A pulley, with a rotational inertia of $\mathbf{1}\mathbf{.0}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{\text{\hspace{0.17em}kg.m}}}^{\mathbf{2}}$ about its axle and a radius of $\mathbf{10}\mathbf{\text{\hspace{0.17em}cm}}$,is acted on by a force applied tangentially at its rim. Theforce magnitude varies in time as$\mathbf{F}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.50}\mathbf{t}\mathbf{}\mathbf{+}\mathbf{}\mathbf{0}\mathbf{.30}{\mathbf{t}}^{\mathbf{2}}\mathbf{}\mathbf{}$ ,with F in newtons and t in seconds. Thepulley is initially at rest. At $\mathbf{t}\mathbf{=}\mathbf{}\mathbf{3.0}\mathbf{\text{\hspace{0.17em}s}}\mathbf{}$ what are its (a) angular acceleration and (b) angular speed?

1. Rotation inertia (M.I) of the pulley about its axle,$\mathrm{I}=1.0\times {10}^{-3}{\text{\hspace{0.17em}kg.m}}^2.$
2. Radius of the pulley is,$\mathrm{r}=0.10\text{\hspace{0.17em}m}$.
3. Force acting on pulley,$\mathrm{F}=0.50\mathrm{t}+0.30{\mathrm{t}}^2$.
4. Initial speed of the pulley is zero.

You can find the torque acting on the pulley from the applied force using the corresponding formula. Then using the relation between angular acceleration and torque, you can find the angular acceleration. Integrating it with respect to t, you can find the angular speed of the pulley at t=3 s. The formulas used are given below.

$\begin{array}{c}\mathbf{\tau }\mathbf{=}\mathbf{r}\mathbf{\times }\mathbf{F}\\ \mathbf{=}\mathbf{I\alpha }\\ \mathbf{\omega }\mathbf{=}\underset{{\mathbf{t}}_{\mathbf{i}}}{\overset{{\mathbf{t}}_{\mathbf{f}}}{\int }}\mathbf{\alpha }\mathbf{}\mathbf{dt}\end{array}$

The torque on the pulley is

$\mathrm{\tau }=\mathrm{r}\times \mathrm{F}$

But, applied force is tangential to r.

$\begin{array}{c}\mathrm{\tau }=\mathrm{rF}\\ =\left(0.10\text{\hspace{0.17em}m}\right)(0.50\mathrm{t}+0.30{\mathrm{t}}^2)\\ =(0.05\mathrm{t}+0.03{\mathrm{t}}^2)\end{array}$

Also,

$\mathrm{\tau }=\mathrm{I\alpha }$

Then, angular acceleration of the pulley is

$\begin{array}{c}\mathrm{\alpha }=\frac{\mathrm{\tau }}{\mathrm{I}}\\ =\frac{(0.05\mathrm{t}+0.03{\mathrm{t}}^2)}{1.0\times {10}^{-3}{\text{\hspace{0.17em}kg.m}}^{\text{2}}}\\ =50\mathrm{t}+30{\mathrm{t}}^2\end{array}$

At $\mathrm{t}=3.0\text{\hspace{0.17em}s},$

$\begin{array}{c}\mathrm{\alpha }=50\left(3\text{\hspace{0.17em}s}\right)+30{\left(3\text{\hspace{0.17em}s}\right)}^2\\ =4.2\times {10}^2\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}\end{array}$

Therefore, angular acceleration of the pulley at $\mathrm{t}=3.0\text{\hspace{0.17em}s},$ is$4.2\times {10}^2\text{\hspace{0.17em}rad}/{\text{s}}^{\text{2}}$.

Angular speed of the pulley at $\mathrm{t}=3.0\text{\hspace{0.17em}s},$is

$\begin{array}{c}\mathrm{\omega }=\underset{0}{\overset{3}{}}\mathrm{\alpha }\mathrm{dt}\\ =\underset{0}{\overset{3}{}}(50\mathrm{t}+30{\mathrm{t}}^2)\mathrm{dt}\\ =(25{\left(3\right)}^2+10{\left(3\right)}^3)\\ =5\times {10}^2\text{\hspace{0.17em}rad}/\text{s}\end{array}$

.

Therefore, angular speed of the pulley at $\mathrm{t}=3.0\text{\hspace{0.17em}s},$is $5\times {10}^2\text{\hspace{0.17em}rad}/\text{s}$.