Question

A 4.00-m length of light nylon cord is wound around a uniform cylindrical spool of radius 0.500 m and mass 1.00 kg. 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 $\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{}{\mathbf{\text{m/s}}}^{\mathbf{2}}$. (a) How much work has been done on the spool when it reaches an angular speed of 8.00 rad/s? (b) How long does it take the spool to reach this angular speed? (c) How much cord is left on the spool when it reaches this angular speed?

Short answer

a) The work done on the spool when it reaches an angular speed of 8.00 rad/s is,$W=4 \mathrm{J}$.

b) The time is taken by the spool to reach the angular speed,$\mathrm{\Delta }t=1.6 \mathrm{s}$.

c) The length left in the cord when it reaches the angular speed,$L=0.8 \mathrm{m}$.

Step-by-step solution

Given Information

We are given the following data for a light nylon cord wound around a uniform cylindrical spool:

$\begin{array}{c}L=4 \mathrm{m}\\ R=0.5 \mathrm{m}\\ m=1 \mathrm{kg}\\ a=2.5 \frac{\mathrm{m}}{{\mathrm{s}}^2}\end{array}$

Step 2: Work done on the spool

a)

It is given that the spoon reaches an angular speed $\omega =8\frac{\mathrm{rad}}{\mathrm{s}}$.

The moment of inertia for a uniform cylindrical is equal to

$\begin{array}{c}I=\frac{1}{2}m{R}^2 \\ =\frac{1}{2}\times 1\times 0.5^2\\ =0.125{\mathrm{kgm}}^2\end{array}$

Calculating how much work has been done:

$\begin{array}{c}W=\frac{1}{2}I{\omega }^2\\ =\frac{1}{2}\times 0.125\times 8^2\\ =4 \mathrm{J}\end{array}$

So, the work done is 4 J.

Calculating the time that takes the spool to reach this angular speed

Substituting the known values into the formula, we get,

$\begin{array}{c}\mathrm{\Delta }t=\frac{R·\omega }{2}\\ =\frac{0.5\times 8}{25}\\ =1.6 \mathrm{s}\end{array}$

So, the time is 1.6 s.

Calculating how much cord is used:

$\begin{array}{c}{L}_{\mathrm{used}}=\frac{1}{2}\alpha ·R{t}^2\\ =\frac{1}{2}\left(\frac{a}{R}\right)\left(R{t}^2\right)\\ =\frac{1}{2}\times \frac{2.5}{0.5}\times 0.5\times 1.6^2\\ =3 \mathrm{m}\end{array}$

Calculating how much cord is left:

$\begin{array}{c}{L}_{left}=L-L_{used}\\ =4 \mathrm{m}-3.2 \mathrm{m}\\ =0.8 \mathrm{m}\end{array}$

So, the required chord length is 0.8 m.