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 cord is left on the spool when it reaches this angular speed?

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

$V=\omega r$

Here,

$V$ is velocity

$\omega$ is angular velocity

$r$ is radius

Now,

$at=\omega rt=\frac{\omega r}{a}$

substitute all value in above equation

$t=\frac{0.5\omega }{2.5}t=\frac{8}{5}=1.6s$

The cord left on the spool is

$R△\theta =r\frac{1}{2}\alpha t^2=r\frac{1}{2}\frac{a}{r}{t}^2$

Substitute all value in above equation

$=\frac{1}{2}a{t}^2=\frac{1}{2}\times 2.5\times {1.6}^2=\frac{2.5\times 1.6\times 1.6}{2}=3.2m$

Hence,

Cord left is $4-3.2=0.8m$