In class I did the yoyo calculation choosing the torque point at the spool center of mass. One can easily choose the contact point of the string with the spool. In fact, if one remembers to use the parallel axis theorem, it is even easier.

QUESTION: An uniform cylinder of radius R and mass M is wrapped with a massless, inflexible string (this means the string doesn't stretch but is able to bend to any radius of curvature without requiring work) about its diameter. What is the magnitude and direction of the linear acceleration.

1.

Since no forces act in the horizontal direction the motion of the yoyo must be perfectly vertical. In the ``real'' world no string has these perfect features and so the actual motion will be measurably different. Still this represents a simple limiting case.

2.

If one chooses contact point of the string as the torque axis then we get

$$\sum \tau = I \alpha = mg R$$

with

$$I = I_{cm}+ MR^2 ~~~~~~ I_{cm} = \frac{1}{2} MR^2$$

and substituting yields

$$\frac{3}{2} MR^2 \alpha= Mg R$$

or

$$\alpha R = \frac{2}{3} Mg = a_{cm} ~~.$$