Question

A small rubber wheel is used to drive a large pottery wheel. The two wheels are mounted so that their circular edges touch. The small wheel has a radius of 2.0 cm and accelerates at the rate of$7.2\mathrm{r}\mathrm{a}\mathrm{d}/{\mathrm{s}}^2$, and it is in contact with the pottery wheel (radius 27.0 cm) without slipping. Calculate (a) the angular acceleration of the pottery wheel, and (b) the time it takes the pottery wheel to reach its required speed of 65 rpm.

Short answer

(a). The angular acceleration of the pottery wheel is $0.53\mathrm{r}\mathrm{a}\mathrm{d}/{\mathrm{s}}^2$. (b) The time taken by the pottery wheel to reach a speed of 65 rpm is 12.84 s.

Step-by-step solution

Identification of the given data

The radius of the rubber wheel is$r_{\mathrm{s}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{l}}=2\mathrm{c}\mathrm{m}=0.02\mathrm{m}$.

The radius of the pottery wheel is$r_{\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}}=27\mathrm{c}\mathrm{m}=0.27\mathrm{m}$.

The angular acceleration of the smaller wheel is${\alpha }_{\mathrm{s}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{l}}=7.2\mathrm{r}\mathrm{a}\mathrm{d}/{\mathrm{s}}^2$.

The linear speed of the pottery wheel is $v=65\mathrm{r}\mathrm{p}\mathrm{m}$.

Understanding angular acceleration and tangential acceleration

The angular acceleration$\alpha$is defined as the time rate of change of angular velocity.

The tangential acceleration$a_{\mathrm{t}\mathrm{a}\mathrm{n}}$is the measure of change in the tangential velocity of a point at a certain radius with time.

The tangential acceleration is related to the angular acceleration as follows:

$a_{\tan }=\alpha r$

Here, r is the radius of the circular path traced.

Determining the angular acceleration of the pottery wheel

The angular accelerations of the rubber (smaller) and the pottery (larger) wheels are different. But, their tangential components of the linear acceleration are equal because the edges of the wheels touch each other without slipping.

$\begin{array}{r}c{\left(a_{\mathrm{t}\mathrm{a}\mathrm{n}}\right)}_{\mathrm{s}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{l}}={\left(a_{\mathrm{t}\mathrm{a}\mathrm{n}}\right)}_{\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}}\\ {\alpha }_{\mathrm{s}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{l}}{r}_{\mathrm{s}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{l}}={\alpha }_{\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}}{r}_{\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}}\\ {\alpha }_{\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}}={\alpha }_{\mathrm{s}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{l}}\frac{r_{\mathrm{s}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{l}}}{r_{\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}}}\end{array}$

Substituting numerical values in the above expression,

$\begin{array}{rl}c{\alpha }_{\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}}& =\left(7.2\mathrm{r}\mathrm{a}\mathrm{d}/{\mathrm{s}}^2\right)\frac{0.02\mathrm{m}}{0.27\mathrm{m}}\\ & =0.53\mathrm{r}\mathrm{a}\mathrm{d}/{\mathrm{s}}^2\end{array}$

Thus, the angular acceleration of the pottery wheel is $0.53\mathrm{r}\mathrm{a}\mathrm{d}/{\mathrm{s}}^2$.

(b) Determining the time taken by the pottery wheel to reach 65 rpm

Suppose the pottery wheel starts from rest. Now, convert the speed into angular speed as follows:

$\begin{array}{r}c\omega =\left(65\frac{\mathrm{r}\mathrm{e}\mathrm{v}}{min}\right)\left(\frac{2\pi \mathrm{r}\mathrm{a}\mathrm{d}}{1\mathrm{r}\mathrm{e}\mathrm{v}}\right)\left(\frac{1\mathrm{m}\mathrm{i}\mathrm{n}}{60\mathrm{s}}\right)\\ =6.807\mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}\end{array}$

Using the following kinematic equation of rotational motion,

$\begin{array}{rl}c\omega & ={\omega }_0+\alpha t\\ t& =\frac{\omega -{\omega }_0}{\alpha }\\ & =\frac{6.807\mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}-0}{0.53\mathrm{r}\mathrm{a}\mathrm{d}/{\mathrm{s}}^2}\\ & =12.84\mathrm{s}\end{array}$

Thus, the time taken by the pottery wheel to reach a speed of 65 rpm is 12.84 s.