Physics - Rotation Question with Solution | TestHub

The angular velocity of the rigid body is given by $\omega =\alpha -\beta t$.

The body comes to rest when its angular velocity becomes zero. Setting $\omega =0$:

$\alpha -\beta t=0\Rightarrow t=\frac{\alpha }{\beta }$

The angular displacement $\theta$ is the integral of angular velocity with respect to time:

$\theta ={\int }_0^t\omega dt={\int }_0^{\alpha /\beta }(\alpha -\beta t)dt$

$\theta ={\left[\alpha t-\frac{\beta t^2}{2}\right]}_0^{\alpha /\beta }$

Substituting the upper limit $t=\frac{\alpha }{\beta }$:

$\theta =\alpha \left(\frac{\alpha }{\beta }\right)-\frac{\beta }{2}{\left(\frac{\alpha }{\beta }\right)}^2$

$\theta =\frac{{\alpha }^2}{\beta }-\frac{{\alpha }^2}{2\beta }=\frac{{\alpha }^2}{2\beta }$

Alternatively, using the kinematic equation ${\omega }^2={\omega }_0^2+2\gamma \theta$:

Initial angular velocity ${\omega }_0=\alpha$ and angular acceleration $\gamma =\frac{d\omega }{dt}=-\beta$.

$0={\alpha }^2-2\beta \theta \Rightarrow \theta =\frac{{\alpha }^2}{2\beta }$

Answer: $\frac{{\alpha }^2}{2\beta }$