Question

A uniform rod of length 2l is placed with one end in contach with the horizontal and is then inclined at an angle to the horizontal and allowed to fall without slipping at contact point when it becomes horizontal, its angular velocity will be

Answer 1

Answer:

Explanation:

From energy conservation, when it is inclined,

U.E = K.E

mg((lsinα)/2) = 0.5*I*w^2

mglsinα = Iw^2

where I is the moment of inertia about the centre of the rod.

I = m(2l)^2/12 => ml^2/3

mglsinα = ((ml^2)*w)/3

gsinα = lw/3

w = (3gsinα)/l rad/s