A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its centre. A tortoise is sitting at the edge of the platform. Now, the platform is given an angular velocity `omega_(0)`. When the tortoise moves along a chord of the platform with a constant velocity (with respect to the platform), the angular velocity of the platform `omega(t)` will vary with time `t` as

Options

A.
B.
C.
D.

Answer

B

Explanation

In absence of external torque, the angular momentum of the system remains conserved.
Let the tortoise move along the chord PQ. When the tortoise moves from P to M its distance from axis point O decreases and so the moment of inertia decrease as `I=mr^(2)`, where r = distance of tortoise from O. When the tortoise moves from M to Q the distance r increases and so I also increases.
`I_(P) = mR^(2) + (MR^(2))/2`..........(i)
`I_(N) = mr^(2) + (MR^(2))/2`.......(ii)

By geometry, `r^(2) =d^(2) + [sqrt(R^(2) -d^(2))-vt]^(2)`........(iii)
Angular momentum is conserved
`I_(p)omega_(0) = I_(N)omega(t)` or `omega(t) =(I_(p)omega_(0))/I_(N)`.........(iv)
`omega(t)` depends on `I_(N'), I_(N)` depends on r and r depends on time t.
The function of t is non-linear.
Hence, `omega(t)` is a non-linear function of t.
`omega` increases when tortoise from P to M and `omega` decreases when tortoise travels from M to Q.
First increases and then decrease of `omega` is revealed in graph (b) and (d). But `omega(t)` is a non-linear function of t, hence graph (b) represents the variation of `omega`(t) w.r.t. time.