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Question

A uniform rod ABC of mass M is placed vertically on a rough horizontal surface. The coefficient of kinetic friction between the rod and the surface is μ. A force F(>μmg) is applied on the rod at point B at distance l/3 below centre of the rod as shown in figure. The initial acceleration of point A is


A
μgFM
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B
FM
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C
4 μg
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D
2μgFM
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Solution

The correct option is D 2μgFM
As F>μMg, so the friction is kinetic in nature and FBD of rod would be like as shown in figure.

Using Newtons second law,
FμMg=Ma
Torque equation about centre of rod
F×L3μMg×L2=ML212α
Solving above equations, we get
a=FMμg and α=4FML6μgL
Initial acceleration of point A is,
aA=aLα2=2μgFM

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