Why is static friction not considered in the work-energy theorem?

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Static friction does not perform work in the context of the work-energy theorem because the point of contact does not move, resulting in zero displacement (Fdx = 0). This characteristic allows for efficient rolling motion without energy loss, as seen in vehicles where tires maintain their rubber. While static friction can affect the motion of objects in contact, it does not contribute to net work done in a system. The discussion highlights that while individual perspectives on work done by friction may vary, the overall conclusion remains that static friction does not result in net work. Understanding this principle is crucial for applying the conservation of energy in mechanics.
phantomvommand
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Homework Statement
A ball rolls down a sphere (without slipping). At what height will the ball lose contact with the sphere?
Ball radius = r, sphere radius = R, ball mass = m, ball I = 2/5mr^2
Relevant Equations
Conservation of energy:
mg(R+r)(1-cos theta) = 1/2mv^2 + 1/2 I w^2.
My question is this:
- Friction exists (for no slipping/pure rolling to occur)
- Why is the work done against friction not accounted for in the conservation of energy equation?

Thank you!
 
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phantomvommand said:
My question is this:
- Friction exists (for no slipping/pure rolling to occur)
- Why is the work done against friction not accounted for in the conservation of energy equation?
For no slipping, you have static friction, which does no work.
 
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PeroK said:
For no slipping, you have static friction, which does no work.
Just to confirm, you mean that it does no work as the point of contact is not moving (ie Fdx = 0, as dx = 0?)
 
phantomvommand said:
Just to confirm, you mean that it does no work as the point of contact is not moving (ie Fdx = 0, as dx = 0?)
Yes, that's why rolling is so efficient. And why you don't lose rubber off your car tyres when drving normally.
 
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phantomvommand said:
Just to confirm, you mean that it does no work as the point of contact is not moving (ie Fdx = 0, as dx = 0?)
It depends how you are defining dx. If as the relative motion of the surfaces then yes, dx=0.
But consider a block resting on an accelerating block, no sliding.
As far as the top block is concerned friction is doing work on it. If that block advances distance x then the work done is Ffx. The lower block takes the opposite view, i.e. the work done by the friction is -Ffx.
In short, no net work is done by static friction.
 
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Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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