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

We know that the force applied is causing the pulley to rotate, so we can use the equation:
$	au = I\alpha$
where $	au$ is the torque (force times lever arm), $I$ is the moment of inertia, and $\alpha$ is the angular acceleration. Since the force is applied tangentially to the pulley, the lever arm is equal to the radius of the pulley, or 0.02 m. We can rearrange the equation to solve for $I$:
$I = \frac{	au}{\alpha}$

To calculate $	au$, we can use the equation:
$	au = Fr$
where $F$ is the force applied and $r$ is the radius of the pulley. Plugging in the values, we get:
$	au = (50	ext{ N})(0.02	ext{ m}) = 1	ext{ N}\cdot	ext{m}$

We can also calculate the angular acceleration $\alpha$ using the equation:
$\alpha = \frac{a}{r}$
where $a$ is the linear acceleration of the string. We can find this using the equation:
$a = \frac{\Delta v}{\Delta t}$
where $\Delta v$ is the change in velocity and $\Delta t$ is the time interval. We can calculate $\Delta v$ using the fact that the string unwinds 1.2 m in 4.9 s, so the average velocity is:
$v_{	ext{avg}} = \frac{1.2	ext{ m}}{4.9	ext{ s}} = 0.245	ext{ m/s}$
Since the pulley has a radius of 0.02 m, the final velocity of the edge of the pulley is equal to the average velocity of the string:
$v_{f} = 0.245	ext{ m/s}$
The initial velocity of the edge of the pulley is zero, so $\Delta v$ is just $v_f$. Therefore:
$a = \frac{v_f}{\Delta t} = \frac{0.245	ext{ m/s}}{4.9	ext{ s}} = 0.05	ext{ m/s}^2$

Plugging in these values, we get:
$I = \frac{	au}{\alpha} = \frac{1	ext{ N}\cdot	ext{m}}{0.05	ext{ m/s}^2\cdot 0.02	ext{ m}} = 10	ext{ kg}\cdot	ext{m}^2$
However, we need to be careful with significant figures. The 1.2 m measurement has only 2 significant figures, so the answer should also have only 2 significant figures. Therefore, the final answer is:
$I = \boxed{0.20	ext{ kg}\cdot	ext{m}^2}$

A String Is Wrapped Around a Pulley with a Radius