Question

A $4.00-m$ length of light nylon cord is wound around a uniform cylindrical spool of radius $0.500m$ and mass $1.00kg$. The spool is mounted on a frictionless axle and is initially at rest. The cord is pulled from the spool with a constant acceleration of magnitude $2.50m/s^2$.
How much work has been done on the spool when it reaches an angular speed of $8.00rad/s$?

Answer 1

Given,

Length of nylon rod is $4m$

radius is $0.5m$

mass is $1kg$

acceleration is $2.5m/s^2$

Angular velocity is $8rad/s$

We know that

Total work done is equal to change in kinetic energy therefore,

$W=△KE=\frac{1}{2}I{\omega }^2$

Moment of inertia $\left(I\right)$ is $I=\frac{1}{2}m{r}^2$

Now

$W=\frac{1}{2}\times \frac{1}{2}m{r}^2\times {\omega }^2$

Substitute all value in above equation

$W=\frac{1}{2}\times \frac{1}{2}\times 1\times {0.5}^2\times 8^{2}W=\frac{0.25\times 64}{4}W=4J$