So the formula for rotation work is

$$\mathrm dW=T\cdot\mathrm d\theta$$ where T is torque.

While solving a question like in case of a body rolling down an incline (pure rolling), we usully equate the change in kinetic energy equal to work done by gravity (just the translational work). Why do we not count the rotational work done when gravity does provide a torque?

In almost all question of mixed motions (translational and rotational) I have come across, none of them have the application of rotational work done, so I would also like to know when do we apply this concept.

  • $\begingroup$ It is not gravity that provides the torque. It's the frictional force as the body slides that provides the torque. $\endgroup$
    – Karthik
    Jan 28, 2019 at 19:04
  • $\begingroup$ How can you say that gravity doesn't provide a torque? It is wrong to say so as according to my understanding torque is dependent upon the point about which it is calculated so what you say might not be true @Karthik.V $\endgroup$ Jan 28, 2019 at 19:12
  • $\begingroup$ Only the frictional force acts along the base of the body. That contributes to torque. While gravity helps in accelerating the body down to the ground (translation). I think this is the idea behind (By the way, I am talking about the torque with respect to the centre of mass). $\endgroup$
    – Karthik
    Jan 28, 2019 at 19:18
  • 1
    $\begingroup$ Possible duplicate of Torque from Gravity $\endgroup$
    – Karthik
    Jan 28, 2019 at 19:21
  • $\begingroup$ @KarthikV I think the question is more about taking into account of the "rotational work" rather than why gravity does not supply it. I think the OP is just mistaken about that one point, and I have clarified this in my answer. $\endgroup$ Jan 28, 2019 at 20:02

1 Answer 1


First, for something like a ball rolling down an incline, gravity has no torque about the center of the ball. The force that causes the ball to start rolling is friction. On a frictionless incline the ball would just slide down the incline without rolling.

Now, with that out of the way, it turns out that we do take into account "rotational work" due to friction. Let's assume a constant friction force $f$ and say that the ball with radius $r$ is released from rest on the incline. We know that the net torque about the center of the ball is given by $$\tau = fr$$ using Newton's second law we also have $$\tau=I\alpha$$ where $I$ is the moment of inertia and $\alpha$ is the angular acceleration.

Now, since our torque is constant (since $f$ and $r$ are both constant) we know two other things. First, the work done by friction is given by $$W=\tau\Delta\theta=I\alpha\Delta\theta$$ and second, we can apply our constant acceleration equations. This means that $$\omega^2=2\alpha\Delta\theta$$ where $\Delta\theta$ is the angle the ball rolls through some time after release, and $\omega$ is the angular velocity at that same point in time.

Combining everything we end up with $$W=\frac12I\omega^2$$

and this result might look familiar. It is what we usually associate "rotational kinetic energy" with. So we do take into account the "rotational work", we just do it implicitly with $\frac12I\omega^2$ rather than explicitly (this is similar to how we use potential energy to implicitly take into account the work done by conservative forces rather than explicitly calculating the work done by said forces).

For example, if the ball starts at a height $h$ above the ground on the incline, using energy conservation at the bottom of the incline we have $$mgh=\frac12mv^2+\frac12I\omega^2$$

  • $\begingroup$ What about when we conserve energy and put K+U = constant. I don't see rotational work up there.?? $\endgroup$ Jan 29, 2019 at 0:39
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    $\begingroup$ You should. Typically the kinetic energy gets broken up into translation and rotation: $K=\frac12mv^2+\frac12I\omega^2$ $\endgroup$ Jan 29, 2019 at 0:48
  • $\begingroup$ I am not asking about the rotational energy where is the rotational work done by f include in conservation statement. I just see the work done by gravity?? $\endgroup$ Jan 29, 2019 at 0:50
  • $\begingroup$ @dawoodmansoor It's in my answer... The work done by $f$ is taken into account by the $\frac12I\omega^2$ term $\endgroup$ Jan 29, 2019 at 0:51
  • $\begingroup$ What about the LHS should it not have work done by friction $\endgroup$ Jan 29, 2019 at 0:53

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