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So if I have an object that is rolling with slipping like a bowling ball, if I use W=F•dr definition, then work done by friction has to be negative. But if I use the W=τ•dθ definition, then work is positive because friction is causing the rolling motion. So does friction do positive or negative work? And how do I find the total distance or theta that friction works over?

More details on system: the original question is if I have an object on an inclined plane, plot the end kinetic energy as a function of theta of the plane if the objects starts a certain height above the ground each time it is rolled

See: https://www.aapt.org/physicsteam/2015/upload/exam1-2015-1-8.pdf number 22

(I was also hoping just for a more generalized answer of W=F•dr vs. W=τ•d if possible)

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closed as unclear what you're asking by Aaron Stevens, Buzz, John Rennie, Jon Custer, user191954 Feb 5 at 8:35

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Student, I have worked this exact problem in great detail, and saved it in a Word document. The document is long. If you want a copy, email me at davidwhite1506@att.net. $\endgroup$ – David White Jan 19 at 17:16
  • $\begingroup$ I think more detail of the system is needed here. Friction can do many things in a problem with rolling objects. $\endgroup$ – Aaron Stevens Jan 19 at 17:45
  • $\begingroup$ Also keep in mind that $\text d W=\tau\cdot\text d\theta$ is not a definition of work. Rather, it is the application of the first definition you give in certain situations. $\endgroup$ – Aaron Stevens Jan 19 at 17:48
  • $\begingroup$ @DavidWhite surely you can provide a digest... $\endgroup$ – ZeroTheHero Jan 19 at 18:43
  • $\begingroup$ @ZeroTheHero, if I could cut and paste from Word into this forum, I would. However, the memo is 5 pages long, with a LOT of derivations in it, and I don't intend to go through the effort to retype such a memo with the amount of text and Latex equations that would be required. $\endgroup$ – David White Jan 19 at 19:54
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You are correct that if the bowling ball is skidding (slipping) down the lane, negative work is done because the friction force is in the opposite direction of the motion of the ball. The dot product between the friction force and displacement is -1.

If the ball is rolling without skidding (slipping) the friction force is static friction. Static friction allows the ball to rotate without slipping. Just like static friction between your shoes and the ground allows you to wall without slipping. But no dissipative (heat producing) work is involved with static friction.

Hope this helps.

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  • $\begingroup$ Static friction can do work on objects that are rolling. The work done by friction goes into changing the rotation of the object. I think you mean to say no energy is lost due to static friction $\endgroup$ – Aaron Stevens Jan 19 at 17:40
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    $\begingroup$ @AaronStevens Thanks Aaron. That is what I meant. I meant no dissipative work. Will revise. $\endgroup$ – Bob D Jan 19 at 18:48
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The power of the actions of a system of force on a solid is $P=\overrightarrow{R}\centerdot \overrightarrow{{{v}_{I}}}+\overrightarrow{{{M}_{I}}}\centerdot \overrightarrow{\omega }$ with $I$ a point of the solid.

If $I$ is the point of contact with the ground, $\overrightarrow{{{M}_{I}}}=\overrightarrow{0}$ and your result is correct.

On the other hand, if $I$ is the center of the wheel, you have to sum the two terms.

More precisely :

If $I$ is the center of mass of the wheel, we have $\overrightarrow{R}=T\overrightarrow{{{e}_{x}}}+N\overrightarrow{{{e}_{y}}}$ and $\overrightarrow{\omega }=\omega \overrightarrow{{{e}_{z}}}$. The power of the actions of contact is $P=T{{v}_{G}}-{{M}_{G}}\omega $. The moment in G is ${{M}_{G}}=-R\omega $.

So the power is $P=T({{v}_{G}}+R\omega )=T{{v}_{g}}<0$ with $({{v}_{G}}+R\omega )={{v}_{g}}$ the sliding speed.

In all cases, the power of the friction actions is negative: $P=T{{v}_{G}}-{{M}_{G}}\omega <0$ But the sign of each term depends on the initial situation.

If you put on the ground a wheel that turns with the center of gravity at rest. The friction will move the wheel forward and slow down the rotation. The reaction of the soil will be forward: $T>0$ We will have $T{{v}_{G}}>0$ and $-{{M}_{G}}\omega <0$

If you put on the ground a wheel that does not turn with the center of gravity thrown forward. The friction will slow down the wheel and accelerate the rotational movement. The reaction of the soil will be backward: $T<0$. We will have $T{{v}_{G}}<0$ and $-{{M}_{G}}\omega >0$

In any case, the slip will stop when the sliding speed is zero $({{v}_{G}}+R\omega )={{v}_{g}}=0$.

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