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
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
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?