Question

Short answer

The results for parts (a) and (b) are $10.1\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^2/\mathrm{s}$ and $-2.5\mathrm{N}\cdot \mathrm{m}$, respectively.

Step-by-step solution

Given data

The mass of the cylinder is$m=48\mathrm{k}\mathrm{g}$.

The radius of the cylinder is$r=15\mathrm{c}\mathrm{m}$.

The skater is spinning at$\omega =3\mathrm{r}\mathrm{e}\mathrm{v}/\mathrm{s}$.

The time is$t=4\mathrm{s}$.

The height is $h=1.5\mathrm{m}$.

Understanding rotational inertia and angular momentum

Consider that the rotational inertia of the cylinder does not change. In this condition, at constant rotational inertia, the variation in angular momentum occurs because of the change in angular velocity.

Determine the angular momentum of the cylinder

The relation of angular momentum can be written as:

$\begin{array}{rl}lL& =I\omega \\ L& =\left(\frac{1}{2}m{r}^2\right)\omega \end{array}$

Here, Iis the moment of inertia.

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

$\begin{array}{rl}lL& =\frac{1}{2}\left(48\mathrm{k}\mathrm{g}\times {\left(15\mathrm{c}\mathrm{m}\times \frac{1\mathrm{m}}{100\mathrm{c}\mathrm{m}}\right)}^2\right)\left(3\mathrm{r}\mathrm{e}\mathrm{v}/\mathrm{s}\times \frac{2\pi \mathrm{r}\mathrm{a}\mathrm{d}}{1\mathrm{r}\mathrm{e}\mathrm{v}}\right)\\ L& =10.1\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^2/\mathrm{s}\end{array}$

Thus, $L=10.1\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^2/\mathrm{s}$ is the required angular momentum.

Determine the torque applied on the cylinder

The relation of angular momentum can be written as:

$\tau =\frac{L^{\prime }-L}{t}$

Here, $L^{\prime }$is the final angular momentum, whose value is zero.

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

$\begin{array}{rl}l\tau & =\left(\frac{0-10.1\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^2/\mathrm{s}}{4\mathrm{s}}\right)\\ \tau & =-2.5\mathrm{N}\cdot \mathrm{m}\end{array}$

Thus, $\tau =-2.5\mathrm{N}\cdot \mathrm{m}$ is the required torque.