A block of mass m slides on a frictionless table. It is constrained to move inside a ring of radius R. At time #t=0#, the block is moving along the inside of the ring with velocity #v_o#. The coefficient of friction between the block and the ring #mu#?

Find the speed of the block at time #t#

1 Answer
Dec 29, 2017

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The above figure presents the situation of sliding of a block of mass m on a frictionless table constrained to move inside a ring of radius R as described in the problem.

Let #v# represents the tangential velocity of the block at an instant of time. At that instant the centrifugal reactionary force on the block is #color(blue)((mv^2)/R)#. So the force of retardation due to friction will be #color(red)((mumv^2)/R)#.

So by Newton's law we can write

#m(dv)/(dt)=-(mumv^2)/R#

#=>(dv)/v^2=-mu/Rdt#

Integrating for #t=0 to t=t# when velocity changes from #v_0to V#

#int_(v_0)^V(dv)/v^2=-mu/Rint_0^tdt#

#=>[v^(-2+1)/(-2+1)]_(v_0)^V=-mu/R[t]_0^t#

#=>[-1/v]_(v_0)^V=-mu/R[t]_0^t#

#=>-1/V+1/v_0=-mu/Rt#

#=>1/V=1/v_0+mu/Rt#

#=>1/V=(R+muv_0t)/(v_0R)#

#=>V=(v_0R)/(R+muv_0t)#

So speed of the block at time t

#color(blue)(V=v_0/(1+(muv_0t)/R))#