A uniform rod of mass `m` and length `l` is in equilibrium under the action of constraint forces, gravity and tension in the string. Find the

a. frictional force acting on the rod.
b. tension in the string.
c. normal reaction on the rod.
Now, the string is cut. Find the
d. angular acceleratiion of the rod just after the string is cut.
e. normal reaction on the rod just after the string is cut.

Explanation

a. Resolving the forces parallel and perpendicular to rod.
Along the length of rod.

`f=mgsin60^@=(sqrt(3mg))/2`
Hence friction force actionon rod at support `P` will be `mg//2`.
b. for finding tension we can take torque about support `P`.
`T(3/4l)=mg(l/4)` or `T=(mg)/3`
c. Just before cutting the string the rod is at equilibrium. considering the forces perpendicular to rod length.
`N+T=mg cos 60^@impliesN=(mg)/2-(mg)/3=(mg)/6`
d. Just after cutting the string equilibrium of the rod will disturbed. the rod will have angular acceleration.
As the rod does not slip point `P` will be rest at the time just after cuttingn the string.
We can apply torque equation about `P:tau_(P)=I_(P)alpha`
`(mgcos60^@).l/4=((ml^(2))/12+m(l/3)^(2))alphaimpliesa=(6g)/(7l)`
e. If we apply torque equation about centre of mass
`N. l/4=((ml^(2))/12)alpha`
`N.l/4=((ml^(2))/12)((6g)/(7l))impliesN=2/7mg`