Rotational dynamics about a fixed axis: A string is wrapped around a pulley with a radius of 2.0 cm and no appreciable friction in its axle. The pulley is initially not turning. A constant force of 50 N is applied to the string, which does not slip, causing the pulley to rotate and the string to unwind. If the string unwinds 1.2 m in 4.9 s, what is the moment of inertia of the pulley? A) 0.17 kg ∙ m² B) 17 kg ∙ m² C) 14 kg ∙ m² D) 0.20 kg ∙ m² E) 0.017 kg ∙ m²

Question

Quizplus · Accepted Answer

The torque on the pulley is given by the product of the force and the radius:
$$	au = F \cdot r = 50	ext{ N} \cdot 0.02	ext{ m} = 1	ext{ N}\cdot	ext{m}$$
The angular acceleration of the pulley is given by:
$$\alpha = \frac{	au}{I}$$
where $I$ is the moment of inertia of the pulley. We can rearrange this equation to solve for $I$:
$$I = \frac{	au}{\alpha}$$
The angular velocity of the pulley is given by:
$$\omega = \alpha t$$
where $t$ is the time taken for the string to unwind 1.2 m. We can rearrange this equation to solve for $\alpha$:
$$\alpha = \frac{\omega}{t}$$
Substituting this into the equation for $I$, we get:
$$I = \frac{	au}{\alpha} = \frac{	au \cdot t}{\omega} = \frac{(1	ext{ N}\cdot	ext{m}) \cdot (4.9	ext{ s})}{\left(\frac{1.2	ext{ m}}{0.02	ext{ m}}ight)} = 0.20	ext{ kg}\cdot	ext{m}^2$$
Therefore, the moment of inertia of the pulley is 0.20 kg·m², which corresponds to choice D.

Rotational Dynamics About a Fixed Axis: a String Is Wrapped