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A long horizontal rod has a bead which can slide along its length and initially placed at a distance 2L from one end A of the rod.
The rod is set in angular motion about A with constant angular acceleration alpha.
If the coefficient of friciton b/w the rod and the bead is mew,and gravity is neglected ,then the time after which the bead starts slipping is

dev kothari , 12 Years ago

Grade 12

1 Answers

shubham sharma

t=underroot(mew/alpha)

See, the bead will start sliding when centripetalforce will be equal to limiting friction. Now there will be two forces on bead one tangential F(t)=ma=2lalpha.m (a=R.alpha).

And second F(c)=2ml(omega)''2 then using omega=alpha.t

Equate F(c)=mew.N where N is provided by F(t). S it becomes

2mL(alpha)''2(t)''2=2mewL(alpha)m

so for t>underroot (mew/alpha) bead starts sliding..

Last Activity: 12 Years ago

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Mechanics> newtons-law...

dev kothari , 12 Years ago

shubham sharma

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