Question

A thin metal rod of length $0.6m$ is in vertical straight position on horizontal floor. This rod is falling freely to a side without slipping. The angular velocity of rod when its top end touches floor, is :

• A. $7rad{s}^{-1}$
• B. $4.2rad{s}^{-1}$
• C. $3.5rad{s}^{-1}$
• D. $2.1rad{s}^{-1}$

Answer 1

Answer: (A)

$7rad{s}^{-1}$

Initially $w_i=0$ so, rotational $K.E_i=0$

$P.E_i=mg\left(\frac{L}{2}\right)$

Let angular velocity of the rod when it touches the ground be $w$.

Thus rotational $K.E_f=\frac{1}{2}I{w}^2=\frac{1}{2}.\frac{1}{3}m{L}^2\left(w^2\right)$

Using conservation of energy, $P.E_i=K.E_f$

$⟹w^2=\frac{3g}{L}=\frac{3\left(9.8\right)}{.6}=49$

$w=7rad/s$