Torque: Yoyo with Two Strings - Direction and Maximum Tension

  • #1
capn awesome
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Homework Statement



So, basic yoyo with center of mass O, moment of inertia Io , and mass m. Has an inner radius of r, and outer of R, and in contact with the ground at point p. Two strings are attached at inner radius r, each pulling with equal tensions, one pulling from the top of r directly to the right, one pulling from the bottom of r directly to the left.

a) in what direction does the yoyo move, assuming no slipping

b) what is the maximum tension T that can be applied before the yoyo begins to slip

Homework Equations



Ʃτ = Iα
a = Rα
I = Icm + mR^2

The Attempt at a Solution



Okay, so I started out by applying the parallel axis theorem to analyze torque around point p.

Ip = Io + mR2

Then summed the torques

Ʃτ = T(R-r) + T(R+r) ± Ffr = (Io+ mR2

α = 2TR ± Ffr/(Io+ mR2)

I am thinking that the yoyo will move clockwise, as the tension applied to the top of inner radius r has a larger numerical value when analyzing torque about p.

I have no idea (if this is correct) how I could find the max tension without slip.

Is it the moment when total torque from the strings overtakes the torque caused by the frictional force?

Thanks
 
Last edited:
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  • #2
Oh, just realized that friction doesn't matter if considering torque around p, as the lever arm = 0...

One other thing that confuses me is what happens if you consider torque around the C.M. Both strings are trying to create a clockwise spin, and friction opposes that meaning that it would act to make the yoyo spin counter clockwise, and then analyzing the torques separately, the bottom string would want to make the yoyo move counter clockwise to the left, and the top string does the opposite.

Would this mean that because there are two forces acting to move it c.c.w, that it would move c.c.w?
 

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