Question

A circular disc of mass $300gm$ and radius $20cm$ can rotate freely about a vertical axis passing through its centre of $O$. A small insect of mass $100gm$ is initially at a point $A$ on the disc (which is initially stationary) the insect starts walking from rest along the rim of the 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 $A$. By what angle has the disc rotated till now; as seen by a stationary earth observer? Also find the time $T$.

Answer 1

Since there is no external no torque on the system.

$\therefore$ Torque produced by the insect $=$ Torque experience by the insect on the disc

$⟹m{r}^2{\omega }^2=M{r}^2{\omega }^{\prime 2}$

where,

$m$ is the mass of insect,

$M$ is mass of disc,

$r$ is radius,

$\omega$ is angular velocity of insect,

${\omega }^{\prime }$ is angular velocity of disc.

$\frac{m}{M}=\frac{100}{300}=\frac{1}{3}$

$\therefore {\omega }^{\prime 2}=\frac{1}{3}{\omega }^2$

Also the angular acceleration of the disc:

$\alpha =2\pi$ $rad/s^2=r{\omega }^{\prime 2}$

$⟹{\omega }^{\prime 2}=\frac{2\pi }{20\times {10}^{-2}}=31.41$

$⟹{\omega }^{\prime 2}=5.6$ $rad/s$

From kinematics:

${\omega }^{\prime }-{\omega }_0=\alpha T$

$⟹T=\frac{{\omega }^{\prime }}{\alpha }[{\omega }_o=0]$

$⟹T=\sqrt{\frac{2\pi }{20\times {10}^{-2}}}\times \frac{1}{2\pi }$

$⟹T=0.89$ $s$

And:

${\omega }^{\prime 2}-{\omega }_o^2=2\alpha \theta$

$⟹\theta =\frac{{\omega }^{\prime 2}}{2\alpha }=\frac{2\pi }{20\times {10}^{-2}}\times \frac{1}{2\times 2\pi }=2.5$ $rad$