Question

73. A hoop of radius R rotates at constant an- gular velocity 12. A small bead of mass m is attached to the hoop, with a frictional force on the bead proportional to the dif- ference in velocity between the bead and edge of the hoop, F = k(RS2 - Rw), where w is the angular velocity of the bead. If the bead begins at angular velocity wo, which of the following describes its subsequent motion? (A) w(t) = woe-kRt/m (B) w(t) = N - woe-kRt/m (C) w(t) = 12 - woe -mRt/ (D) wat) = (1 - e-kRt/m) (E) w(t) = 12 - (12 - wo)e-kRt/m

Answer 1

frictional force on the bead

$f_c = kR(\Omega -\omega)$

So,

$mR\dot{\omega}= kR(\Omega -\omega)$

$\dot{\omega} + \frac{k}{m}\omega = \frac{k}{m}\Omega$

which can be solved with I.F

$I.F = e^{\frac{k}{m}t}$

so, the solution is

$\omega e^{\frac{k}{m}t} = \Omega(e^{\frac{k}{m}}-1)+C$

at t = 0 ,  

$C = \omega_0$

$\omega = \Omega -(\Omega -\omega_0)e^{-\frac{k}{m}t}$

option E