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A uniform thin rod of mass $M$ and length $L$ is standing vertically along the $y$-axis on a smooth horizontal surface, with its lower end at the origin $(0,0)$. A slight disturbance at $t=0$ causes the lower end to slip on the smooth surface along the positive $x$-axis, and the rod starts falling. The acceleration vector of the centre of mass of the rod during its fall is: [R is the reaction from the surface]

1. $\vec a_{CM}=\dfrac{Mg+R}{M}$
2. $\vec a_{CM}=\dfrac{Mg-R}{M}$
3. $\vec a_{CM}=\vec 0$
4. None of these

Correct answer: $\vec a_{CM}=\dfrac{Mg-R}{M}$

Solution

The acceleration of the centre of mass is determined by the net external force on the rod. Here the external forces are $Mg$ downward and reaction $R$ upward, so the net force is $Mg-R$ downward. Therefore, $\vec a_{CM}=\dfrac{Mg-R}{M}$ in the vertical direction.

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