Explanation

(c) As there is no external torque, angular momentum will remain constant. When the tortoise moves from $A$ to $C$, figure, moment of inertia of the platform and tortoise decreases. Therefore, angular velocity of the system increases. When the tortoise moves from $C$ to $B$, moment of inertia increases. Therefore, angular velocity decreases
[image:7_image130.emf]
If, $M$=mass of platform
$R =$radius of platform
$m =$mass of tortoise moving along the chord $AB$
$a =$perpendicular distance of $O$ from $AB$
Initial angular momentum, $I_{1} = mR^{2} + \frac{MR^{2}}{2}$
At any time $t$, let the tortoise reach $D$ moving with velocity $v$
$\therefore\ AD = vt$
$AC = \sqrt{R^{2} - a^{2}}$
As $DC = AC - AD = (\sqrt{R^{2} - a^{2}} - vt)$
$\therefore\ \ OD = r = a^{2} + \left\lbrack \sqrt{R^{2} - a^{2}} - vt \right\rbrack^{2}$
Angular momentum at time $t$
$I_{2} = mr^{2} + \frac{MR^{2}}{2}$
As angular momentum is conserved
$\therefore\ \ I_{1}\omega_{0} = I_{2}\omega(t)$
This shows that variation of $\omega(t)$ with time is nonlinear. Choice (c) is correct