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I have a question regarding wheels.

Suppose I have a skateboard with perfect bearings, and I live in a world without air. At $t=0$, it is on the ground and I give it some initial speed, $\vec{v}_0$. Perfect contact of the wheels on the ground is assumed. Since there is no friction (perfect bearings, no air so no fluid friction due to it, perfect contact between the wheels and the ground...), the skateboard will keep rolling forever at constant speed.

Now, notice that on each of the wheels, the ground exerts a contact force which has a tangent component to the wheel. This force must thus have some torque with respect to the center of the wheel. How comes it does not provide any work then ? I mean, it should modify the rotational kinetic energy of the wheels, $\frac{1}{2} I \omega^2$ ?

(To argue that no work is provided, my point is that both $\vec{g}$ and the normal part of the ground response force are orthogonal to the movement of the skateboard, so they cannot provide work).

I initially assumed that the problem is that my tangent force is never applied to the same physical point of the wheel, and that since the time during which it is applied to a given point is infinitesimal, so is the associated work. But I have another thought experiment invalidating this : say now that I flip the skateboard upside down, so that its wheels are in the air. Using the flat of my hands, I ``brush'' one of the wheels, so as to get it to rotate. I just gave it energy (since it now rotates, while before it was overall motionless), using the exact same kind of force as the previously mentioned tangent action.

Can anyone please help me get a clear inside on that? I looked everywhere on the internet, but couldn't find any answer that helped me.

PS: Do not hesitate to ask me to clarify the question, English is not my mother tongue nor the tongue in which I studied all these, so my vocabulary could be flawed!

Qmechanic
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Kal8578
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2 Answers2

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This friction does do work on the wheel. It does negative work on the skateboard as one whole object, which is why translational acceleration is smaller when an object rolls down a ramp than when it slides without friction. The friction also does positive rotational work on the wheels. Some of what would be translational kinetic energy of the skateboard is transferred into the rotation of the wheels. There is no standard formula for this force, as it is static friction. The wheels do not slide along the ground. The friction force will be just big enough to keep the wheels turning fast enough to roll rather than slide. If the forward push on the skateboard, or any rolling object, gets too large, then the needed friction will be larger than what the static friction coefficient allows. In such a case, the wheels will slide until the kinetic friction can bring the wheels up to speed.

Ken Mellendorf
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  • I don't understand why this force would do negative work on the skateboard as a whole. In my first thought experiment, the skateboard will roll forever at constant speed, so its total kinetic energy (translational of everything + rotational of the wheels) must remain constant. Yet, negative work would decrease it ? – Kal8578 Feb 21 '23 at 14:12
  • You are right that the total work done is zero. If there is no outside force, if the wheels and skateboard are already completely up to speed, then the wheels do not need any torque to keep them turning. As static friction is only as large as it has to be, the friction drops to zero when the wheels no longer need torque. If anything tries to speed up the skateboard, then the friction builds to the correct value to keep the wheels spinning in time with the motion of the skateboard. – Ken Mellendorf Feb 22 '23 at 19:09
  • But how can it be that "the total work done is zero" and yet "It does negative work on the skateboard as one whole object" ? What do you call "total work", if it's not the work on the skateboard as one whole object ? – Kal8578 Feb 23 '23 at 21:09
  • Translational work affects the motion of the center of mass of the skateboard (including the wheels). Rotational work affects the motion of each wheel around its individual center of mass but does affect the center of mass of the entire skateboard. All work combined, translational and rotational, is what I refer to as total work. It is a transfer of energy with no change to the total kinetic energy (translational plus rotational) of the object. – Ken Mellendorf Feb 23 '23 at 21:44
  • Ok, so "negative work one the skateboard as a whole object" referred to translational KE work ? – Kal8578 Feb 24 '23 at 12:14
  • You are correct.. – Ken Mellendorf Feb 27 '23 at 18:10
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The force of friction is part of the total external force; the total external force changes the translational kinetic energy (KE) of the center of mass (CM) of the body. With respect to the CM, the force of friction is part of the total external torque; the total external torque changes the rotational KE of the body with respect to the CM.

For rolling without slipping, the work done by friction for translational motion plus the work done by friction for rotational motion is zero. For slipping, this is not true.

Most physics mechanics texts assume a rigid body for which there is no "heating" (internal energy is constant). In this case there is no heating even if there is slipping.

See Is work done by torque due to friction in pure rolling? on this exchange.

John Darby
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  • Thanks for the link and the answer, I think I got it. Just to make sure, let me rephrase something: when there is friciton, as compared to when there is not, part of the total kinetic energy isn't given to translational kinetic energy but to rotational kinetic energy. This is also why if my skateboard had its wheels locked on their axes (no rotation) and perfectly slipping on the ground, it would go faster than it does now. Am I correct ? – Kal8578 Feb 21 '23 at 14:37
  • Yes, you have the idea now. – John Darby Feb 27 '23 at 19:03
  • Perfect, thanks ! – Kal8578 Mar 03 '23 at 12:23