Question

A rod has mass $M$ and length $l$. Equation of motion of the rod , which is hinged at one end, when displaced by small angular displacement $\theta$, can be expressed as (Assume air friction is zero, Moment of inertia of rod about axis passing through hinged point is $I$ and acceleration due to gravity is $g$)

• A. $I\frac{d\theta }{dt}=Mgl\theta$
• B. $I\frac{d^2\theta }{d{t}^2}=-Mg\frac{l}{2}\theta$
• C. $I\frac{d^2\theta }{d{t}^2}=-Mgl\theta$
• D. $I\frac{d^2\theta }{d{t}^2}=-\frac{Mg}{l}\theta$

Answer 1

Answer: (B)

$I\frac{d^2\theta }{d{t}^2}=-Mg\frac{l}{2}\theta$

Torque $\tau =I.\alpha$
$\alpha$ is the angular acceleration
$\alpha =\frac{d^2\theta }{d{t}^2}$
Torque is given by $=Mg\frac{l}{2}\theta$ (Force times length)
$I\frac{d^2\theta }{d{t}^2}=-Mg\frac{l}{2}\theta$