Question

A tire 0.500 m in radius rotates at a constant rate of $\mathbf{200}\mathbf{ }\mathbf{\text{rev}}\mathbf{/}\mathbf{min}$. Find the speed and acceleration of a small stone lodged in the tread of the tire (on its outer edge).

Short answer

The speed of the small stone is $10.5 \text{m/s}$and the acceleration is $219 {\text{m/s}}^2$.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

The rate of rotation is,$200 \text{rev}/\min$ .

The radius of tire is, 0.500 m.

The formula to calculate speed

Write the formula to calculate speed

$\mathit{v}\mathbf{=}\mathit{r}\mathit{\omega }$

Here, $\mathit{v}$is the speed,$\mathit{r}$is the radius of tire and $\mathit{\omega }$is the rate of rotation.

Find the speed and acceleration of a small stone

Substitute values above equation to find v .

$\begin{array}{rcl}v& =& (0.500 \text{m})(200 \text{rev/min})\left(\frac{1 \text{min}}{60 \text{s}}\right)\left(\frac{2\pi \text{rad/s}}{1 \text{rev/s}}\right)\\ & =& 10.47 \text{m/s}\\ & \approx & 10.5 \text{m/s}\\ & & \end{array}$

Write the formula to calculate centripetal acceleration

$a_{\text{c}}=\frac{v^2}{r}$

Substitute values above equation to find a_c .

$\begin{array}{rcl}{a}_{\text{c}}& =& \frac{{(10.47 \text{m/s})}^2}{0.500 \text{m}}\\ & =& 219.24 {\text{m/s}}^2\\ & \approx & 219 {\text{m/s}}^2\\ & & \end{array}$

Therefore, the speed of the small stone lodged in the tread of the tire is $10.5 \text{m/s}$and the acceleration is $219 {\text{m/s}}^2$.