First slide

Torque

Question

A heavy uniform bar is carried by two men on their shoulders. The weight of the bar is W If one man lets it fall from the end carried try him, the weight experienced by the other at the instant the rod is falling is

Moderate

A

W
B

W/2
C

W/4
D

2W

Solution

See fig.
First of all we consider the acceleration of C.M. Here, the bar turns about A with an angular acceleration α after withdrawing the support of one man.
Hence $Iα = W (\frac{l}{2}) or α = \frac{Wl}{2 I}$
$For a bar, I = {Ml}^{2} / 3$
$∴ α = \frac{mgl \times 3}{2 \times {Ml}^{2}} = \frac{3 g}{2 l}$
The linear acceleration of C.M.
$a = \frac{l}{2} \times α = \frac{l}{2} \times \frac{3 g}{2 l} = \frac{3 g}{4}$
From figure, W - R = m a
$\begin{aligned} mg - R = M \times (3 g / 4) \\ R = \frac{mg}{4} = \frac{W}{4} \end{aligned}$

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