Question

Moment of inertia of a body about a given axis is $1.5\, kg\, m^2$. Initially the body is at rest. In order to produce a rotational kinetic energy of $1200\, J$, the angular accleration of $20\, rad/s^2$ must be applied about the axis for a duration of :

• A. 2s
• B. 5s
• C. 2.5 s
• D. 3s

Answer 1

Answer: (A)

Given moment of inertia 'I' = 1.5 kgm$^2$
Angular Acc. "a" = $20\, Rad/s^2$
$KE =\frac{1}{2} I\omega^{2} $
$ 1200 = \frac{1}{2} 1.5 \times\omega^{2} $
$ \omega^{2} = \frac{1200\times2}{1.5} = 1600 $
$ \omega = 40 \text{rad}/s^{2} $
$ \omega =\omega_{0} +\alpha t $
$ 40=0+20t $
$ t=2\sec$