Question

A long horizontal rod has a bead which can slide along its length and is initially placed at a distance L from one end A of the rod.The rod starts from rest in angular motion about A with a constant angular acceleration a .If the coefficientof friction between the rod and the bead is u and gravity is neglected then what is the time after which the bead starts slipping

Answer 1

Final Answer : t = √u/α

Assuming the given motion takes place in Horizontal plane.

Steps:
1) The bead will start sliding when centripetal force is equal to limiting friction.
i. e.
mw^2L = f(max)
=> mw^2 L= u N ----(1)

2) We know that,
Here. Normal Reaction is provided by Tangential force.
F(t) = mLα =N ---(2)

3 ) Substituting value of N in eq. (1),
Now,
At time t,.
w = 0 + αt
w = αt

Now,
mw^2L = umLα
=> m(αt) ^2 L = umLα
=> αt^2 = u
=> t =√(u/α) .

At time t , bead will start sliding.

Answer 2

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