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Question

A small collar of mass m is given an initial velocity of magnitude vo on the horizontal circular track fabricated from a slender rod. If the coefficient of kinetic friction is μk, determine the distance travelled before the collar comes to rest. (Recognize that the frictional force depends on the net normal force).


A
rμkln⎢ ⎢v20+v40+r2g2rg⎥ ⎥
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B
r4μkln⎢ ⎢v20+v40+r2g2rg⎥ ⎥
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C
2rμkln⎢ ⎢v20+v40+r2g2rg⎥ ⎥
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D
r2μkln⎢ ⎢v20+v40+r2g2rg⎥ ⎥
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Solution

The correct option is B r2μkln⎢ ⎢v20+v40+r2g2rg⎥ ⎥
FBD of collar:


where mv2r= centripetal force (acting towards centre),
v= velocity of particle at any instant}
Nx=mv2r and Ny=mg
So, N=N2x+N2y=(mg)2+(mv2r)2
N=mg2+v4r2 ... (1)

Here, frictional force is acting opposite to the direction of motion. Hence, f=ma
f=μkN (μk= kinetic friction coefficient)
So, μkN=ma
a=μkNm ... (2)
From eq. (1)
a=μkmm.g2+v4r2 [a=dvdt]
dvdx.dxdt=μkg2+v4r2 [v=dxdt]
v.dvdx=μkg2+v4r2

Let v2r=k
By differentiation, 2vr.dv=dk
or v.dv=rdk2

So, r.dk2dx=μkg2+k2
Integrating both sides,
r.dk2g2+k2=μkdx
r2[ln[k+k2+g2]]=μk[x]
(k=v2r)
r2lnv2r+(v2r)2+g20vo=μk[x]s0
[ initial velocity is v0 and final velocity =0]

r2ln(g2)lnv20r+ (v20r)2+g2=μks
s=r2μk⎢ ⎢lnv20+v40+g2r2rg⎥ ⎥

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