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I have four doubts regarding friction during rolling.

  1. Does slipping mean zero angular velocity, or is it just the $v$ velocity not being equal to $\omega r$?

  2. If a wheel is initially given the perfect velocity $v=\omega r$ (the wheel is on an smooth incline), will it continue rolling or will it start slipping at some point?

  3. Same as 2 but the incline has friction.

  4. Can I use conservation of angular momentum when friction is present?

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  • $\begingroup$ slipping mean $\omega\,r-v >0$ $\endgroup$
    – Eli
    Commented Apr 5, 2021 at 6:44
  • $\begingroup$ Slipping does not mean zero angular velocity as the wheel is still rotating. It means that the speed of the rim, $\omega r$ is greater than the translational velocity $v$. $\endgroup$
    – Mozibur Ullah
    Commented Apr 5, 2021 at 10:32
  • $\begingroup$ This earlier answer should help you. physics.stackexchange.com/questions/607942/… $\endgroup$
    – John Darby
    Commented Apr 5, 2021 at 14:52
  • $\begingroup$ @JohnDarby I have seen them , Howe I couldn’t understand $\endgroup$
    – user287374
    Commented Apr 6, 2021 at 5:32

3 Answers 3

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To point 1

Does slipping mean zero angular velocity or is it just the v velocity not being equal to wr?

The geometric bond $v=\omega r$ tells the speed $v$ of a particle a distance $r$ from the wheel centre. Within the wheel, this equation must hold true for every particle - otherwise the wheel would break apart because particles would "skew" relative to each other.

A particle on the rim, the periphery, must also obey this equation. At the contact point with the ground, the ground particle should follow this rim particle and should thus also follow this equation in order to not "skew" and move differently relative to the rim particle. If the ground particle does not fullfil this equation, then you have sliding.

The speed of such ground particle (the speed of the ground) is relative to the centre of the wheel (rotational centre or centre-of-mass). This is equivalent to the speed of the centre-of-mass (so the linear speed of the wheel as a whole) relative to the ground.

So, sliding and slipping for a wheel means that the the centre-of-mass' speed, $v\neq \omega r$ is not fulfilled. But $\omega$ does not have to be zero - the wheel can rotate just fine regardless of whether it matches the motion relative to the grond (think a spinning ball on ice).

To points 2 and 3

Without friction there is no (tangential) tie between wheel and ground. There is thus no reason to expect the equation - the geometric bond - to be fulfilled.

  • So, when rolling up an incline, the wheel will ideally continue spinning even as it slows down linearly - and it will continue spinning as before even while speeding up downwards.

  • With friction we would expect the ground to "grap onto" the rim and keep the geometric bond fulfilled, as long as the friction is strong enough.

To point 4

Can I use conservation of angular momentum when friction is present?

The angular momentum conservation law always holds true. Whether you can use it only depends on whether your system is not too complicated to calculate for. Whith friction slowing down the wheel rotation, angular momentum conservation tells us that the Earth obtains some oppositely directed angular momentum. This is obviously easily lost as insignificant, and so this law might rarely be useful in such case.

But it depends on what you need to find with this law. Maybe friction isn't relevant for a particular scenario. So this question is too broad to be answered generally.

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Here are answers to your questions in order:

  1. Slipping means $v \neq w r $.

  2. As soon as it accelerates, $v > w r $ and so it will not roll past the starting position, as $w$ will remain the same.

  3. If there's enough friction to keep it from slipping, yes.

  4. It is not conserved relative to the center of mass of the wheel if there is friction. You can use Newton's $2^{\text{nd}}$ law in its rotational form.

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  • $\begingroup$ can u elaborate the 4 th point , wht is the rotational form of the 2nd law $\endgroup$
    – user287374
    Commented Apr 5, 2021 at 10:42
  • $\begingroup$ The rate of change of the angular momentum of the wheel (relative to its CM) is equal to the torque (again, rel to CM). $\endgroup$
    – Mohammad Alhejji
    Commented Apr 5, 2021 at 10:57
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Here are the answer to your doubts in order:-

  1. Slipping means there is relative motion between two surfaces. Only when $v=r\omega$, there is no relative motion between the surface of the rolling object and the surface on which it is rolling and hence no slipping. In other cases, when $v\not=r\omega$, there is relative motion between the surfaces and hence slipping as well.

  2. If the wheel starts rolling on the smooth incline, the wheel will be accelerated linearly due to the force of gravity on it without its angular velocity being affected (as there is no friction to exert the required torque). As a result, after the starting point, $v\not=r\omega$ and the wheel will stop rolling and start slipping.

  3. If there is SUFFICIENT friction on the incline, the friction force will be able to exert the required torque and the wheel will continue rolling on the incline.

  4. If some net external force other than friction happens to act on the rolling body such that the external force vector is at some perpendicular distance from the point at which the friction is acting as in the case of rolling on incline,enter image description here

Then in that case, we can't apply conservation of angular momentum. However, if no external force other than friction acts or if the net external force other than friction acts in such a manner that it is directed towards the point at which the friction acts (perpendicular distance of the force from the point of application of friction=0) like in the figure,

enter image description here

Then we can apply conservation of angular momentum about the point of application of the friction (since torque due to friction and the external force about the point will be zero).

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