Question

Short answer

The result for part (a) is$1.44\times {10}^{-4}\mathrm{r}\mathrm{a}\mathrm{d}/{\mathrm{s}}^2$ , and those of part (b) are$0.011\mathrm{m}/{\mathrm{s}}^2$ and $6.12\times {10}^{-4}\mathrm{m}/{\mathrm{s}}^2$,respectively.

Step-by-step solution

Given data

The revolutions per minute are$N=1\mathrm{r}\mathrm{e}\mathrm{v}/\mathrm{m}\mathrm{i}\mathrm{n}$.

The time interval is$t=12\mathrm{m}\mathrm{i}\mathrm{n}$.

The diameter of the cylinder is$D=8.5\mathrm{m}$.

The time after acceleration starts at $t_0=6\mathrm{m}\mathrm{i}\mathrm{n}$.

Understanding acceleration

In this problem, for the evaluation of the components of the acceleration on the outer skin, use the kinematic relation between angular speed and the radius of the cylinder.

Determine the angular acceleration

The relation to find the angular acceleration is given by:

$\alpha =\frac{{\omega }_2-{\omega }_1}{t}$

Here, ${\omega }_1$ and ${\omega }_1$ are the initial and final angular velocities, where the initial angular velocity is equivalent to zero.

On plugging the values in the above relation, you get:

$\begin{array}{rl}l\alpha & =\left(\frac{\left(1\mathrm{r}\mathrm{e}\mathrm{v}/\mathrm{m}\mathrm{i}\mathrm{n}\times \frac{2\pi \mathrm{r}\mathrm{a}\mathrm{d}}{1\mathrm{r}\mathrm{e}\mathrm{v}}\times \frac{1\mathrm{m}\mathrm{i}\mathrm{n}}{60\mathrm{s}}\right)-0}{\left(12\mathrm{m}\mathrm{i}\mathrm{n}\times \frac{60\mathrm{s}}{1\mathrm{m}\mathrm{i}\mathrm{n}}\right)}\right)\\ \alpha & =\left(\frac{\left(0.104\mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}\right)}{\left(720\mathrm{s}\right)}\right)\\ \alpha & =1.44\times {10}^{-4}\mathrm{r}\mathrm{a}\mathrm{d}/{\mathrm{s}}^2\end{array}$

Thus, $\alpha =1.44\times {10}^{-4}\mathrm{r}\mathrm{a}\mathrm{d}/{\mathrm{s}}^2$ is the angular acceleration.

Determine the instantaneous angular velocity

The relation to find the angular velocity is given by:

$\begin{array}{rl}l{\omega }_0& ={\omega }_1+\alpha t_0\\ {\omega }_0& =\left(0\right)+\left(1.44\times {10}^{-4}\mathrm{r}\mathrm{a}\mathrm{d}/{\mathrm{s}}^2\right)\left(6\mathrm{m}\mathrm{i}\mathrm{n}\times \frac{60\mathrm{s}}{1\mathrm{m}\mathrm{i}\mathrm{n}}\right)\\ {\omega }_0& =0.051\mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}\end{array}$

The relation to find the radial acceleration is given by:

$\begin{array}{r}l{a}_{\mathrm{r}}={{\omega }_0}^{2}r\\ a_{\mathrm{r}}={{\omega }_0}^2\left(\frac{D}{2}\right)\end{array}$

Determine the radial and tangential components of acceleration

On plugging the values in the above relation, you get:

$\begin{array}{rl}l{a}_{\mathrm{r}}& ={\left(0.051\mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}\right)}^2\left(\frac{8.5\mathrm{m}}{2}\right)\\ a_{\mathrm{r}}& =0.011\mathrm{m}/{\mathrm{s}}^2\end{array}$

The relation to find the tangential acceleration is given by:

$\begin{array}{r}l{a}_{\mathrm{t}}=\alpha r\\ a_{\mathrm{t}}=\alpha \left(\frac{D}{2}\right)\end{array}$

On plugging the values in the above relation, you get:

$\begin{array}{rl}l{a}_{\mathrm{t}}& =\left(1.44\times {10}^{-4}\mathrm{r}\mathrm{a}\mathrm{d}/{\mathrm{s}}^2\right)\left(\frac{8.5\mathrm{m}}{2}\right)\\ a_{\mathrm{t}}& =6.12\times {10}^{-4}\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, $a_{\mathrm{r}}=0.011\mathrm{m}/{\mathrm{s}}^2$ and $a_{\mathrm{t}}=6.12\times {10}^{-4}\mathrm{m}/{\mathrm{s}}^2$ are the radial and tangential acceleration.