A yo-yo is a physics lab that you can fit in your pocket. If a yo-yo rolls down its string for a distance h, it loses potential energy in amount mgh but gains kinetic energy in both translational 
and rotational 
forms. As it climbs back up, it loses kinetic energy and regains potential energy.
In a modern yo-yo, the string is not tied to the axle but is looped around it. When the yo-yo “hits” the bottom of its string, an upward force on the axle from the string stops the descent. The yo-yo then spins, axle inside loop, with only rotational kinetic energy. The yo-yo keeps spinning (“sleeping”) until you “wake it” by jerking on the string, causing the string to catch on the axle and the yo-yo to climb back up. The rotational kinetic energy of the yo-yo at the bottom of its string (and thus the sleeping time) can be considerably increased by throwing the yo-yo downward so that it starts down the string with initial speeds vcom and ω instead of rolling down from rest.
To find an expression for the linear acceleration acom of a yo-yo rolling down a string, we could use Newton’s second law just as we did for the body rolling down a ramp in Fig. 11-8. The analysis is the same except for the following:
1. Instead of rolling down a ramp at angle θ with the horizontal, the yo-yo rolls down a string at angle θ = 90° with the horizontal.
2. Instead of rolling on its outer surface at radius R, the yo-yo rolls on an axle of radius R0 (Fig. 11-9a).
3. Instead of being slowed by frictional force 
, the yo-yo is slowed by the force 
on it from the string (Fig. 11-9b).
Fig. 11-9 (a) A yo-yo, shown in cross section. The string, of assumed negligible thickness, is wound around an axle of radius R0. (b) A free-body diagram for the falling yo-yo. Only the axle is shown.
The analysis would again lead us to Eq. 11-10. Therefore, let us just change the notation in Eq. 11-10 and set θ = 90° to write the linear acceleration as
where Icom is the yo-yo’s rotational inertia about its center and M is its mass. A yo-yo has the same downward acceleration when it is climbing back up the string, because the forces on it are still those shown in Fig. 11-9b.
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