    Question

# A uniform cylinder of mass M and radius R is placed on a rough horizontal board, which in turn is placed on a smooth surface. The coefficient of friction between the board and the cylinder is μ. If the board starts accelerating with constant acceleration a, as shown in the figure, then A
For pure rolling motion of the cylinder, the direction of frictional force is forward and its magnitude is Ma3
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B
The maximum value of a, so that the cylinder performs pure rolling is 3μg
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C
The acceleration of the center of mass of the cylinder under pure rolling condition for the given a is 2a3
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D
None of the above
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Solution

## The correct options are A For pure rolling motion of the cylinder, the direction of frictional force is forward and its magnitude is Ma3 B The maximum value of a, so that the cylinder performs pure rolling is 3μg C The acceleration of the center of mass of the cylinder under pure rolling condition for the given a is 2a3From the frame of reference, attached to the horizontal board, there is a pseudo force Ma on the cylinder in the backward direction.So to prevent slipping, the frictional force acts in the forward direction.Let this force be f.So the linear acceleration of the cylinder is, A=(Ma−f)/M in the backward direction.Torque due to this frictional force is fR and corresponding angular acceleration is given as, α=fR/I=fR/0.5MR2=2f/MRFor pure rolling, A=αR⟹(Ma−f)/M=2f/M⟹f=Ma/3Now the maximum value of frictional force is, μMgSo the maximum value of a is given by, μMg=Mamax/3⟹amax=3μg, for pure rolling to happen.Now acceleration of the CoM is, A=(Ma−f)/M=2a/3  Suggest Corrections  0      Related Videos   Rolling
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