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Mechanics

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Joined:	9 Oct 2007
Post:	73

21 Oct 2007 18:18:53 IST

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Plz help yaar/.......HC VERMA 28 CIRCULAR MOTION

None

a table with smooth horizontal surface fixed in cabin rotates with angular velocity

in a circular path of radius R.A smooth groove AB of length L(<<R) the groove makes an angle teeta with radius OA of circle in which cabin rotates.A small particle is kept at point A in the grrovve and released to move along AB .Fin d the time it takes to reach B.......

PLZ help..............

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Comments (2)

Rahul Sharma

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Joined: 29 Aug 2007

Posts: 222

21 Oct 2007 19:21:13 IST

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Here, the particle will experience a force equal to m w^2 r in the right direction but since it can only move along the groove, only a component of m w^2 r will accelerate it and the other will not play any part in accelerating it along the groove. Also, L<
So, considering all forces on the particle,

Vertical N (from table) = mg

Horizontal : 1. N (from groove) = m w^2 r sin theta
2. m w^2 r cos theta = ma (as particle will
accelerate along the groove)

So, a = w^2 r cos theta

Using, s = ut + 1/2 at^2, where u=0 since particle is released from rest and s = L and a = w^2 r cos theta

So, you get the answer as

t = root (2L/ w^2 r cos theta).

Plz rate me for my efforts!

Niteesh Mehra

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Joined: 28 Oct 2007

Posts: 207

15 Aug 2008 16:22:47 IST

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thanks for the help..

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