Is work done in rolling friction? I am confused by rolling friction.
Suppose you have a cylinder rolling which starts at rest at the top of an incline plane and begins to roll down the plane without slipping. Is work done by the incline on the cylinder? 
I know from doing some problems that the total kinetic energy (translational and rotational) is $mgh$, which is true only if the only work done on the cylinder is by gravity. But also, the cylinder must have nonzero angular acceleration, so friction must be exerting torque on the cylinder, so work must be done by friction. One of these statements is wrong.
 A: In the case of rolling down the slope without slipping, static friction force holds the contact point between the incline and the cylinder. It does rotational work. We can view this work as a medium between gravitational potential energy and rotational kinetic energy. So when it accelerates down the incline, friction force takes some of the gravitational potential energy and turns it into rotational kinetic energy for the cylinder. And in the end, the amount of gravitational potential energy decrease is equal to the amount of translational kinetic energy plus rotational kinetic energy increase.
A: Work is force times distance. If there is no slip, the force of friction acts over a distance of 0. There is no work. 
Gravity does work. As the cylinder rolls down the hill, it accelerates. It gains kinetic energy in two forms: translation and rotation. 
Gravity would do the same work on an identical cylinder that slide down the same slope without friction. The kinetic energy of the two would be the same at each position. 
The rolling cylinder would travel more slowly than the sliding cylinder. But it would also spin. 
A: The work is done by gravity. Friction simply holds the instantaneous contact point stationary to the ramp, so it doesn't do any work. The rotation is around the contact point, and is cause by gravity acting through the centre of mass of the roller.
A: If you take the perspective that "work is the increase in total mechanical energy of a system", then there really no work being done. The potential energy is just being converted to kinetic energy (of some sort). It's not really correct to say that " total kinetic energy (translational and rotational) is mgh". It would be more correct to say that any increase in kinetic energy:
$$E[total] = E[translationa]+E[rotational] = \frac{1}{2}mv^2+\frac{1}{2} I \omega^2 $$
is associated with an equal decrease in $$mgh =E[potential]$$. 
So it's the exchange of potential energy for kinetic energy that is doing hte work. The incline is just a constraint that limits the path over which this exchange takes place in position. Friction is actually causing an increase in heat energy. It is a "dissapative" exchange of potential energy for thermal energy.
A: Here it is more appropriate to say that the net work done by friction is zero, because friction here support rotation and at the same time opposes translation motion.so amount of work done by friction in rotation is cancelled out by work done in translation.So the loss of potential energy is equal to the Gain in total kinetic energy.and here friction acting at lowest point is static.
A: Pure rolling friction does no work because there is no relative motion between the rolling object and the surface it rolls on.  Sliding friction does work due to relative motion.
This is similar to fluid flowing in a pipe assuming the fluid sticks to the pipe walls; the pipe does no work on the fluid since there is no relative motion of the fluid relative to the pipe surface.
The constraint from rolling friction does provide a torque that causes rotation of the body as it moves.  Force and Torque, yes: work, no.
