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If a yo-yo's string is not attached to anything and the yo-yo is dropped, it is obvious it will fall at $g$. In this scenario, Tension of string = 0.

If a yo-yo's string is attached to a cieling and it is dropped, what is $R$ (path of center of mass)? Well, we know from experience that the yo-yo is going to fall at least, but is it going to fall at $g$? We know it is rotating as well. What would $T$ (tension of string) be? Would $T$ = yo-yo's mass * gravity?

Assume that there is no friction.

A related question is: if there is a stick in space and a force is applied to the center of mass, and for another stick in space an equivalent force is applied to the edge of the stick, will both sticks' center of mass move the same? Would the 2nd stick be rotating?


marked as duplicate by ja72, ACuriousMind, Ali, Danu, Kyle Kanos Sep 13 '14 at 18:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Applying Newton's 2nd law will get you part of the way in the yo-yo case. And it will get you all of the way in the stick case. $\endgroup$ – BMS Sep 11 '14 at 6:31
  • $\begingroup$ See physics.stackexchange.com/a/29160/392 $\endgroup$ – ja72 Sep 13 '14 at 16:08

For the yo-yo question:

what is R (path of center of mass)?

Think about how a yo-yo works - there's a string attached to a center axis and wound around itself. This means that your string attached to the ceiling will be applying tension T at some distance r away from the yo-yo's center of mass. But because the string is not fixed but wound around itself, what happens to the yo-yo?

Hint: your location of tension T changes (r decreases) as the yo-yo unwinds.

Well, we know from experience that the yo-yo is going to fall at least, but is it going to fall at g?

Draw yourself a torque/moment diagram with your forces applied at the correct location (tension T up from wherever the string attaches to the yo-yo in the wound state, and mg down from the center of mass). Where is your acceleration happening as a result of the tension T? Think of a ball rolling on the ground with an acceleration applied tangent to the top surface. How much does the COM accelerate?

Hint: if your acceleration is applied away from your COM, the COM acceleration will always be less.

What would T (tension of string) be? Would T = yo-yo's mass * gravity?

You can find this using a free body diagram and solving your force balance equation.


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