Question

11.
A uniform metal rod of 2 kg is hinged at one end and is rotated by 90° anticlockwise as shown find
the workdone by gravity in this process is
final
Initial
(1) -10J
(2) +10J
(3) 10
(4) -20J​

Answer 1

Answer:

Work done by gravity=-10Joules

Explanation:

Given:

The rod will rotate about the end A.

Let a be the linear acceleration of COM of rod

$\alpha$ = angular acceleration.

$\begin{aligned}&\alpha=\frac{\tau}{\mathrm{I}}=\frac{\mathrm{mg} \frac{1}{2}}{\frac{\mathrm{ml}^{2}}{3}}=\frac{3 \mathrm{~g}}{2 \mathrm{l}} \\&\mathrm{a}=\frac{1}{2} \alpha=\frac{1}{2} \frac{3 \mathrm{~g}}{2 \mathrm{l}}=\frac{3 \mathrm{~g}}{4}\end{aligned}$

Torque= $\tau}$

From force equation:

$\begin{aligned}&m g-F=m a \\&F=m g-m \frac{3 g}{4}=m \frac{g}{4}=\frac{20}{4}=5Kgms^{-2} \end{aligned}$

workdone=  $\tau}\theta$

                   = $2*10*\frac{1}{2} =10J\\\theta is in anticlockwise so \\workdone=-10J$