A circular disc of mass `300 gm` and radius `20 cm` can roate freely about a vertical axis passing through its centre of `O.A` small inset of mass `100 gm` 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

`t = 2// sqrt(5) sec, theta = 4 pi//5 rad`.

Explanation

From conservation of angular momentum

`I_("disc") omega_("disc") = I_("insect") omega_("insect") rArr I_("disc") alpha_("disc") = I_("insect") alpha_("insect")`
`(1)/(2) xx(0.3) xx(0.2)^(2) xx 2 pi = (0.1) xx (2)^(2) xx alpha_("insect")`
`alpha_("insect") = 3 pi rad//sec^(2)`
Angular covered by disc `= theta`,
so angle covered by insect `= 2 pi - theta`
For disc `theta = (1)/(2) alpha_(disc) T^(2)`
`theta = (1)/(2) xx 2 pi T^(2) rArr T^(2) = theta//pi`
For insect `2 pi - theta = (1)/(2) alpha_(insect) T^(2)`
From equation (1) and (2)
`2pi - theta = (1)/(2) xx 3 pi xx (theta)/(pi) rArr 2 pi -theta = (3 theta)/(2)`
`theta = (4 pi)/(5) rad`.