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NCERT Exemplar · Q28

Q.A uniform disc of mass mm and radius RR stands vertically on its rim on a horizontal table; the coefficient of friction between the disc and the table is μ\mu. A horizontal force F⃗\vec{F} is applied at the centre (the axle) of the disc, in the plane of the disc. Find the maximum value of FF for which the disc rolls without slipping.

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The applied force at the centre exerts no torque, so friction alone must spin the disc up to roll. Newton's second law for translation and rotation, combined with the rolling condition and I=12mR2I=\tfrac12 mR^2, gives the needed friction f=F/3f=F/3. Since static friction cannot exceed μN=μmg\mu N=\mu mg, rolling without slipping survives only up to F=3μmgF=3\mu mg.

Set-up

Let the disc have mass mm, radius RR, and moment of inertia about its centre I=12mR2I = \tfrac{1}{2}mR^2. The horizontal force FF acts at the centre; friction ff acts at the contact point at the bottom. The normal reaction is NN.

Equations of motion

  1. Vertical balance: N=mgN = mg (the force FF is horizontal).
  2. Translation: the net horizontal force accelerates the centre,

F−f=ma.F - f = m a.

  1. Rotation about the centre: FF passes through the centre, so only friction gives a torque,

fR=Iα=12mR2 α.fR = I\alpha = \frac{1}{2}mR^2\,\alpha.

  1. Rolling condition: a=αRa = \alpha R.

Solve

From (3) and (4): fR=12mR2⋅aR=12mRafR = \tfrac12 mR^2\cdot\dfrac{a}{R} = \tfrac12 mRa, so

f=12ma.f = \frac{1}{2}m a. …

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