Question

A torque of 100Nm is applied to a body, capable of rotating about a given axis. If the body starts from rest and acquires kinetic energy of 10,000J in 10 second. Find (a) its moment of inertia about the given axis (b) angular momentum at the end of 10 second.

Answer 1

Hint: Here the kinetic energy and torque of the body is given. Hence, in the case of a rotating body the kinetic energy is given by half the product of moment of inertia and the square of angular momentum. Equating the both equations we get the moment of inertia and the square of angular momentum. The torque changes the change of angular momentum with time. Thus by substituting the given values we will get the angular momentum at the end of 10 second. Also angular momentum is the product of moment of inertia and angular momentum.

Complete step by step solution:
(b) Given that kinetic energy,
$KE=10000J$ ………(1)
t=10sec
$\tau =100Nm$
In the case of a rotating body the kinetic energy is given by,
$KE=\dfrac{1}{2}I{{\omega }^{2}}$ ………….(2)
Equating equation (1) and (2) we get,
$10000=\dfrac{1}{2}I{{\omega }^{2}}$
$\Rightarrow 20000=I{{\omega }^{2}}$
Also given that,
$\tau =100Nm$
We know that the torque which the change of angular momentum with time.Hence,
$\dfrac{dL}{dt}=100Nm$
where, dL is the change in angular momentum.
Also,
$dL={{L}_{f}}-{{L}_{i}}$
Thus, $\dfrac{{{L}_{f}}-{{L}_{i}}}{dt}=100Nm$
Since dt=10sec
Substituting this we get,
$\dfrac{{{L}_{f}}-{{L}_{i}}}{10}=100$
$\Rightarrow {{L}_{f}}-{{L}_{i}}=1000$
Here, the body is initially at rest.
That is, ${{L}_{i}}=0$
Hence, ${{L}_{f}}=1000Nm/s$
Thus, the angular momentum at the end of 10 second is 1000Nm/s.
(a) We know that,
$KE=\dfrac{1}{2}I{{\omega }^{2}}$
Also, $L=I\omega $
Substituting for $I\omega $ we get,
$KE=\dfrac{1}{2}L\omega $
$\Rightarrow \omega =\dfrac{2\times KE}{L}$
$\Rightarrow \omega =\dfrac{2\times 10000}{1000}$
$\begin{align}
  & \Rightarrow \omega =\dfrac{20000}{1000} \\
 & \therefore \omega =20 \\
\end{align}$
 $L=I\omega $
$I=\dfrac{L}{\omega }$
$\begin{align}
  & I=\dfrac{1000}{20} \\
 & \therefore I=50kg/{{m}^{2}} \\
\end{align}$
Thus, the moment of inertia about the given axis is $50kg/{{m}^{2}}$.

Note:
Thus when the value of angular velocity is doubled, then its kinetic energy will increase by four times. The rotational kinetic energy can be described as the total kinetic energy of a rotating object. According to the law of conservation of angular momentum when the torque acting on the body is zero, then the kinetic energy is conserved. The angular momentum depends upon mainly two factors. That is, the rotational velocity and moment of inertia.