Question

A circular disc of mass 300 gm and radius 20 cm can rotate freely about a vertical axis passing through is center O. A small insect of mass 100 gm is initially at a point A on the rim. The insect initially stationary starts walking from rest along the rim of disc with such a time varying relative velocity that the disc rotates in the opposite direction with a constant angular acceleration = $2\pi rad/s^2$. After some time T, the insect is back at the point A

. What is the time taken by insect to reach the original position?

• A. $\frac{2}{\sqrt{5}}$
• B. $\frac{2}{\sqrt{3}}$
• C. $\frac{\sqrt{5}}{2}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is A

$\frac{2}{\sqrt{5}}$

Let the force applied by insect on the disc be F.

It is clear from the contact,

$R\times F=I_d{\alpha }_d$

$R\times F=\frac{M{R}^2}{2}\times 2\pi$

$F=\pi M_{d}R$ ..........(1)

The same amount of force will be applied to the insect by the disc,

Considering its torque about the centre of disc,

$\pi MdR\times R=I_i{\alpha }_{i}i=insect$

$M_i{R}^2{\alpha }_i=M_d\pi R^2$

${\alpha }_i=\frac{M_d}{M_i}\times \pi \{\frac{M_d}{M_i}=3\}$

${\alpha }_i=3\pi$

now with respect to disc's frame of reference, ${\alpha }_{\frac{i}{d}}=5\pi$ {Both α are opposite in direction}

now relative angular displacement = $2\pi$

$\Rightarrow 2\pi =\frac{1}{2}{\alpha }_{\frac{i}{d}}{t}^2\Rightarrow t=\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}s$