = (Q_0 - Q_1) e^{-t / C R' }$$ </p>
<p>After shorting the cell the circuit can be drawn as on the right above, because B and C are at the same potential. The resistance between C and D is $R$ because this is the result of two resistors $2R$ in parallel. Then the resistance along ADC is $2R$. This is the same as the resistance along ABC. ADC and ABC are in parallel, and have an equivalent resistance $R$. This is in series with the resistor in branch AC. So the effective resistance in series with the capacitor is $R'=2R$. The time constant for the circuit is $CR'=2CR$.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3311/final-charge-the-capacitor-charge-capacitor-function-time&show=3318#a3318Tue, 18 Dec 2018 03:46:46 +0000Answered: A projectile is ejected with vertical speed $v$ from the surface of a planet of mass $M$ and radius $R$
http://physics.qandaexchange.com/?qa=3295/projectile-ejected-with-vertical-speed-surface-planet-radius&show=3316#a3316
<p>Your mistake is that you assumed that the acceleration due to gravity $g$ is constant. This is only approximately true over distances which are very much smaller than the average distance of the object from the centre of the planet. The equation $y=vt-\frac12 gt^2$ only applies for constant acceleration $g$.</p>
<p>You need to use the conservation of energy. The initial KE plus gravitational PE at the surface of the planet is equal to the final KE plus gravitational PE at any other point such at that where it comes to rest instantaneously.</p>
<p>If the projectile were not fired vertically but instead at an angle to the vertical it would be necessary also to apply the conservation of angular momentum. In this problem the angular momentum is always zero.</p>
<p>If $v=c$ then Special Relativity Theory ought to apply instead of Newtonian Mechanics, so the kinetic energy would not be $\frac12 mv^2$. However this question assumes that Newtonian Mechanics is still valid when $v=c$. The tags for Special and General Relativity are not needed for this question.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3295/projectile-ejected-with-vertical-speed-surface-planet-radius&show=3316#a3316Sun, 16 Dec 2018 00:00:02 +0000Answered: Prove that the vector of acceleration always points towards the origin
http://physics.qandaexchange.com/?qa=3312/prove-that-vector-acceleration-always-points-towards-origin&show=3313#a3313
<p>This is a very mixed-up question.</p>
<p>Planets move in elliptical orbits. which can be described by the <strong>parametric equations</strong> $$x(t)=a\cos(kt), y(t)=b\sin(kt)$$ However, in these equations <strong>in general t is NOT time</strong>. It is just a parameter with no physical meaning. These equations correctly describe the path ($x,y$) taken by the planet, but not the relations ($x, t$) and ($y, t$) between its position and time. That relation is much more complex, and cannot be described by any simple function.</p>
<p>The only case in which parameter $t$ represents time for a gravitational orbit, for which the force has the form $-K/r^2$, is when the orbit is a circle, ie when $a=b$. Then $k$ is the angular velocity. In general (ie when $a\ne b$) the angular velocity is not $k$.</p>
<p>The equations do give the correct relation between position and time for an orbit of a mass on a spring with zero natural length, ie <strong>a Hooke's Law central force</strong> which has the form $-Kr$. However, even for such an orbit $k$ is not the angular velocity, which is not constant - unless the orbit is circular. </p>
<p><strong>How to show that the vector acceleration always points towards the origin</strong></p>
<p>The functions you have obtained for acceleration are the components of a vector : $$a_x=-k^2x, a_y=-k^2y$$ The magnitude of this vector is $$a=\sqrt{a_x^2+a_y^2}=-k^2r$$ where $r$ is the distance of the object from the centre of the ellipse. This confirms that the central force is proportional to $r$ - ie it is a Hooke's Law force. You need to find the direction of this vector. The gradient of the vector wrt the x axis is $$m=\frac{a_y(t)}{a_x(t)}=\frac{y}{x}$$ This vector passes through the point $(x, y)$ and its gradient is $y/x$. From this information you can conclude that <strong>the force is always directed towards the origin</strong> which is at the centre of the ellipse.</p>
<p>Although the orbit is an ellipse, <strong>it has no relation to a gravitational orbit</strong>, for which the force is always directed towards one focus of the ellipse, where the Sun is located, not towards the centre of the ellipse. </p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3312/prove-that-vector-acceleration-always-points-towards-origin&show=3313#a3313Sat, 15 Dec 2018 22:49:42 +0000Answered: System of two ideal Fermi gases in three dimensions.
