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I read the following scenario about a yo-yo:

enter image description here

The translational and rotational forms of Newton’s second law for this yo-yo is $Mg - T=Ma_y$ and $TR_0=I\alpha_z$, respectively.

It was also said that "We consider the string to be of a negligible thickness and assume that it does not slip as it is unwinding. Thus, the point where the string contacts the shaft is instantaneously at rest".

I have always thought that rolling without slipping happens due to the friction between the surface and the wheel (i.e. we must have friction to have rolling without slipping). However, it's clear that there isn't friction between the string and the yo-yo's shaft in this scenario. Thus, I am wondering if I was mistaken and that friction is not what results in rolling without slipping?

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  • $\begingroup$ How are you drawing the conclusion that there isn't friction between the string and the yo-yo's shaft? $\endgroup$
    – David White
    Commented Jul 17, 2021 at 23:25
  • $\begingroup$ @DavidWhite Hi! From the Newton’s second law statements that are given to describe this system, I don't think they included friction? $\endgroup$
    – Claire
    Commented Jul 17, 2021 at 23:49
  • $\begingroup$ This might help - Is work done in rolling friction? $\endgroup$
    – mmesser314
    Commented Jul 18, 2021 at 0:43
  • $\begingroup$ @Cheryl, the fact that the string is not slipping on the yo-yo axis means that there is enough static friction to keep the string from slipping. I think that you have mis-interpreted the problem statement. $\endgroup$
    – David White
    Commented Jul 18, 2021 at 2:12
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    $\begingroup$ Rolling without slipping could also happen for other reasons. For example, consider a wheel covered in tiny teeth that match the slope (a rack and pinion). $\endgroup$
    – Alex K
    Commented Mar 13, 2023 at 12:18

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You're partly right, party not.

When we think of common rolling/slipping/friction problems, the most common is a round object on an inclined slope. In that situation, if there is no friction, the object slides down the slope. Friction is needed to create the torque that leads to rolling - the force of gravity acts parallel to the slope through the centre of the object, and friction acts parallel to the slope in the opposite direction through the edge of the object. Hence torque, and rolling. If maximum friction is sufficient, the object will not slide, only roll.

But your yoyo is different. In a yoyo, the string is tied to the middle and wrapped round many times. If the string can't stretch (inelastic), then literally, sliding simply can't happen. The only way the yoyo can descend at all, is if it unwinds on the string. So it is completely independent of friction in that sense.

Alternatively even if it wasn't tied, we might think that the string windings create enough friction to stop the string sliding on the axis anyway. That doesn't have to be friction with the axis alone - if there is friction between string and string, then the string "locks" itself on the axis too. You can see this by wrapping string round a tree branch many times and trying to pull it free - the friction between string and branch, as well as string and overlaying string, locks it so it can't slip.

But now imagine a yoyo where the string wasn't tied,but was just wrapped many times. We could imagine a frictionless yoyo, or one that had almost unwound and the last part of the string was not secured by a knot. Then the string could slide around the yoyo, to release some length. That would alter the outcome.

So the question rules out that scenario by stating that at no.time, is there a situation where the string can slide around on the yoyos axis. Its either tied, or there is enough friction to prevent that. That's to ensure you solve the problem they intend, and practice that specific skill set, and don't complicate it.

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  • $\begingroup$ Hi! Thanks for the answer! Just to clarify a point - you said that "Friction is needed to create the torque that leads to rolling - the force of gravity acts parallel to the slope through the centre of the object, and friction acts parallel to the slope in the opposite direction through the edge of the object." As the rolling cylinder rotates around an axis through its CoM, I am wondering how would the force of gravity, which acts on the object's CoM, contribute to the rolling? $\endgroup$
    – Claire
    Commented Jul 20, 2021 at 14:59
  • $\begingroup$ In simple physics examples like these, a torque (rotational force) occurs when an object experiences a net force that doesn't act through its CoM. The classic case is two forces acting in opposite directions through different points on the object. If you imagine I (in front of you) pull your left shoulder, and Jane (behind you) pulls your right shoulder, if we pull with the same force you wouldn't move (zero net force) but you would turn (nonzero net torque). In this case, gravity acts vertically downwards through the CoM.... $\endgroup$
    – Stilez
    Commented Jul 20, 2021 at 18:09
  • $\begingroup$ We can resolve gravity into a force normal to the slope (which does nothing, you cant fly off the slope or get pulled into it!), and a force parallel to the slope down the slope. There is also a friction force which resists the sliding motion of gravity downslope... so friction is acting parallel to the slope and upwards along the slope. But the friction force occurs at the cylinder RIM, where it touches the slope, not through the CoM. So you have 2 opposite forces, acting through DIFFERENT points within the cylinder (axis/CoM and rim). So there's a net torque, and it begins to rotate. $\endgroup$
    – Stilez
    Commented Jul 20, 2021 at 18:12

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