Question bank/JEE Main/Physics

JEE MainPhysicsToughRigid Body Dynamics

A uniform rod AB of mass m and length I at rest on a smooth
horizontal surface. An impulse P is applied to the end B. The time taken
by the rod to turn through a right angle is

Show answer and solution
Impulse = change in momentum
∴       P⋅l2=mc212⋅ω\therefore\ \ \ \ \ \ \ P \cdot \frac{\mathcal{l}}{2} = \frac{mc^{2}}{12} \cdot \omega
(about centre of AB)
⇒ω=6Pml\Rightarrow \omega = \frac{6P}{m\mathcal{l}}
π2=ωt⇒t=π2ω=πml2×6p\frac{\pi}{2} = \omega t \Rightarrow t = \frac{\pi}{2\omega} = \frac{\pi m\mathcal{l}}{2 \times 6p}
⇒t=πml12p\Rightarrow t = \frac{\pi m\mathcal{l}}{12p}

Source: Grass-G / P-07 Physics paper 17 October.docx

More JEE Main questionsChapter practice

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.

Scan a doubtView plans