Question

A long string is wrapped around a $6.0\mathrm{cm}$diameter cylinder, initially at rest, that is free to rotate on an axle. The string is then pulled with a constant acceleration of $1.5\mathrm{m}/{\mathrm{s}}^2$ until $1.0\mathrm{m}$ of string has been unwound. If the string unwinds without slipping, what is the cylinder’s angular speed, in $\mathrm{rpm}$, at this time?

Short answer

$\mathrm{\omega }=550\mathrm{rpm}$

Step-by-step solution

Step 1. Given Information

We are given with:

$$\begin{gathered} \mathrm{D}=6\mathrm{cm} \\ \mathrm{a}=1.5\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{d}=1\mathrm{m} \end{gathered}$$

We have to find the cylinder’s angular speed in rpm.

Step 2. Calculation

The linear speed of string is:

$$\begin{gathered} {\mathrm{v}}^2=2\mathrm{ad} \\ \mathrm{v}=\sqrt{2\times 1.5\times 1} \\ \mathrm{v}=1.73\mathrm{m}/\mathrm{s} \end{gathered}$$

The angular speed is:

$$\begin{gathered} \mathrm{\omega }=\frac{\mathrm{v}}{\mathrm{r}} \\ \mathrm{r}=\mathrm{D}/2 \\ \mathrm{r}=0.06/2 \\ \mathrm{r}=0.03\mathrm{m} \\ \mathrm{\omega }=\frac{1.73}{0.03} \\ \mathrm{\omega }=57.74\mathrm{rad}/\mathrm{s} \end{gathered}$$

To convert rad/s to rpm

role="math" localid="1647938262943" $$\begin{gathered} \mathrm{\omega }=57.74\times \frac{60}{2\mathrm{\pi }}\mathrm{rpm} \\ \mathrm{\omega }=550\mathrm{rpm} \end{gathered}$$