Question 72 of 130

A uniform rod $AB$ of mass $2kg$ and Length $30cm$ at rest on a smooth horizontal surface. An impulse of force $0.2Ns$ is applied to end B. The time taken by the rod to turn through at right angles will be $\dfrac{\pi }{x}s,$ where $x=$ ____.

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Answer: 4

Impulse $J=0.2Ns$
$\Rightarrow J=\int Fdt=0.2Ns$
Now, angular impulse $(\vec{M})$ will be
${M}_{c}=\int \tau dt$
$=\int F\dfrac{L}{2}dt$
$=\dfrac{L}{2}\int Fdt=\dfrac{L}{2}\times J$
$=\dfrac{0.3}{2}\times 0.2$
$=0.03$
Moment of inertia of the rod about the centre of mass, ${I}_{cm}=\dfrac{M{L}^{2}}{12}=\dfrac{2\times (0.3{)}^{2}}{12}=\dfrac{0.09}{6}$
Angular impulse will be equal to the change in angular momentum.
$M={I}_{cm}\left({\omega }_{f}-{\omega }_{i}\right)$
$\Rightarrow 0.03=\dfrac{0.09}{6}\left({\omega }_{f}\right)$
$\Rightarrow {\omega }_{f}=2rad{s}^{-1}$
For angular displacement, we can write $\theta =\omega t$
$\Rightarrow t=\dfrac{\theta }{\omega }=\dfrac{\pi }{2\times 2}=\dfrac{\pi }{4}s$.
Therefore, $x=4$.