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Work done in slowly pulling the thread upwards

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A bead of mass m is attached to lower end of a light inextensible thread that passes through a small frictionless hole in a horizontal table top . The upper end of the thread is held motionless . The bead is pulled aside and projected horizontally to move on a circular path of radius R with a speed $v_0$ at a particular depth below the table top . Find the work done in slowly pulling the thread upwards to reduce the depth of circular path of the bead to half of its previous value .

Let the angle between the thread and horizontal be $\theta$ then $tan \theta = gR/v_0^2$
then work done by gravity would be $mgh = mg^2R^2/2V_0^2$.
But I am not able to find the final velocity to find the change in kinetic energy .

asked Mar 17, 2018 in Physics Problems by koolman (4,286 points)
Sorry the image is not coming properly . The image is rotated .
The angular momentum of the bead is conserved. So the speed of the bead will increase as the string is shortened. That will help you work out the final velocity. Apply the work-energy theorem : the work done equals the increase in potential and kinetic energies. This is the energy method.

The force method (which should give the same answer) is to write an expression for the tension F in the string, and integrate it as the string is pulled through the hole (W=integral of F x vertical distance moved by the hand).
$mv_0 R = mv'R'$
In this $v'$ and R' both are variable and I am not getting the second equation to solve it .
You already have another equation in your question. You can eliminate $\theta$ because $\tan\theta=H/R$ where $H$ is the depth of the bead below the hole.
How can we say that after Pulling it by H/2 will the angle would be same ?
The angle will not be the same.

1 Answer

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Best answer

The thread is pulled up slowly. This ensures that the orbit of the bead remains circular and horizontal. Small impulses from the thread perturb the orbit, making it elliptical and inclined to the horizontal. However, if these impulses are delivered equally at all points around the orbit, the orbit remains circular and horizontal.

We are given the initial radius $R_0$ and speed $v_0$ of the bead's orbit. The tension in the thread is always directed towards the axis about which the bead rotates, so the angular momentum of the bead is conserved. The mass of the bead is constant, so the quantity $Rv=R_0v_0$ is constant.

The other relation which you are looking for is the one you have already found from balancing forces :
Using both equations, and the geometry of the conical pendulum, you can find the work done.

Energy Method

The work done on the bead equals its increase in PE+KE.

The depth of the orbit below the hole decreases from $H_0$ to $H_1=\frac12 H_0$ so the increase in PE is $\frac12 mgH_0$.

From your calculations
The initial depth is

The increase in KE of the bead is
$\frac12 m(v_1^2-v_0^2)=\frac12 mv_0^2[(v_1/v_0)^2-1]=\frac12mv_0^2[(R_0/R_1)^2-1]$
The ratio of the initial to final depth of orbit is
so $(R_0/R_1)^2=\sqrt2$

The work done to pull the thread upwards is therefore
$W=\frac12 m[(gR_0/v_0)^2+(\sqrt2-1)v_0^2]$

Force Method

The work done in pulling the bead up from depth $H_0$ to $H_1=\frac12 H_0$ below the hole is
$W=\int_{H_0}^{H_1} T(- dL)=\int_{H_1}^{H_0} TdL$
where $L$ is the length of the thread below the hole and $T$ is the variable tension in the thread.

The minus sign is used because $L$ and $H$ increase downwards whereas $W$ increases upwards. Assuming that the hole is frictionless, the tension above the hole is the same as it is below the hole. Like a frictionless pulley, the hole merely changes the direction of the tension.

Note that the work done is not $\int TdH$ because the distance moved by the pulling force above the hole is not equal to the distance through which the bead rises below the hole. Neither is the work done $\int T\sin\theta dH = \int mg dH$. This accounts for the increase in PE of the bead but not the increase in KE.

$T$ can be found from balancing the forces on the bead :
so $W=\int TdL=mg \int (L/H)dL$

From geometry :
so $(L/H)dL=dH+(R/H)dR$

From the Energy Method (above)
$R/H=R_0^2 v_0^2/gR^3$
$\int (R/H)dR = -\frac12 R_0^2v_0^2/gR^2$
$W=mg[(H_0-H_1)-\frac12 R_0^2v_0^2(1/gR_0^2-1/gR_1^2)]$
$=\frac12 m[gH_0+(R_0^2/R_1^2-1)v_0^2]$
$=\frac12 m[(gR_0/v_0)^2+(\sqrt2-1)v_0^2]$

Note that $H \propto R^4$ therefore $\tan\theta = H/R \propto H^{\frac34}$. The exponent is +ve so as $H$ decreases then $\theta$ also decreases.

answered Apr 1, 2018 by sammy gerbil (28,466 points)
edited Oct 23, 2018 by sammy gerbil