Question

The moment of inertia of a body about a given axis is $1.2kg/{{m}^{2}}$. Initially, the body is at rest. In order to produce a rotational kinetic energy of 1500 J, an angular acceleration of $25rad/{{s}^{2}} $ must be applied about that axis for the duration of
A. 4 s
B. 2 s
C. 8 s
D. 10 s

Answer 1

Hint: There are two methods of solving this question. Both methods will be covered in the solution part. Before solving the question, try to remember the formula of rotational kinetic energy. Also, we will be using another formula which you can remember through an analogy of the first equation of the motion.

Formula used:
For solving the given question, we will be using the formula for the rotational kinetic energy, i.e.,
Rotational kinetic energy $=KE=\dfrac{1}{2}I{{\omega }^{2}}$
And, \[\omega ={{\omega }_{0}}+\alpha t\]

Complete step by step answer:
Before going through the solution, let’s look at the given parameters
Moment of inertia $=I=1.2kg/{{m}^{2}}$
Angular acceleration $=\alpha =25rad/{{s}^{2}}$
In the question, the kinetic energy is already given to be 1500J
So,
$\Rightarrow 1500=\dfrac{1}{2}\times 1.2\times {{\omega }^{2}}$
$\Rightarrow {{\omega }^{2}}=\dfrac{1500\times 2}{1.2}$
$\Rightarrow {{\omega }^{2}}=2500$
$\Rightarrow \omega =50rad/s$
Now, as we know,
\[\Rightarrow \omega ={{\omega }_{0}}+\alpha t\]
\[\Rightarrow 50=0+25t\]
$\Rightarrow t=2s$
So, angular acceleration of $25rad/{{s}^{2}}$ must be applied about that axis for the duration of 2s

Note:
Another method of solving this question is, by making a simple replacement
As we know,
$\omega =\alpha t$
By directly using the above value in the rotational kinetic energy,
$KE=\dfrac{1}{2}I{{(\alpha t)}^{2}}$
$\Rightarrow 1500=\dfrac{1}{2}\times 1.2\times 25\times 25\times {{t}^{2}}$
$\Rightarrow \dfrac{1500\times 2}{625\times 1.2}={{t}^{2}}$
$\Rightarrow {{t}^{2}}=4$
$\Rightarrow t=2s$