A boy sits in a revolving chair that rotates at an angular speed of...

A boy sits in a revolving chair that rotates at an angular speed of

120

revolutions per minute. Two heavy balls are part of the revolving system, and the boy can pull the balls closer to himself or push them apart. If the boy decreases the moment of inertia of the system from

6 \text{ kg-m}^2

2 \text{ kg-m}^2

by pulling the balls closer, determine the new angular speed.

Options

A $180 \text{ revolutions per minute}$
B $40 \text{ revolutions per minute}$
C $240 \text{ revolutions per minute}$
D $360 \text{ revolutions per minute}$

Correct answer

D. $360 \text{ revolutions per minute}$

Step-by-step solution

According to the principle of conservation of angular momentum, if no external torque acts on a system, its total angular momentum remains constant.

I_1 \omega_1 = I_2 \omega_2

Given:

I_1 = 6 \text{ kg-m}^2

\omega_1 = 120 \text{ rpm}

I_2 = 2 \text{ kg-m}^2

Substituting the values into the equation:

6 \times 120 = 2 \times \omega_2

\omega_2 = \dfrac{6 \times 120}{2}

\omega_2 = 360 \text{ rpm}

Practice Rotational Mechanics on Quantrex Academy →

More from Rotational Mechanics