JEE Advanced level
A JEE Advanced-level example on Rotational Motion and Centre of Mass
Question. A uniform solid cylinder of mass M = 4 kg and radius R = 0.2 m rests on a rough horizontal surface. A light inextensible string is wound around the cylinder and pulled horizontally with force F = 24 N, with the string leaving tangentially from the top of the cylinder. The cylinder rolls without slipping. Find (a) the acceleration of the centre of mass, and (b) the magnitude and direction of the friction force on the cylinder.
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- Set up: Let acceleration of centre = a (in direction of F). For pure rolling, angular acceleration α = a/R. Let friction = f acting forward (same direction as motion, which can happen when force is applied above centre).
- Translation equation: F + f = Ma gives 24 + f = 4a ... (1)
- Rotation about centre: Torque by F (at top) = FR clockwise. Torque by f (at bottom) = fR anticlockwise. Net torque = FR − fR = Iα = (1/2)MR²(a/R) = (1/2)MRa. So F − f = (1/2)Ma gives 24 − f = 2a ... (2)
- Solve (1) and (2): Adding gives 48 = 6a, so a = 8 m/s². From (2): f = 24 − 16 = 8 N.
- Friction is positive, confirming it acts forward (in the direction of motion). This happens because the applied force at the top creates excess clockwise torque, and friction must act forward to maintain pure rolling.
Answer: (a) Acceleration = 8 m/s². (b) Friction = 8 N acting forward (in the direction of motion of the cylinder).
This example was solved twice independently and both solutions agree.