Question

A stationary horizontal uniform circular disc of mass m and radius R can freely rotate about its axis. Now, its starts experiencing a torque $\tau =a\theta +b$ , where $\theta$ is the angular displacement of disc and a and b are positive constants. Find the angular velocity of the disc as a function of $\theta$.

Answer 1

The torque acting on the circular disc is given by :

$ı=I\alpha$ ( where I = moment of inertia of the disc $\alpha$ angular accelaration )

$\Rightarrow \alpha =\frac{ı}{I}=\frac{a\Theta +b}{\frac{m{R}^2}{2}}$( Moment of inertial of disc$=\frac{m{R}^2}{2})$

$\Rightarrow \alpha =\frac{2(a\Theta +b)}{m{R}^2}$

Now for small angular dispacement $d\Theta$ the angular velocity $(\omega ) is given by :

$\omega =0+ad\Theta$( intial anugular velocity = 0)

So for angular displacement $\Theta$ the angular velocity as a function of $\Theta$ is :

$\omega ={\int }_0^{\Theta }\frac{2(a\Theta +b)}{m{R}^2}d\Theta =\frac{2}{m{R}^2}(\frac{a{\Theta }^2}{2}+b\Theta )$