Problem :
A popular yo-yo trick is to have the yo-yo "climb" the string. A yo-yo
with mass .5 kg and moment of inertia of .01 begins by rotating at an
angular velocity of 10 rad/s. It then climbs the string until the
rotation of the yo-yo stops completely. How high does the yo-yo get?
We solve this problem using conservation of energy. Initially the yo-
yo has purely rotational kinetic energy, as it rotates in place at the
bottom of the string. As it climbs the string some of this rotational
kinetic energy is converted to translational kinetic energy, as well as
gravitational potential energy. Finally, when the yo-yo reaches the
top of its climb, the rotation and translation stops, and all of the
initial energy is converted to gravitational potential energy. We may
assume the system conserves energy, and equate
initial and final energy, and solve for h:
| Ef | = | Eo |
|
| mgh | = |  Iσ2 |
|
| h | = |  |
|
| | = |  |
|
| | = | .102 meters |
|
Problem :
A ball with moment of inertia of 1.6, mass of 4 kg, and radius of 1 m
rolls without slipping down an incline which is 10 meters high. What
is the speed of the ball when it reaches the bottom of the incline?
Again, we use conservation of energy to solve this problem of combined
rotational and translational motion. Fortunately, since the ball rolls
without slipping, we can express the kinetic energy in terms of only
one variable, v, and solve for v. If the ball did not roll without
slipping, we would also have to solve for σ, which would imply
that the problem would not have a solution. Initially, the ball is at
rest and all energy is stored in
gravitational potential energy. When the ball reaches the bottom of
the incline, all the potential energy is converted to both rotational
and translational kinetic energy. Thus, like any conservation problem,
we equate initial and final energies:
| Ef | = | Eo |
|
Mv2 + I | = | mgh |
|
(4)v2 + (1.6) | = | (4g)(10) |
|
| 2v2 + .8v2 | = | 40g |
|
| v | = |  = 11.8 m/s |
|
