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in Laws of motion by (15.2k points)
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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 bead is µ, and gravity is neglected, then the time after which the bead starts slipping is

(a) \(\sqrt{\frac{\mu}{\alpha}}\)

(b) \(\frac{\mu}{\sqrt{a}}\)

(c) \(\frac{1}{\sqrt{\mu\, \alpha}}\)

(d) Infinitesimal

2 Answers

+1 vote
by (39.0k points)
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Best answer

Correct option is (a) \(\sqrt {\frac \mu \alpha}\)

Tangential force (F1) of the bead will be given by the normal reaction (N), while centripetal force (Fc) is provided by friction (fr). The bead starts sliding when the centripetal force is just equal to the limiting friction.

Therefore, Ft = ma = mαL = N

∴ Limiting value of friction

(fr)max = µN = µmαL   …(i)

Angular velocity at time t is ω = at

∴ Centripetal force at time t will be

Fc = mLw2 = mLa2 t2  …(ii)

Equating equation (i) and (ii), we get t = \(\sqrt {\frac \mu \alpha}\)

For t > \(\sqrt {\frac \mu \alpha}\), Fc > (fr)max i.e. , the bead starts sliding.

In the figure Ft is perpendicular to the paper inwards.

+1 vote
by (15.7k points)

Correct answer is (a)\(\sqrt{\frac{\mu}{\alpha}}\)

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