Question

Moment of inertia of a body about a given axis is $1.5kg{{m}^{2}}$. Initially the body is at rest. In order to produce a rotational kinetic energy $1200J$, the angular acceleration of $20rad/{{s}^{2}}$ must be applied about the axis for a duration of?
$A.2s$
$B.5s$
$C.2.5s$
$D.3s$

Answer 1

Hint: We have to apply the relation of kinetic energy in terms of moment of inertia and angular velocity. The rate of change of angular position of a rotating body is called angular velocity. Moment of inertia describes how easily a body can be rotated about a given axis.
Formula Used:
We are going to use the following formula of kinetic energy in terms of moment of inertia and angular velocity and second equation for angular motion respectively:-
$KE=\dfrac{I{{\omega }^{2}}}{2}$and $\omega ={{\omega }_{o}}+\alpha t$

Complete answer:
In finding the time duration in the given process, we first have to find out the angular velocity for kinetic energy $1200J$ with help of the following relation:-
$KE=\dfrac{I{{\omega }^{2}}}{2}$ Where, $I$ is moment of inertia, $\omega $ is angular velocity and $KE$ is kinetic energy.
Putting values we get,
$1200=\dfrac{1.5{{\omega }^{2}}}{2}$
${{\omega }^{2}}=\dfrac{1200\times 2}{1.5}$
${{\omega }^{2}}=1600$
$\omega =40$.
We get the value of angular velocity, $\omega $for the kinetic energy $1200J$.
Now using equation of motion in terms of angular velocity which is given below:-
$\omega ={{\omega }_{o}}+\alpha t$………….. $(i)$Where, ${{\omega }_{o}}$is the initial angular velocity, $\alpha $ is angular acceleration and $t$ is time duration.
Putting values in $(i)$, we get
$40=0+20t$
${{\omega }_{o}}=0$As the body was in rest initially and $\alpha =20rad/{{s}^{2}}$as given in the problem, so
$20t=40$
$t=2s$

Hence, option $(A)$ is correct among the given options.

Note:
We have to apply the concept of angular motion in this problem. We should know the difference between the basics of linear motion and angular motion. Don’t be confused between the angular velocity or angular acceleration and linear velocity or linear acceleration. In these angular motion problems, expression for $KE=\dfrac{m{{v}^{2}}}{2}$is not applicable but $KE=\dfrac{I{{\omega }^{2}}}{2}$ is used for accurate results.