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  • A uniform rod of mass 1.2 kg and length 1m is resting on a smooth horizontal surface. An impulse of  N-m is ap

A uniform rod of mass 1.2 kg and length 1m is resting on a smooth horizontal surface. An impulse of \pi N-m is applied to one of its end as shown in the figure. Then the time taken by the rod to turn by \pi/2 radiation will be 

Option: 1

0.1s


Option: 2

0.2s


Option: 3

0.3


Option: 4

0.4


Answers (1)

best_answer

angular impulse= change in angular momentum

\pi \times \frac{l}{2}=I _{cm} \omega

\pi \times \frac{l}{2}=\frac{m l^2}{12} \times \omega

\begin{aligned} & \omega=\frac{6 \pi}{m l} \\ & \theta=\omega t \end{aligned}

\begin{aligned} & t=\frac{\theta}{\omega} \\ & t=\frac{\pi / 2}{6 \pi / m l} \end{aligned}

\begin{aligned} & t=\frac{m l}{12}=\frac{1.2 \times 1}{12} \\ & t=0.1 \mathrm{~s} \end{aligned}

 

Posted by

Pankaj Sanodiya

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