A particle of mass m is attached to an end of a uniform rod of mass M $$=$$ $$2m$$ and length ‘l’ which is suspended through its mid point by an inextensible string as shown. Initially the rod is in horizontal position and at rest. The system is released from this position. Just after the release.

Question

Infinity Learn · Accepted Answer

The FBD of the rod and the ball are shown. Applying  τ$$=I$$α about the C.M. of the rod, we have,  $$N(l2)=(Ml212)$$αWriting Newton’s II law in the vertical direction on the CM of the rod we have $$2mg+N-T=0$$  and writing Newton’s II law in the vertical direction on the ball we have, mg−$$N=m(l2)$$α Solve these equations.α$$=6g5l$$ and $$T=12mg5$$

Column-A		Column-B (Torque of normal reaction on body B about centre of mass)
(a)	(F is directed towards\| centre of mass)	(p)	Clockwise
(b)		(q)	Anticlockwise
(c)		(r)	Zero
		(s)	Conclusion cannot be drawn with the information given

A particle of mass m is attached to an end of a uniform rod of mass M = 2m and length ‘l’ which is suspended through its mid point by an inextensible string as shown. Initially the rod is in horizontal position and at rest. The system is released from this position. Just after the release.

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