Question

The moment of inertia of the body about a given axis is 1.2kgm2.Initially,the body is at rest. In order to produce rotational kinetic energy of 1500 j, an angular acceleration of 25rad/s2 must be applied about that axis for a duration of

Answer 1

Rotational Mechanics

We are given the following data:

$\textsf{Moment of inertia} = I = 1.2\ kg\, m^2 \\\\ \textsf{Rotational Kinetic Energy} = K = 1500\, J \\\\ \textsf{Angular acceleration} = \alpha = 25\ rad/s^2 \\\\ \textsf{Initial Angular Velocity} = \omega_{\circ} = 0\, rad/s$

The initial angular velocity is zero, because we are given that the body is initially at rest.

The body is to be rotated about its axis so that it attains a rotational kinetic energy of 1500 J. For this, it must attain a specific angular velocity. Let it be $\omega$.

We have:

$\displaystyle K = \frac{1}{2}I\omega^2 \\\\\\ \implies 1500 = \frac{1}{2} \times 1.2 \times \omega^2 \\\\\\ \implies \omega^2 = \frac{1500\times 2}{1.2} \\\\\\ \implies \omega^2 = 2500 \\\\\\ \implies \omega = \sqrt{2500}\ rad/s \\\\\\ \implies \Large \omega = 50\ rad/s$

The body is initially at rest. It must be brought to an angular velocity of 50 radians per second by a constant angular acceleration of 25 $rad/s^2$.

Suppose the time required for this is $t$. By the Rotational Equations of Motion, we have:

$\omega = \omega_{\circ} + \alpha t \\\\\\ \implies 50 = 0 + 25 \times t \\\\\\ \implies t = \dfrac{50}{25}\ s\\\\\\ \implies \huge \boxed{\sf t = 2\ s}$

Thus, The angular acceleration must be applied for 2 seconds so that the body attains the required kinetic energy.

Answer 2

K=1/2 Iw*2

1500= 1/2×1.2× w*2

W*2= 2500

W=50

W=w. +at

50=0+25×t

t=2sec.