Question bank/JEE Main/Physics

JEE MainPhysicsModerateRigid Body Dynamics

A thin rod having mass m and length 4[IMAGE] is
free to rotate about horizontal axis passing through a point distant
[IMAGE] from one of its end, as shown in figure. It is
released, from the horizontal position as shown:
[IMAGE] What will be acceleration of centre of mass at this instant?


Show answer and solution
Torque equation
τHinge =IHinge ∝\tau_{Hinge\ } = I_{Hinge\ } \propto
mgl=(m(4l)212+ml2)∝mg\mathcal{l =}\left( \frac{m(4\mathcal{l})^{2}}{12} + m\mathcal{l}^{2} \right) \propto
3g7l=∝\frac{3g}{7\mathcal{l}} = \propto
Tangential acceleration=∝l=3g7l= \propto \mathcal{l =}\frac{3g}{7\mathcal{l}}
Radial acceleration=ω2l=0= \omega^{2}\mathcal{l} = 0

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

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