Question

A body having moment of inertia about its axis of rotation equal to 3 kg - m² is rotating with angular velocity equal to 3 rad/s. Kinetic energy of this rotating body is the same as that of body of mass 27 kg moving with a speed of

• A. 1.0 m/s
• B. 0.5 m/s
• C. 1.5 m/s
• D. 2.0 m/s

Answer 1

Answer: (A)

1.0 m/s

Answer 2

The rotational kinetic energy of a body is given by the formula $KE_{rot} = \frac{1}{2}I\omega^2$, where $I$ is the moment of inertia and $\omega$ is the angular velocity.

The translational kinetic energy of a body is given by the formula $KE_{trans} = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the linear velocity.

According to the problem, the kinetic energy of the rotating body is equal to the kinetic energy of the translating body.

Given: Moment of inertia of the rotating body, $I = 3 \, \text{kg-m}^2$. Angular velocity of the rotating body, $\omega = 3 \, \text{rad/s}$. Mass of the translating body, $m = 27 \, \text{kg}$. Let the speed of the translating body be $v$.

The rotational kinetic energy is: $KE_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} \times (3 \, \text{kg-m}^2) \times (3 \, \text{rad/s})^2$ $KE_{rot} = \frac{1}{2} \times 3 \times 9 \, \text{Joule}$ $KE_{rot} = \frac{27}{2} \, \text{Joule}$

The translational kinetic energy is: $KE_{trans} = \frac{1}{2}mv^2 = \frac{1}{2} \times (27 \, \text{kg}) \times v^2$

Equating the two kinetic energies: $KE_{rot} = KE_{trans}$ $\frac{27}{2} = \frac{1}{2} \times 27 \times v^2$

Multiply both sides by 2: $27 = 27 \times v^2$

Divide both sides by 27: $v^2 = \frac{27}{27}$ $v^2 = 1$

Taking the square root of both sides: $v = \sqrt{1} \, \text{m/s}$ Since speed is a scalar and non-negative quantity, we take the positive root: $v = 1 \, \text{m/s}$

The speed of the body of mass 27 kg is 1.0 m/s.