Question

A rod of mass M and length $L$ is standing on frictionless horizontal surface vertically. With slight disturbance it starts to fall. When rod is making angle of ${37}^{\circ }$ with vertical,

• A. Speed of center of mass is $\frac{gL}{3}$
• B. Angular velocity of rod is $\sqrt{\frac{15g}{13L}}$
• C. Velocity of center of mass is vertically downwards
• D. Speed of point of contact with ground is $\sqrt{\frac{12gL}{65}}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B Angular velocity of rod is $\sqrt{\frac{15g}{13L}}$
C Velocity of center of mass is vertically downwards
D Speed of point of contact with ground is $\sqrt{\frac{12gL}{65}}$


By conservation of energy,
$\Rightarrow mg\frac{L}{2}(1-\frac{4}{5})=\frac{1}{2}I{w}^2$

where $I=\frac{13}{75}m{L}^2$ is about $IAR$

$mg\frac{L}{10}=\frac{1}{2}.\frac{13}{75}m{L}^2{w}^2$

$\Rightarrow w=\sqrt{\frac{15g}{13L}}$

Since there is no horizontal force COM moves in vertical direction.

and $V_{com}=r_{com}\omega \Rightarrow V_{com}=\frac{L}{2}\cos {53}^{\circ }\omega =\frac{3}{10}\sqrt{\frac{15gL}{13}}$

Velocity of point of contact is $V=\frac{L}{2}\sin {53}^{\circ }\omega$

$V=\sqrt{\frac{12gL}{65}}$