As shown in the figure rod of mass m and length ℓ is hinged at O at one end and free to rotate in vertical plane. A force F of magnitude mg is applied on another end of rod which always acts perpendic

Question

As shown in the figure rod of mass m and length $ℓ$ is hinged at O at one end and free to rotate in vertical plane. A force F of magnitude mg is applied on another end of rod which always acts perpendicular to the rod. Force starts acting at time t = 0 (as shown). Find vertical component of normal reaction acting on rod when rod becomes horizontal for 3rd time :- (given m = 2 kg, $ℓ$ = 1 m, g = 10 m/s $^{2}$ )

TestHub · Accepted Answer

At the horizontal position F is vertical (upward) and weight acts down

Torque about the hinge:

$m g ℓ - m g \frac{ℓ}{2} = \frac{m ℓ ^{2}}{3} α$

$α = \frac{3 g}{2 ℓ}$

Vertical motion of the centre of mass is tangential, so:

$R_{y} + F - m g = m \frac{ℓ}{2} α$

$R_{y} = m \frac{ℓ}{2} \times \frac{3 g}{2 ℓ} = \frac{3 m g}{4} = 15 N$

Answer: C

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