A 5-kg block is attached to a pulley by a light rope. The rope is wound around the pulley, and when the block is released, the rope unspools without slipping. The pulley is a solid disk of radius 0.2 m and mass 2 kg. What is the approximate angular speed of the pulley when the block has fallen 1 meter?
A) 0.7 rad/s
B) 7.0 rad/s
C) 15.7 rad/s
D) 22.2 rad/s
E) The answer cannot be determined without knowing the final speed of the block.

Question

Quizplus · Accepted Answer

The potential energy lost by the block is converted into kinetic energy as it falls. This kinetic energy is transferred to the pulley and can be described in terms of rotational kinetic energy:

$K = \frac{1}{2} I \omega^2$

Where $K$ is the kinetic energy of the pulley, $I$ is the moment of inertia of the pulley, and $\omega$ is the angular speed of the pulley.

Assuming the rope is light and the pulley is a solid disk, we can use the moment of inertia formula $I = \frac{1}{2} MR^2$.

The potential energy lost by the block is $mgh = 5 \cdot 9.8 \cdot 1 = 49$ J.

This energy is equal to the kinetic energy of the pulley:

$K = 49$ J

$I = \frac{1}{2}(2)(0.2)^2 = 0.02$ kg m$^2$

Solving for $\omega$:

$\omega = \sqrt{\frac{2K}{I}} = \sqrt{\frac{2(49)}{0.02}} = 15.7$ rad/s

Therefore, the approximate angular speed of the pulley when the block has fallen 1 meter is 15.7 rad/s.

A 5-Kg Block Is Attached to a Pulley by a Light