Question

The moment of inertia of a body about a given axis is $1.2$ kg-$m^2$. To produce a rotational kinetic energy of $1500$J an angular acceleration of $25$ rad/$s^2$ must be applied for.

• A. $8.5$s
• B. $5$s
• C. $2$s
• D. $1$s

Answer 1

Answer: (C)

$2$s

Kinetic energy of rotation is half the product of the moment of inertia(l) of the body and the square of the angular velocity $\left(\omega \right)$ of the body.
Kinetic energy of rotation$=\frac{1}{2}\times$ moment of inertia $\times$ angular velocity
i.e., $K=\frac{1}{2}t{\omega }^2$
${\omega }^2=\frac{2K}{l}$
Given $l=1.2kg{m}^2,K=1500$J
${\omega }^2=\frac{2\times 1500}{1.2}$
$\Rightarrow \omega =50rad/s$
From the equation of angular motion, we have
$\omega ={\omega }_0+\alpha t$
where ${\omega }_0$ is initial angular velocity, $\alpha$ is angular acceleration and t is time
given ${\omega }_0=0$
$\omega =50rad/s$,
$\alpha =25rad/s^2$
$t=\frac{\omega }{\alpha }=\frac{50}{25}=2s$.