# Work done by friction in pure rolling

I just want a clarity about work done by friction in pure rolling when external force is acting. So suppose we have a ring and an external force is applied at the centre, the ground has friction. After some time pure rolling starts such that $$a = r* alpha$$. After this is work done by friction still zero. Because although the point of contact is at rest but rotational speed is increasing due to angular acceleration so there should be a rotational work done by friction? Please if anyone can clarify

• Oct 17, 2023 at 4:10
• Because pure rolling happens when point of contact is not getting displaced with respect to ground if body is pure rolling on ground which means displacement of point of application of force is zero and hence work done by friction is zero Oct 18, 2023 at 13:04

If you want to write work in two parts, $$W_{tot} = \int \vec{F}\cdot d\vec{s}_{trans} + \int \vec{\tau}\cdot d\theta$$ Then yes, the part due to torque will be non zero. However, first term will be negative to make sure that the total work is zero.
To evaluate the first term, recall that we only have to consider translational motion. Accounting for translational motion of the ball and assuming constant friction force, $$dW_{trans} = - f_{fric} ds$$ Where ds is small translational displacement of the ball. Now, for the rotational work, $$dW_{rot} = f_{fric} r d\theta$$ But remember that for pure rolling, $$ds=rd\theta$$. So, $$dW_{rot}= f_{fric} ds$$ Thus, $$dW_{tot} = dW_{trans} + dW_{rot} = 0$$