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 $\mu$. A force $F(>\mu mg)$ 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. $\mu g-\frac{F}{M}$
• B. $\frac{F}{M}$
• C. $4\mu g$
• D. $2\mu g-\frac{F}{M}$

Answer 1

The correct option is D $2\mu g-\frac{F}{M}$

As $F>\mu Mg$, so the friction is kinetic in nature and FBD of rod would be like as shown in figure.

Using Newtons second law,
$F-\mu Mg=Ma$
Torque equation about centre of rod
$F\times \frac{L}{3}-\mu Mg\times \frac{L}{2}=\frac{M{L}^2}{12}\alpha$
Solving above equations, we get
$a=\frac{F}{M}-\mu g$ and $\alpha =\frac{4F}{ML}-\frac{6\mu g}{L}$
Initial acceleration of point $A$ is,
$a_A=a-\frac{L\alpha }{2}=2\mu g-\frac{F}{M}$