A uniform 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 centre of mass of the rod when it makes an anglethetawith horizontal.

A uniform rod of mass m and length L is standing on a smooth horizonta

1.	A uniform 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 centre of mass of the rod when it makes an anglethetawith horizontal.
Answer» SOLUTION :As the surfac is smooth ,loss in gravitational potential ENERGY of the rod=gain in rotational kinetic energy about IC. `mgh=(I)/(2)Iomega^(2)`(I is MOMENT of INERTIA about IC) `h=(L)/(2)(1-sintheta)` and `I=((mL^(2))/(12)+(mL^(2))/(4)COS^(2)theta)` On substituting these values we get `omega=sqrt((12g(1-sintheta))/(L(1+3 cos^(2)theta)))` and `V_(cm=((L)/(2)costheta)omega`