A long horizontal rod has a bead which can slide along its length and

1.	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 is set in angular motion about A with a constant angular acceleration α. If the coefficient of friction between the rod and the bead is μ and the gravity is neglected, the time after which the bead starts slipping is(a) √(μ/α)(b) μ/√α(c) 1/√(μα)(d) infinitesimal
Answer» Correct Answer is: (a) √(μ/α). The linear acceleration of the bead is a = Lα. ∴ the reaction force on the bead due to the rod is N = ma = mLα. After time t, the angular velocity of the bead is ω = αt . ∴ the centripetal acceleration of the bead is ω²L = α²t²L. ∴ the force of friction at limiting position is μN = μmLα. ∴ for slipping, μmLα = mα²t²L or t = √(μ/α)

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 is set in angular motion about A with a constant angular acceleration α. If the coefficient of friction between the rod and the bead is μ and the gravity is neglected, the time after which the bead starts slipping is(a) √(μ/α)(b) μ/√α(c) 1/√(μα)(d) infinitesimal

Answer»

Correct Answer is: (a) √(μ/α).

The linear acceleration of the bead is a = Lα.

∴ the reaction force on the bead due to the rod is N = ma = mLα.

After time t, the angular velocity of the bead is ω = αt .

∴ the centripetal acceleration of the bead is ω²L = α²t²L.

∴ the force of friction at limiting position is μN = μmLα.

∴ for slipping,

μmLα = mα²t²L

or t = √(μ/α)

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