91Ó°ÊÓ

The revolutions per minute are\(N = 1\;{\rm{rev/min}}\).

The time interval is\(t = 12\;{\rm{min}}\).

The diameter of the cylinder is\(D = 8.5\;{\rm{m}}\).

The time after acceleration starts at \({t_0} = 6\;{\rm{min}}\).

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.

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:

\(\begin{aligned}{l}\alpha &= \left( {\frac{{\left( {1\;{\rm{rev/min}} \times \frac{{2\pi \;{\rm{rad}}}}{{1\;{\rm{rev}}}} \times \frac{{1\;{\rm{min}}}}{{60\;{\rm{s}}}}} \right) - 0}}{{\left( {12\;{\rm{min}} \times \frac{{60\;{\rm{s}}}}{{1\;{\rm{min}}}}} \right)}}} \right)\\\alpha &= \left( {\frac{{\left( {0.104\;{\rm{rad/s}}} \right)}}{{\left( {720\;{\rm{s}}} \right)}}} \right)\\\alpha &= 1.44 \times {10^{ - 4}}\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}\end{aligned}\)

Thus, \(\alpha = 1.44 \times {10^{ - 4}}\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}\) is the angular acceleration.

The relation to find the angular velocity is given by:

\(\begin{aligned}{l}{\omega _0} &= {\omega _1} + \alpha {t_0}\\{\omega _0} &= \left( 0 \right) + \left( {1.44 \times {{10}^{ - 4}}\;{\rm{rad/}}{{\rm{s}}^2}} \right)\left( {6\;{\rm{min}} \times \frac{{60\;{\rm{s}}}}{{1\;{\rm{min}}}}} \right)\\{\omega _0} &= 0.051\;{\rm{rad/s}}\end{aligned}\)

The relation to find the radial acceleration is given by:

\(\begin{aligned}{l}{a_{\rm{r}}} = {\omega _0}^2r\\{a_{\rm{r}}} = {\omega _0}^2\left( {\frac{D}{2}} \right)\end{aligned}\)

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

\(\begin{aligned}{l}{a_{\rm{r}}} &= {\left( {0.051\;{\rm{rad/s}}} \right)^2}\left( {\frac{{8.5\;{\rm{m}}}}{2}} \right)\\{a_{\rm{r}}} &= 0.011\;{\rm{m/}}{{\rm{s}}^2}\end{aligned}\)

The relation to find the tangential acceleration is given by:

\(\begin{aligned}{l}{a_{\rm{t}}} = \alpha r\\{a_{\rm{t}}} = \alpha \left( {\frac{D}{2}} \right)\end{aligned}\)

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

\(\begin{aligned}{l}{a_{\rm{t}}} &= \left( {1.44 \times {{10}^{ - 4}}\;{\rm{rad/}}{{\rm{s}}^2}} \right)\left( {\frac{{8.5\;{\rm{m}}}}{2}} \right)\\{a_{\rm{t}}} &= 6.12 \times {10^{ - 4}}\;{\rm{m/}}{{\rm{s}}^2}\end{aligned}\)

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