Physics - Circular Motion Question with Solution | TestHub

If angular acceleration is constant, we have

$\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$

The given$\theta ={15}^o$

${\omega }_0=0$

For the first conditions (time = t sec)

${15}^o=0+\frac{1}{2}\alpha t^2$

$\Rightarrow$${15}^o=\frac{1}{2}\alpha t^2$......(i)

For the second conditions (time = t+2t sec)

${\theta }_1=\frac{1}{2}\alpha {\left(3t\right)}^2=\frac{1}{2}\left(\alpha \right)9t^2$.......(ii)

So for 2t sec,$\mathrm{\Delta }\theta ={\theta }_1-\frac{1}{2}\alpha t^2$

$\mathrm{\Delta }\theta =9\times \frac{1}{2}\alpha t^2-\frac{1}{2}\alpha t^2$

$=8\frac{1}{2}\alpha t^2$

$=8\times {15}^o={120}^o$(from equation(i))