Question

A uniform thin rod of mass $m$ and length $l$ is standing on a smooth horizontal surface. A slight disturbance causes the lower end to slip on the smooth surface and the rod starts falling. Find the velocity of the centre of mass of the rod at the instant when it makes an angle $\theta$ with horizontal.

Answer 1

Given: A uniform thin rod of mass m and length is standing on a smooth horizontal surface. A slight disturbance causes the lower end to slip on the smooth surface and the rod starts falling.

To find the velocity of centre of mass of the rod at the instant when it makes angle $\theta$ with the horizontal.

Solution:

Since There is No Horizontal Force on Rod So Centre of mass of Rod will fall vertically Downward only (Does not have any Horizontal Velocity) So moment of inertia of this Rod is Intersection Point of perpendicular Lines drawn to velocity as Shown in figure !

And In ICR frame Rod is in Pure Rotational Motion So we can use Energy Conservation in ICR frame!

$\frac{1}{2}mgL(1-\sin \theta )=\frac{1}{2}{I}_{icr}{\omega }^2⟹\frac{1}{2}mgL(1-\sin \theta )=\frac{1}{2}m(\frac{L^2}{12}+\frac{L^2}{4}{\cos }^2\theta ){\omega }^2⟹{\omega }^2=\frac{gL(1-\sin \theta )}{\frac{L^2+3{\cos }^2\theta L^2}{12}}⟹{\omega }^2=\frac{12gL(1-\sin \theta )}{L^2(1+3{\cos }^2\theta )}⟹\omega =\sqrt{\frac{12g(1-\sin \theta )}{L(1+3{\cos }^2\theta )}}$

Hence the velocity of centre of mass of the rod is

$v=\omega \frac{L}{2}\cos \theta ⟹v=\sqrt{\frac{12g(1-\sin \theta )}{L(1+3{\cos }^2\theta )}}\times \frac{L}{2}\cos \theta ⟹v=\sqrt{\frac{12g(1-\sin \theta )\times L^2{\cos }^2\theta }{L(1+3{\cos }^2\theta )\times 4}}⟹v=\sqrt{\frac{3gL(1-\sin \theta ){\cos }^2\theta }{(1+3{\cos }^2\theta )}}$

is the velocity of centre of mass of the rod at the instant when it makes angle $\theta$ with the horizontal.