Question

Question #d6245

Answer 1

Suppose,the torque created by frictional force is `tau`,so `tau = I alpha` (where, `I` is moment of inertia of the disk and `alpha` is angular acceleration)

Now,time period (`T`) of the disk initially was `5/28.9 s`

So,initial angular velocity = `omega _o = (2pi)/T=(2pi)/(5/28.9)`

So,we can apply , `omega = omega _o - alpha t`

when,it comes to rest `omega =0` and ,that time taken is `t=12s` (given)

So, `alpha =(omega_o)/t=3.03 (rad)/s^2`

so, `tau = 4*3.03 N.m=12.12 N.m` (given `I=4 Kg.m^2`)