# Friction on a flat rotating surface

• I
If I push an object such as a cylinder of wood along a flat table (flat face of cylinder in contact with the table) through it's center of mass, the friction or energy required is not dependent of the surface area the block makes with the table, Friction = μ N, correct? And the energy required = friction force x distance.

However, if I now rotate the cylinder about the center of mass (axis of rotation normal to the table) then does the friction (and energy required to rotate the block) depend on the surface area of the block in contact with the table (block weight and coefficient of friction are constant)?
Obviously the mathematics is going to be a lot more complicated for the rotating block but I suspect the diameter of the block will affect the friction and energy require to turn the block one revolution. Correct?

haruspex
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the mathematics is going to be a lot more complicated for the rotating block
If rotating on the spot it is easy. Consider a small area dA of a cylinder radius R, mass M. The normal force is proportional to dA, so the frictional force is too: μgM.dA/(πR2). All the frictional forces are tangential. The torque they exert varies as radius. So total torque is μgM∫r2.drdθ/(πR2) = μMg(2πR3/3)/(πR2) = 2μMgR/3.
Since the MoI is MR2/2, the angular acceleration is 4μg/(3R).

If rotating and sliding linearly it becomes extremely nasty. I believe it would not follow a straight line. Ever watched the Scots/Irish/Canadian/NZ sport of curling?

waverider
Great! thanks for the confirmation and equations.

Yeah, I have seen the sports curling...fun to watch, and I agree the math could get very nasty.

A.T.