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MHT CET 2024 (9 May, Shift 2) · Centre of Mass of Discrete Particles

Centre of Mass of Discrete Particles question from MHT CET 2024 (9 May, Shift 2)

Q9MHT CET 2024 · 9 MAY · SHIFT 2Medium

Centre of Mass of Discrete Particles

A metal rod of weight ' $W$ ' is supported by two parallel knife-edges A and B. The rod is in equilibrium in horizontal position. The distance ' between two knife-edges is ' $r$ '. The centre of mass of the rod is at a distance ' $x$ ' from $A$ . The normal reaction on A is

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Answer: (d)

Weight $W$ acts at the CM, distance $x$ from A and $(r - x)$ from B.
Torque about B is zero: $N_A \cdot r = W(r - x)$
$N_A = \frac{W(r - x)}{r}$
So the answer is (d) $\frac{W \cdot (r - x)}{r}$ .

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