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There is no change in rotational or linear velocity during the process.

Can you explain the work done by the person rolling the ball, in the two cases, ideally. I would prefer an explanation without using conservation of energy theorem, but that is also fine. I am getting confused in both my approaches.

ACB
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Holy cow
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2 Answers2

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A ball on a hill is pulled on by gravity, making a force (parallel to the ground, pointing down the hill) of $F=mg \sin(\theta)$. Acting at its center-of-gravity.

To roll a ball up a hill requires a force of half that much, if applied at the top of the ball, half because of leverage if you push on the top. The distance we need to apply this force is twice the length of the hill. Work done is $$Fd= (\tfrac{1}{2}mg \sin(\theta))~(2L)= mgL \sin(\theta)$$ The height of the hill is $$h= L \sin(\theta) \implies \text{Work}=mgh$$

So we expend work energy of $mgh$ and increase the ball’s height potential energy by $mgh$. But the leverage let us use half the force over twice the distance. Assumes no friction. Wasn’t sure exactly what you were looking for maybe that helps.

Al Brown
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    Good answer Al. – joseph h Sep 02 '21 at 10:16
  • I'm not sure about this solution, for two reasons: (1) If there is no friction and the external force applied is $\frac 1 2 mg \sin \theta$ up the slope then the net force on the ball is $\frac 1 2 mg \sin \theta$ down the slope so how does it move up the slope ? (2) how do you conclude that the force is applied for "twice the length of the hill" ? – gandalf61 Sep 02 '21 at 10:42
  • When pushing on the top of a ball you have leverage. The instantaneous motion if a ball is rotation about the point where it touches the ground. The point where it touches the ground is the only part not moving. So the gravity at center is at a distance $r$ from there $\tau_g = mg \sin ~r$ and the pushing on top is at a distance $d=2r$ from that point, $~,~F=0.5 mg\sin~$ $~,~ \tau= 0.5 mg\sin ~d ~~= \tau_g$ – Al Brown Sep 02 '21 at 10:52
  • @AlBrown Mechanical advantage from a lever arm requires a fixed pivot. Without friction you have no pivot. Without friction, applying a force $mg \sin \theta$ at the top of the ball spins it in place and applying a smaller force at the top of the ball spins it while it slides down the slope. – gandalf61 Sep 02 '21 at 11:10
  • I see what you mean. I should have clarified. By “no friction” I actually just meant no rolling friction and no energy loss due to friction. I did not mean “absence of the static friction necessary to allow rolling to occur”. We have an instantaneous fixed pivot. – Al Brown Sep 02 '21 at 11:21
  • @gandalf61 You may already be familiar with rolling friction vs static vs kinetic. If not this answer i made reviews briefly: https://physics.stackexchange.com/a/661328/307354 – Al Brown Sep 02 '21 at 11:23
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A person rolling a ball up hill does positive work to increase the potential energy and overcome friction. The force is in the direction of motion. To keep the ball moving at a constant speed coming down the hill, he exerts a force opposite to the direction of motion and does negative work. He may have help from friction in preventing an increase in kinetic energy.

R.W. Bird
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