First slide

Rotational motion

Question

A horizontal heavy uniform bar of weight W is supported at its ends by two men. At the instant, one of the men lets go off his end of the rod, the other feels the force on his hand changed to

Difficult

A

W
B

$\frac{W}{2}$
C

$\frac{3 W}{4}$
D

$\frac{W}{4}$

Solution

Let the mass of the rod is $M \Rightarrow Weight (W) = M g$
Initially for the equilibrium $F + F = M g \Rightarrow F = M g / 2$
When one man withdraws, the torque on the rod $τ = I α = M g \frac{l}{2}$
$\Rightarrow \frac{M l^{2}}{3} α = M g \frac{l}{2} [As I = M l^{2} / 3]$
$\Rightarrow Angular acceleration,$ $α = \frac{3}{2} \frac{g}{l}$
and linear acceleration,
$a = \frac{l}{2} α = \frac{3 g}{4}$
Now if the new normal force at A is $F^{'} then M g - F^{'} = M a$
$\Rightarrow F^{'} = M g - M a = M g - \frac{3 M g}{4} = \frac{M g}{4} = \frac{W}{4}$

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