A string is wrapped tightly around a fixed frictionless pulley that has a moment of inertia of 0.0352 $\mathrm { kg } \cdot \mathrm { m } ^ { 2 }$ and a radius of 12.5 cm. The string is pulled away from the pulley with a constant force of 5.00 N, 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) 4.95 m/s
B) 2.36 m/s
C) 2.09 m/s
D) 1.18 m/s
E) 3.18 m/s

Question

A string is wrapped tightly around a fixed frictionless pulley that has a moment of inertia of  0.0352  $\mathrm { kg } \cdot \mathrm { m } ^ { 2 }$ and a radius of  12.5 cm. The string is pulled away from the pulley with a constant force of  5.00 N, 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) 4.95 m/s
B) 2.36 m/s
C) 2.09 m/s
D) 1.18 m/s
E) 3.18 m/s

Quizplus · Accepted Answer

The speed of the string can be found using the work-energy principle. The work done by the force on the string is equal to the change in rotational kinetic energy of the pulley. The work done by the force is $W = F \cdot d = 5.00 \, \text{N} \cdot 1.25 \, \text{m} = 6.25 \, \text{J}$. The rotational kinetic energy is given by $\frac{1}{2} I \omega^2$, where $I$ is the moment of inertia and $\omega$ is the angular velocity. Since $v = r\omega$, where $v$ is the linear speed of the string and $r$ is the radius of the pulley, we can solve for $v$ using the work-energy principle: $6.25 \, \text{J} = \frac{1}{2} \cdot 0.0352 \, \text{kg} \cdot \text{m}^2 \cdot \left(\frac{v}{0.125 \, \text{m}}\right)^2$. Solving for $v$ gives approximately $2.36 \, \text{m/s}$.

A String Is Wrapped Tightly Around a Fixed Frictionless Pulley kg⋅m2\mathrm { kg } \cdot \mathrm { m } ^ { 2 }kg⋅m2

A String Is Wrapped Tightly Around a Fixed Frictionless Pulley $\mathrm { kg } \cdot \mathrm { m } ^ { 2 }$