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Question

A uniform disc of radius R is spinned to the angular velocity ω and then carefully placed on a horizontal surface. How long will the disc be rotating on the surface if the friction coefficient is equal to k ? The pressure exerted by the disc on the surface can be regarded as uniform.

A
t=5Rω04kg
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B
t=3Rω04kg
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C
t=3Rω08kg
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D
None of these
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Solution

The correct option is B t=3Rω04kg
It is the moment of friction force which brings the disc to rest. The force of friction is applied to each section of the disc, and since these sections lie at different distance from the axis, the moments of the forces of friction differ from section to section.
To find Nz, where z is the axis of rotation of the disc let us partition the disc into thin rings (figure shown below). The force of friction acting on the considered element dfr=k(2πrdrσ)g, (where σ is the density of the disc)
The moment of this force of friction is dNz=rdfr=2πkσgr2dr
Integrating w.r.t. r from zero to R, we get
Nz=2πkσgR0r2dr=23πkσgR3.
For the rotation of the disc about the stationary axis z, from the equation Nz=Iβz
23πkσgR3=(πR2σ)R22βz or βz=4kg3R
Thus from the angular kinematical equation
ωz=ω0z+βzt
0=ω0+(4kg3R)t or t=3Rω04kg
229168_141511_ans_ee7dc7521727443584dd4f793ea71cfa.png

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