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A uniform thin rod of mass $M$ and length $L$ stands vertically on a frictionless horizontal surface, initially along the $y$-axis with its lower end at the origin. A tiny disturbance causes it to begin toppling. If $R$ is the normal reaction force from the surface on the rod, what is the acceleration vector of the rod's centre of mass as it falls?

1. $\vec a_{cm}=\dfrac{Mg+R}{M}$ (upward)
2. $\vec a_{cm}=\dfrac{Mg-R}{M}$ (downward)
3. $\vec a_{cm}=Mg-R$ (resultant vector)
4. None of these

Correct answer: $\vec a_{cm}=\dfrac{Mg-R}{M}$ (downward)

Solution

For the rod as a whole, the centre-of-mass acceleration is determined by the net external force. Since weight $Mg$ acts downward and the normal reaction $R$ acts upward, the net downward force is $Mg-R$, so $a_{cm}=(Mg-R)/M$ downward.

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