http://physics.qandaexchange.com/?qa=3294/system-of-two-ideal-fermi-gases-in-three-dimensions&show=3310#a3310
<p>It is all about knowing that the ground-state pressure of the ideal Fermi gas is:</p>
<p>$$P_o = \frac{2}{5} n \epsilon_f$$</p>
<p>And actually (after some dimensional analysis), I realised that <strong>equation 19 is wrong</strong> as it should be:</p>
<p>$$p_o = \frac{2}{5}( \frac{3 h^3}{(2s+1)8 \sqrt{2} \pi m^{3/2}})^{2/3} n^{5/3}$$</p>
<p>From that point on, it drags that mistake.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3294/system-of-two-ideal-fermi-gases-in-three-dimensions&show=3310#a3310Thu, 13 Dec 2018 12:13:41 +0000Answered: Radial distribution of particle separation in a liquid at small distances
http://physics.qandaexchange.com/?qa=3233/radial-distribution-particle-separation-liquid-distances&show=3308#a3308
<p><strong>1. Preference</strong></p>
<p>The spatial distribution of particles settles into a configuration which minimises potential energy. This is achieved when there is some regular structure - for example, <strong>hexagonal close packing</strong>. Gravitational potential energy is far too small on the scale of microscopic particles. The balance is between kinetic energy and electrostatic potential energy as defined by the Lennard-Jones potential.</p>
<p>Ideal gas particles are assumed to have no attraction at all, only "hard sphere" repulsion when they come into contact. The potential function for ideal gas interactions is therefore $U(r)=0$ for $r>2r_0$ and $U(r)=∞$ for $r≤2r_0$, where $r_0$ is the radius of particles so $2r_0$ (twice particle radius) is the minimum distance between the centres of particles.</p>
<p><strong>2. The Nature of the Oscillations</strong></p>
<p>These are not <strong>dynamic</strong> oscillations in the motion of particles (periodic variations with time), like a mass on a spring. So the question about over/under-damping is not relevant - although it is possible some analogy could be made with dynamic oscillations. </p>
<p>These are <strong>static</strong> oscillations in the distribution of inter-particle distances (oscillations in space), like the ripples of a wave function in quantum mechanics, or the ripples in sand on a beach. </p>
<p>$g(r)$ is a probability distribution function. The "oscillations" indicate that there are periodic values of particle separation which are more likely than average (peaks) or less likely than average (troughs). These "oscillations" are not necessarily sinusoidal (harmonic).</p>
<p>High density causes the "oscillations" by restricting the space which particles can move about in. When particles are forced close together they slide into relative positions where they can keep as much motion as possible - ie they form structures such as hexagonal close packing. At low densities particles have plenty of freedom to move about and can occupy all relative positions equally. All values of separation $r$ are equally likely - the probability distribution function is uniform, flat.</p>
<p>The oscillations are not necessarily sinusoidal. The graph you linked does <strong>look</strong> like a damped harmonic oscillation. But it is not an oscillation in time and space like a pendulum, it is a periodic variation in the probability of finding a particle at distance $r$ from an arbitrary particle. It is probably not easy to see it if you look at the particles. It is a statistical variation which shows up when you calculate average the distances, because averaging removes random variations.</p>
<p>The explanation you quoted suggests that the graph is statistical. It has been constructed by taking a snapshot of the jiggling particles, and measuring the distance of every particle from one aribitrary particle, which is chosen as the origin. (Perhaps this is repeated with every particle in turn being used as the origin. This is the same as measuring the distance between every pair of particles.) The distances of every particle from the origin are measured, and a histogram is plotted of the <strong>density</strong> of particles at a distance in the range $r−Δr$ to $r+Δr$ (vertical axis) vs r (horizontal axis).</p>
<p>For a gas you would expect a uniform density distribution. </p>
<p>For a 'cold' 3D cubic crystalline solid you find sharp peaks at separations r which are multiples of $1,\sqrt2, \sqrt3, 2, \sqrt5, \sqrt6, ...$ units. In fact for all $r^2=l^2+m^2+n^2$ with all possible combinations of integers $l, m, n$. In between are ranges in $r$ which have zero frequency - eg 1.25 units. These peaks and troughs extend over a large range in $r$ - ie there is <strong>long range order</strong>. </p>
<p>For a 'hot' crystal these peaks and the troughs in between them are broader and more rounded, like rolling hills and valleys, giving the impression of oscillations. (The unit of spacing also increases as the solid expands.) </p>
<p>For a liquid or amorphous solid there are peaks (and troughs) only at small values of r - ie there is <strong>short range order</strong>. At large values of r the distribution is more uniform like a gas.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3233/radial-distribution-particle-separation-liquid-distances&show=3308#a3308Tue, 11 Dec 2018 20:14:41 +0000A projectile is ejected with vertical speed v from the surface of a planet of mass M and radius R.
