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A homogeneous solid cylindrical roller of radius R and mass M is pulled on a cricket pitch by a horizontal force F applied at its centre. Assuming rolling without slipping, the angular acceleration of the cylinder is

Asked in JEE Main 10th Jan 1st Shift 2019 · Rolling dynamics and slipping

Answer: (3) (2F)/(3MR)

Step-by-step solution

Newton: F-f=Ma; torque about the axis: fR=Iα=1/2 MR²/aR, so f=1/2 Ma.

F=3/2 Ma⇒ a=(2F)/(3M), and α=/aR=(2F)/(3MR).

Why the other options are wrong

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