Question

The moment of inertia of a body about a given axis is \[1.2\,{\rm{ }}kg{m^2}\]. Initially, the body is at rest. For what duration on the angular acceleration of \[25\,{\rm{ }}radian/{s^2}\] must be applied about that axis in order to produce rotational kinetic energy of 1500 joule?

Answer 1

Hint:When the angular acceleration is constant then we can use the 1st equation of motion to find the angular velocity after any interval of time.

Formula used:
\[{\omega _f} = {\omega _i} + \alpha t\]
Here, \[{\omega _f}\]is the final angular velocity after time t when the angular acceleration is \[\alpha \]and \[{\omega _i}\]is the initial angular velocity.
\[K = \dfrac{{I{\omega ^2}}}{2}\]
Here, K is the rotational kinetic energy of the body with a moment of inertia I and angular velocity \[\omega \].

Complete step by step solution:
To find the angular velocity of the body when the rotational kinetic energy is 1500 joule, we use the formula of the rotational kinetic energy.
\[K = \dfrac{{I\omega _f^2}}{2} \\
\Rightarrow {\omega _f} = \sqrt {\dfrac{{2K}}{I}} \]
Putting the value, we get
\[{\omega _f} = \sqrt {\dfrac{{2 \times 1500}}{{1.2}}} = 50\,radian/\sec \]

Let after time t we get the obtained angular velocity so that rotational kinetic energy of the body is 1500 joules. To find the time t, we use the 1st equation of rotational motion. As the body is at rest initially, so the initial angular velocity of the body is 0.0 radian/sec. The final angular velocity is 50.0 radian/sec. The constant angular acceleration is 25 radian per Second Square.

Putting the numerical values and solving, we get
\[{\omega _f} = {\omega _i} + \alpha t \\
\Rightarrow t = \dfrac{{{\omega _f} - {\omega _i}}}{\alpha }\]
\[\therefore t = \left( {\dfrac{{50 - 0}}{{25}}} \right)\sec = 2.0\,\sec \]
Hence, after 2.0 sec the rotational kinetic energy of the body is 1500 joules.

Therefore, the correct answer is 2.0 sec.

Note: We can use the work energy theorem to solve this question. The applied torque will produce the angular acceleration in the body. In rotating the body the applied torque will do work. The change in rotational kinetic energy will be equal to the work done by the applied torque.