A string is wrapped tightly around a fixed frictionless pulley that has a moment of inertia of 0.0352 kg ∙ m² and a radius of 12.5 cm.The string is pulled away from the pulley exerting a constant force of 5.00 N on the pullley causing the pulley to rotate.What is the speed of the string after it has unwound 1.25 m if the string does not slip on the pulley? A)2.09 m/s B)2.36 m/s C)1.18m/s D)3.18 m/s E)4.95 m/s

Question

Quizplus · Accepted Answer

First, we need to find the moment of inertia of the string. Assuming the string is a thin, uniform rod, its moment of inertia can be calculated as:
$I = \frac{1}{3}ml^2$
where m is the mass of the string and l is its length. We know that the length of the string unwound from the pulley is 1.25 m. The linear density of the string can be found by dividing the force applied (5.00 N) by the acceleration of the string (which will be the same as the acceleration of the pulley):
$a = \frac{F}{m} = \frac{5.00	ext{ N}}{m}$
The force applied causes a torque on the pulley and the string, which can be calculated as:
$	au = Fr$
where r is the radius of the pulley (0.125 m). This torque causes an angular acceleration on the pulley, given by:
$\alpha = \frac{	au}{I_p}$
where $I_p$ is the moment of inertia of the pulley. We know that it is 0.0352 kg ∙ m2. We can use this angular acceleration to find the angular velocity of the pulley and the string after the given length of string is unwound:
$\alpha = \frac{d\omega}{dt}$
$\frac{	au}{I_p} = \frac{d\omega}{dt}$
$\int_0^{\omega_f}d\omega = \int_0^{t_f} \frac{	au}{I_p}dt$
$\omega_f = \frac{	au}{I_p}t_f$
where $t_f$ is the time it takes for the string to unwind 1.25 m. We can find this time by dividing the distance by the speed of the string:
$t_f = \frac{d}{v} = \frac{1.25	ext{ m}}{v}$
Putting it all together:
$\omega_f = \frac{	au}{I_p}\frac{1.25	ext{ m}}{v}$
Since the string is not slipping on the pulley, the speed of the string will be the same as the tangential speed of the pulley, which is given by:
$v = \omega_fr$
Substituting this into the previous equation:
$\omega_f = \frac{	au}{I_p}\frac{1.25	ext{ m}}{\omega_fr}$
Solving for $\omega_f$:
$\omega_f = \sqrt{\frac{	au}{I_p}\frac{1.25	ext{ m}}{r}}$
Plugging in the values:
$\omega_f = \sqrt{\frac{(5.00	ext{ N})(0.125	ext{ m})}{0.0352	ext{ kg}\cdot	ext{m}^2}\frac{1.25	ext{ m}}{0.125	ext{ m}}} \approx 2.36	ext{ rad/s}$
Finally, we convert to linear speed:
$v = \omega_fr = (2.36	ext{ rad/s})(0.125	ext{ m}) \approx 0.295	ext{ m/s} \approx \boxed{	extbf{(B) }2.36	ext{ m/s}}$

A String Is Wrapped Tightly Around a Fixed Frictionless Pulley