A uniform rod of mas m and length l is kept vertical with the lower end clamed. It is slightly pushed to let it fall down under gravity. Find its angular speed when the rod is passing through its lowest positon. Neglect any friction at the clamp. What will be the linear speed of the free end at this instant?

Explanation

As the rod reaches its lowest position , the centre of mass is lowered by the distannce l . Its gravitational potential energy is decreased by mgl . As no energy . As the rotation occurs about the horizontal axis through the clamped ennd , the moment of inertia is ` I = ml^(2)//3`

`1/2 I omega^(2) = mgl`
`1/2 ((ml^(2))/3)omega^(2) = mgl or omega = sqrt((6g)/l)`
The linear speed of the free end is `v = l omega = sqrt(6gl)`
` [ 2m(a)^(2) +m(2a)^(2) +8m xx (6a)^(2) //12 ] omega `