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 & the man are at rest. The man starts moving along the rim of the board at constant speed vo relative to the board. Find the angle of board’s rotation when the man passes his starting point on the disc first time.

Key Concepts Needed

1. Moment of Inertia ($I$)
   For a uniform disc of mass $M$ and radius $R$:

   $$I_{\text{disc}} = \frac{1}{2} M R^2$$

   For a point mass $m$ on the rim:

   $$I_{\text{man}} = m R^2$$

2. Angular Momentum ($L$)
   For any object:

   $$L = I \omega$$

   where $\omega$ is the angular velocity about the chosen axis.

3. Conservation of Angular Momentum
   If no external torque acts about the axis, the total angular momentum remains constant.

4. Relative vs Absolute Angular Velocity
   The man’s speed $v_0$ is relative to the board.

   • Relative angular velocity: $\dfrac{v_0}{R}$ (with respect to the board)
   • Absolute angular velocity of the man (seen from the ground): $$\Omega_{\text{man}} = \omega_{\text{board}} + \frac{v_0}{R}$$
   • $\omega_{\text{board}}$ will be opposite in sign to keep total $L$ zero.

5. Time for One Relative Revolution
   The man must sweep an angle $2\pi$ relative to the board to reach his starting point on the board.
   Thus time required:

   $$t = \frac{2\pi}{\frac{v_0}{R}} = \frac{2\pi R}{v_0}$$

Logical Order to Tackle the Problem

1. Write angular momentum balance right after he starts.
2. Solve for the board's angular velocity $\omega$.
3. Compute the time taken for the man to complete one lap relative to the board.
4. Multiply $\omega$ and time $t$ to get the board’s angular displacement $\theta$.
5. Mind the sign: board turns opposite to the man; final answer is asked as an angle (magnitude usually suffices).

Imagine a Big Lazy Susan and a Boy

Think of a large round dining-table Lazy Susan that can spin freely but is not spinning at the start. A boy is standing on its edge.

• When the boy starts walking around the edge, the surface under his feet (the Lazy Susan) feels a push in the opposite direction.
• Because nothing outside is pushing on the whole system, the total twistiness (scientists call it angular momentum) has to stay zero — whatever twist the boy creates, the board must create an opposite twist so they cancel.
• The boy keeps walking until he returns to the same spot on the board. How much has the board itself turned while he was walking?

We solve it by saying, "Total twist at the start is zero, so it must stay zero" and use the simple rule:

mass × how-far-out × spin-speed

We add the boy's share and the board's share and set the sum to zero, then see how much time it takes the boy to go all the way around. Multiply the board’s spin-speed by that time, and we get the angle the board turned.

Step 1: Set Up Angular Momentum Conservation

Initially, total angular momentum about the vertical axis through the centre is zero.

After the man starts running, let the board’s angular velocity (w.r.t ground) be $\omega$ (take counter-clockwise positive).
The man’s absolute angular velocity becomes

Total angular momentum must still be zero:

Substitute the moments of inertia:

Step 2: Solve for $\omega$

Divide through by $R^2$:

Combine terms:

Hence

Negative sign: board spins opposite to the man.

Step 3: Time for One Lap Relative to the Board

Relative angular speed between man and board:

Time to cover $2\pi$ relative angle:

Step 4: Angular Displacement of the Board

Simplify (cancel $v_0$ and $R$):

Magnitude of rotation:

Direction: opposite to the man’s motion.