Question

A pulley is rotated about its axis by a force $F=(20t-5t^2)\text{N}$ (where $t$ is measured in seconds) applied tangentially. Find the magnitude of angular impulse [in $\text{Nm-s}$] on the pulley in the initial $2\text{seconds}$. Given radius of pulley $=0.5\text{m}$.

• A. $10/3$
• B. $20/3$
• C. $40/3$
• D. zero

Answer 1

The correct option is C $40/3$

As we know,
$\text{Angular Impulse}\left(J\right)=\underset{t_1}{\overset{t_2}{\int }}\tau .dt$
where $\overrightarrow{\tau }=\overrightarrow{d}\times \overrightarrow{F}$
where $d$ is the perpendicular distance of force vector from the axis of rotation.
i.e $\tau =Fd\sin \theta =Fd$ (since force acts tangentially, $\theta ={90}^{\circ }$)
$d=R=0.5\text{m}$

Hence, $J=\underset{0}{\overset{2}{\int }}(20t-5t^2)\times 0.5dt$
$J={[\frac{10t^2}{2}-\frac{2.5t^3}{3}]}_0^2$
$J=[5\times 4-\frac{2.5\times 8}{3}]$
Or, $J=[20-\frac{20}{3}]$
Or, $J=\frac{40}{3}$ $\text{Nm-s}$