Question

A thin bar of mass $M$ and length $L$ is free to rotate about a fixed horizontal axis through a point at its end. The bar is brought to a horizontal position and then released. The axis is perpendicular to the rod. The angular velocity when it reaches the lowest point is

• A. directly proportional to its length and inversely proportional to its mass
• B. independent of mass and inversely proportional to the square root of its length
• C. dependent only upon the acceleration due to gravity and the length of the bar
• D. directly proportional to its length and inversely proportional to the acceleration due to gravity

Answer 1

Answer: (B), (C)

The correct options are
B independent of mass and inversely proportional to the square root of its length
C dependent only upon the acceleration due to gravity and the length of the bar

Centre of mass of the rods lie at C and C'

Let $w$ be the final angular velocity of the rod.

As $A{C}^{\prime }=\frac{L}{2}$, thus weight of the rod lowers by $\frac{L}{2}$.

Thus loss of gravitational P.E is $\frac{MgL}{2}$

Work-energy theorem, $\frac{MgL}{2}=\frac{1}{2}{I}_A{w}^2-0$

$\frac{MgL}{2}=\frac{1}{2}\times \frac{M{L}^2}{3}{w}^2$

$⟹w=\sqrt{\frac{3g}{L}}$