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Suppose you let a solid roll down an incline without slipping, from height $h$. My textbook gives the following conservation of energy relation $$mgh = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\omega^2.$$

Why do we not have to include the work done by the static friction (nonconservative force) on the left side? I know it is supposed to do zero work, as there is no motion where it acts, but it is the only force providing a torque and $\omega$ is obviously increasing, so in my opinion it should be doing (rotational) work.

Qmechanic
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  • Static friction supplies the necessary torque to produce the rotational KE, i.e., it is r responsible for the second term on the right. The equation you are using is a conservation of energy equation. The equations breaking down the forces responsible for the work can be found in the accepted answer that BioPhysicist gave in the link he provided. – Bob D May 10 '22 at 20:31
  • I think that answer deals exactly with that issue.https://physics.stackexchange.com/a/707449/195949 – Claudio Saspinski May 10 '22 at 20:47

2 Answers2

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The solid is assumed to be a rigid body. Friction causes rotation and does do rotational work with respect to the center of mass. But, for no slipping of a rigid body, the net work from friction is zero because the decrease in translational kinetic energy of the center of mass due to friction is exactly matched by the increase in rotational energy with respect to the center of mass due to friction. Said another way, the net work from friction is zero because the point where friction acts is instantaneously at rest in the inertial frame of reference. For a detailed discussion of both of these reasons see Consistent Approach for Calculating Work By Friction for Rigid Body in Planar Motion and Is work done by torque due to friction in pure rolling?. An answer by @Dale in the second reference provides a very simple way to determine whether or not friction does net work; this is a much clearer answer than many confusing answers given elsewhere.

With slipping, the work done by friction is not zero, it is negative; the negative increase in translational kinetic energy is greater in magnitude than the positive increase in rotational energy. Said another way, the point where friction acts is not instantaneously at rest in the inertial frame of reference. And, in the limit with no rotation (a box just sliding) the net work from friction is its most negative.

Note: the total work on the body is that from friction and gravity. For a rigid body there is no increase in the internal energy of the body (no "heating"). (In reality, no body is truly rigid, so heating cannot be ignored.)

John Darby
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  • the decrease in translational kinetic energy of the center of mass due to friction How can translational kinetic energy decrease if friction does no work though? Or, is friction doing negative translational work, but positive rotational work? If that's the case, how can the former be possible (again, there is no displacement at the point where friction acts)? –  May 11 '22 at 11:56
  • Approach (1) Kinetic energy = translational KE of CM + rotational KE about CM. Yes: friction does negative translational work on CM, but positive rotational work with respect to CM; with respect to the moving CM where there is displacement at point of friction. (CM is special; with respect to CM change in angular momentum = external torque even if CM accelerating, as here.) Approach (2) Viewed another way in the lab frame there is no displacement at point of friction if no slip. Same result using two approaches. (1) addresses your question: torque does work but net work is 0 for no slip. – John Darby May 11 '22 at 13:27
  • To be clear. (1) With respect to CM (accelerating frame) point of friction is rotating. (2) With respect to lab (inertial) frame point of friction is instantaneously at rest if no slip. With slip, with respect to lab frame point of friction is not instantaneously at rest – John Darby May 11 '22 at 13:31
  • For (2), does friction to translational work too then? –  May 11 '22 at 15:55
  • @Wernher In (2), the inertial lab frame, with no slip friction does no translational work and no rotational work since the point where friction acts is instantaneously stationary. (1) uses the fact that the motion can also be viewed as translation of the CM plus rotation about the moving CM. – John Darby May 11 '22 at 16:37
  • If no rotational work is done by the friction, which force provides the torque which increases the angular velocity of the body rolling down the incline? –  May 11 '22 at 19:02
  • @Wernher With respect to the moving CM friction provides torque. With respect to the instantaneous stationary point of contact, the component of gravity down the incline provides torque. Friction does no net work, gravity does net work. – John Darby May 12 '22 at 09:05
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You're correct it does zero work because before the motion and after the motion the incline is in the same location. There can be no energy extracted from it. The only energy available is the gravitational position energy from the height of the ball.

The rotational energy added to the solid comes at the expense of the kinetic energy it would otherwise have. On a frictionless plane, the solid would have no rotational energy at the bottom, but more kinetic energy. The frictional forces in this problem can affect the energy balance but not the energy total.

BowlOfRed
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