A yo-yo's inertia...?

Question

A yo-yo has a rotational inertia of 950 g cm^2 and a mass of 120 g. Its axle radius is 3.2 mm, and its string is 120 cm long. The yo-yo rolls from rest down to the end of the sting.

(a) What is the magnitude of its translational acceleration?

(b) How long does it take to reach the end of the string?

As it reaches the end of the string, what are its (c) translational speed, (d) translational kinetic energy, (e) rotational kinetic energy, and (f) rotational speed?

What equations did you use to solve this, and can you explain your answers?

Answer 1

If you can answer (a) the rest should be easy. You can define an "effective mass" Me as the sum of the mass and inertia/r^2. This is the mass that is accelerated by m*g. Then a=m*g/Me. This can be checked by computing the forces needed for the translational and rotational accelerations; their sum should equal m*g:
f1=m*a
ang.accel=a/r
torque=ang.accel*I
f2=torque/r
f=f1+f2
My result was I=0.000095 kg-m^2, Me=9.3973 kg, a=0.12523 m/s^2.
Checking, f1=0.015027, f2=1.1618, f1+f2=m*g=1.1768