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Question

A block of mass m moves on a horizontal circle against the wall of a cylindrical room of radius R. The floor of the room on which the block moves is smooth but the friction coefficient between the wall and the block is μ. The block is given an initial speed v0. As a function of the speed v write (a) the normal force by the wall on the block, (b) the frictional force by wall and (c) the tangential acceleration of the block. (d) Integrate the tangential acceleration dvdt=vdvds to obtain the speed of the block after one revolution.

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Solution




Given:
Radius of the room = R
Mass of the block = m
(a) Normal reaction by the wall on the block = N = mv2R
(b) Force of frictional by the wall = μN=μmv2R
(c) Let at be the tangential acceleration of the block.
From figure, we get:
-μmv2R=matat=-μv2R
(d)
On using a=dvdt=vdvds, we get:vdvds=μv2 Rds=-RμdvvIntegrating both side, we get:s=-RμInv+cAt, s=0, v=v0So, c=RμInv0s=-RμInvv0 vv0=e-μsRv=v0e-μsRFor one rotation, we have:s=2πr v=v0e-2πμ

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