Question

A uniform round board of mass M and radius R is placed on a fixed smooth horizontal plane and is free to rotate about a fixed axis which passes through its centre. A man of mass m is standing on the point marked A on the circumference of the board. At first the board and the man are at rest. Then the man starts moving along the rim of the board at constant speed ${{v}_{0}}$ relative to the board. Find the angle of the board's rotation when the man passes his starting point on the disc for the first time.

Answer 1

Use the law of conservation angular momentum about the point of rotation. Find the speed of the board at the circumference and calculate its angular velocity. Later find the time in which the man reaches the starting point. Then you can find the angle of rotation of the board.

Formula used:
$\omega =\dfrac{u}{R}$,
where $\omega$ is angular velocity, u is the speed at a distance R from the point of rotation.
$L=muR$,
where L is angular momentum of mass m moving in circular motion of radius R and with speed u.
$L=I\omega$,
where L is angular momentum of a body rotation about is fixed point with momentum of inertia I.
$t=\dfrac{d}{v}$,
t is time, d is distance covered and v is speed.
$\theta =\omega t$,
where $\theta$ is angle of rotation.

Complete step by step answer:
It is given that the round board is placed on a fixed smooth horizontal plane with a fixed centre such that the board is free to rotate about its centre. It is said that a man is standing at a point on the circumference of the board. Initially, the man and the board are at rest and then the man starts moving. When the man starts moving, the frictional force between the board and the man will come into play. The friction will push the man forward and the board backward. Both forces will be of equal magnitude but opposite direction. This means the net force on the system is zero.This means that the net torque on the system is zero.Hence, the angular momentum of the system is conserved at point of rotation.

We know that both the bodies were at rest initially, which means that the initial angular momentum of the system is zero. Therefore, the angular momentum of the system when they start moving is also zero. We know that the man and the board are rotating in opposite directions. Therefore, the magnitudes of the two must be equal.Let the velocity of the man and the board at the circumference be ${{u}_{1}}$ and ${{u}_{2}}$. (both of them are in opposite directions).Let the angular velocity of the board be $\omega$ and $\omega =\dfrac{{{u}_{2}}}{R}$.
The angular momentum of the man is ${{L}_{1}}=m{{u}_{1}}R$ …. (i) and that of the board is ${{L}_{2}}=I\omega =I\dfrac{{{u}_{2}}}{R}$ ….. (ii)
The board can be considered as a circular disc and the momentum of inertia of a disc about an axis perpendicular to its plane and passing to its centre is $I=\dfrac{M{{R}^{2}}}{2}$.
Substitute the value of I in (ii).
${{L}_{2}}=\dfrac{M{{R}^{2}}}{2}\dfrac{{{u}_{2}}}{R}\\\ \Rightarrow {{L}_{2}} =\dfrac{MR{{u}_{2}}}{2}$
Now equate (i) and (ii).
$m{{u}_{1}}R=\dfrac{MR{{u}_{2}}}{2}$
$\Rightarrow m{{u}_{1}}=\dfrac{M{{u}_{2}}}{2}$ ….. (iii).
But it is given that the relative speed of the man with respect to the board is ${{v}_{0}}$.
${{v}_{0}}={{u}_{1}}-(-{{u}_{2}})={{u}_{1}}+{{u}_{2}}$.
$\Rightarrow {{u}_{1}}={{v}_{0}}-{{u}_{2}}$
Substitute this value in (iii).
$\Rightarrow m\left( {{v}_{0}}-{{u}_{2}} \right)=\dfrac{M{{u}_{2}}}{2}$
$\Rightarrow m{{v}_{0}}-m{{u}_{2}}=\dfrac{M{{u}_{2}}}{2}$
$\Rightarrow {{u}_{2}}=\dfrac{m{{v}_{0}}}{\dfrac{M}{2}+m}=\dfrac{2m{{v}_{0}}}{M+2m}$.
And we know that $\omega =\dfrac{{{u}_{2}}}{R}$.
$\Rightarrow \omega =\dfrac{2m{{v}_{0}}}{M+2m}\times \dfrac{1}{R}$.
Now, with respect to the board the man will rotate by an angle of $2\pi$ with a speed of ${{v}_{0}}$. This means the man will cover is distance equal to the circumference, i.e.$2\pi R$
Therefore, time taken to reach the starting point is $t=\dfrac{2\pi R}{{{v}_{0}}}$.
Then the angle by which the board rotates is equal to,
$\theta =\omega t\\\ \Rightarrow\theta =\dfrac{2m{{v}_{0}}}{M+2m}\times \dfrac{1}{R}\times \dfrac{2\pi R}{{{v}_{0}}}\\\ \therefore\theta =\dfrac{4\pi m}{M+2m}$.

Hence, the angle of the board's rotation when the man passes his starting point on the disc fist time is $\dfrac{4\pi m}{M+2m}$.

Note: Some students may make a mistake by applying conservation of linear momentum. It is not so that linear momentum is not conserved. It is conserved but we cannot write any equation for it as the bodies are in rotational motion. Hence, we equate the angular momentum.