(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A Yo-Yo of massMhas an axle of radiusband a spool of radiusR. Its moment of inertia can be taken to beMR. The Yo-Yo is placed upright on a table and the string is pulled with a horizontal force F as shown. The coefficient of friction between the Yo-Yo and the table is^{2}/2mu.

What is the maximum value ofFfor which the Yo-Yo will roll without slipping? (See thumbnail below)

2. Relevant equations

[tex]F=Ma[/tex]

[tex]\tau=I\alpha[/tex]

3. The attempt at a solution

I have no idea. Neverhteless I can show the equations. Firstly I have the Newtons second law [tex]F-\mu Mg=Ma,[/tex] where [tex]F\ge \mu Mg,[/tex] so that it does not move backwards with a forward force. Secondly I have the equation [tex]\mu MgR-Fb=I\alpha,[/tex] where [tex]Fb\ge \mu MgR,[/tex] so that it does not roll forward with a backward pull.

Now using the moment of inertia and [tex]a=\alpha R,[/tex] I have [tex]\mu Mgr-Fb=I\alpha=\frac{RMa}{2}[/tex] and putting the equations together [tex]F-\mu Mg=2 \mu Mg-\frac{2Fb}{R}.[/tex]

But this is a contradiction because in the constraint that i used the right hand side of this equation is never negative whereas the right hand side is never positive. The only solution is that both are zero. So the condition that the Yo-Yo should not slide is [tex]F=\mu Mg.[/tex] But this contradicts my second constraint that [tex]Fb\ge \mu MgR,[/tex] becauseb<R. The solution must then be that the pull should not exceed the friction force, which is variable. But this is not correct. So where does my thinking go bananas?

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# Homework Help: Yo-Yo on a table contradiction

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