A uniform solid cylindrical roller of mass m is being pulled on a horizontal surface with a force F parallel to the surface and applied at its centre. If the acceleration of the cylinder is a and it is rolling without slipping, then the value of F is

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(3) 3/2ma. The force is applied at the centre, so it has no moment about the centre. Only friction can spin the roller up. Centre of mass: F-f=ma, with f the friction at the contact, acting backwards. Torque about the centre: fR=Iα=1/2mR²·a/R, so f=1/2ma. Substituting: F=ma+1/2ma=3/2ma. The same result comes straight from torques about the contact point: FR=(1/2mR²+mR²)a/R gives F=3/2ma. In general F=ma(1+I/(mR²)) for a force applied at the centre — the extra term is the price of spinning the body up.

A uniform solid cylindrical roller of mass m is being pulled on a horizontal surface with a force F parallel to the surface and applied at its centre. If the acceleration of the cylinder is a and it is rolling without slipping, then the value of F is

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