Question

A thin hoop initially at rest with a mass of $2.5kg$ and a radius of $2.5m$ is subject to a net torque $\tau \left(t\right)=500t+50$, where $\tau$ is measured in $Nm$.
Which of the following best represents the angular velocity of the cylinder after $3$ secods?

• A. $465rad/s$
• B. $144rad/s$
• C. $154.6rad/s$
• D. $2250rad/s$
• E. $2400rad/s$

Answer 1

The correct option is C $154.6rad/s$

Given : $m=2.5$ kg $r=2.5$ m

Moment of inertia of the hoop $I=m{r}^2=2.5\times (2.5{)}^2=15.625$ $kg{m}^2$

Using $\tau =I\alpha$

$\therefore$ $\alpha =\frac{\tau }{I}=\frac{500t+50}{15.625}$

Integrating w.r.t time we get angular velocity $w\left(t\right)=\int \alpha dt$

$\therefore$ $w\left(t\right)=\frac{1}{15.625}\times \int (500t+50)dt=\frac{250t^2+50t}{15.625}$ rad /s

$⟹$ $w{|}_{t=3s}=\frac{250\times 3^2+50\times 3}{15.625}=154.6$ $rad/s$