http://physics.qandaexchange.com/?qa=3274/projectile-ejected-vertical-speed-surface-planet-mass-radius
<p>Show that it comes to rest at a distance r from the centre of the planet where $$r=R/(1−Rv^2/2GM)$$ <br>
If v = c in the above example what will be the condition between M and R such that the planet acts like a Newtonian black hole.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3274/projectile-ejected-vertical-speed-surface-planet-mass-radiusSun, 09 Dec 2018 11:01:33 +0000Check whether the followings are null, spacelike or timelike
http://physics.qandaexchange.com/?qa=3275/check-whether-the-followings-are-null-spacelike-or-timelike
<p>(a) <br>
$x^0=\int r^2+z^2d\tau$<br>
$x^1=\int r\sin\theta d\tau$<br>
$x^2=\int r\cos\theta d\tau$ <br>
$x^3=\int z d \tau$<br>
(b)<br>
$x^0= \sqrt3 ct$<br>
$x^1=ct$<br>
$x^2=ct_0 \sin(t/t_0)$<br>
$x^3=ct_0 \cos(t/t_0)$<br>
where $t_0$ is constant.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3275/check-whether-the-followings-are-null-spacelike-or-timelikeSun, 09 Dec 2018 11:01:29 +0000Answered: Rotation of L shaped rods after impulse
http://physics.qandaexchange.com/?qa=3304/rotation-of-l-shaped-rods-after-impulse&show=3305#a3305
<p>The hardest part of these questions is working out what they mean! Here it is not clear (i) whether joint B is free to translate or is fixed to the ground, and (ii) whether the rods are joined rigidly at B (so that they remain at $90^{\circ}$ at all times), or pinned loosely so the angle between them can change freely.</p>
<p>If the rods are joined rigidly at B then there will be a <strong>torque</strong> exerted on AB by rod BC as well as an impulsive force, and <em>vice versa</em> for AB acting on BC. This is because for two 2D objects to be attached rigidly together, and maintain the same relative positions, they must be pinned at 2 distinct points. The diagram suggests there is only one pin at joint B. Also, Part 2 of the question does not mention any torque. So I think the rods must be pinned loosely at B by a single pin. If this pin is also fixed to the ground then the impulse on AB will have no effect on BC because the reaction at the pin will be transmitted to the Earth, which is so massive that it does not move. </p>
<p>So I think<strong> joint B must be pinned loosely with a single pin and is free to translate</strong>. This makes sense because rods AB and BC can then have different angular velocities (see options in Part 1).</p>
<p>Your attempt assumes the two rods are rigidly connected, which is a reasonable interpretation of the question, but I think for the above reasons it must be wrong.</p>
<p>You are correct about the position of the COM of the 2 rods. But you have made an error in applying the <a rel="nofollow" href="http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html">Parallel Axis Theorem</a>. This always starts with an axis through the COM and gives you the MI about an axis parallel to that through the COM. You have started at end B of the rod, not at the COM, so your calculation is incorrect. You should get $\frac{5}{12}m\ell^2$. </p>
<hr>
<p><strong>Solution</strong></p>
<p>Rod AB is struck at its COM. If it were not pinned to rod BC it would move upwards on the page, without rotating. This tells us that the reaction impulse $K$ at the pin is purely downwards on the page when acting on AB, and purely upwards when acting on BC.</p>
<p>Impulse $K$ acts through the COM of rod BC so this rod does not rotate : $\omega_2=0$. It moves upwards with velocity $v_2$ such that $K=mv_2$.</p>
<p>The impulses on AB can be resolved into a linear impulse $J-K$ acting upwards at the COM and a clockwise couple $K\frac{\ell}{2}$ about the COM. See <a rel="nofollow" href="https://physics.stackexchange.com/q/312523">Is net torque is not zero about all points on the rod for a linearly accelerating rod?</a> The COM of rod AB moves upwards with velocity $v_1$ such that $J-K=mv_1$. It also rotates with angular velocity $\omega_1=\frac{v_1-v_2}{ \ell/2}$ such that $K\frac{\ell}{2}=I_1\omega_1$ where $I_1=\frac{1}{12}m\ell^2$ is the moment of inertia of AB about its COM.</p>
<p>We now have enough equations so that we can find the following unknowns : $$\omega_1=\frac{6J}{5m\ell}, \omega_2=0$$ $$v_1=\frac{4J}{5m}, v_2=\frac{J}{5m}$$ $$K=\frac15 J$$</p>
<p>In Part 1 options A, B, C are correct, and in Part 2 option A is correct.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3304/rotation-of-l-shaped-rods-after-impulse&show=3305#a3305Sat, 08 Dec 2018 12:57:11 +0000Answered: The lean of a motorcyclist
http://physics.qandaexchange.com/?qa=3273/the-lean-of-a-motorcyclist&show=3276#a3276
<p>The back/front view should look like this :</p>
<p><img src="https://i.imgur.com/YGLurfc.png[/img]" alt=""></p>
<p>The vertical forces $N$ and $W$ cancel out. The resultant force is a radial friction force $f$ acting towards the centre of the turning circle. This is the centripetal force.</p>
<p>Another friction force acts tangentially, increasing the tangential velocity. This force is coming out of the page of the back/front view Free Body Diagram.</p>
Physics Problemshttp://physics.qandaexchange.com/?qa=3273/the-lean-of-a-motorcyclist&show=3276#a3276Sun, 25 Nov 2018 00:56:03 +0000Answered: 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 +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 +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 density at boundary surface
http://physics.qandaexchange.com/?qa=3216/induced-charge-density-at-boundary-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-density-at-boundary-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 +0000