## Monday, 15 September 2014

### ROTATIONAL KINETIC ENERGY PROBLEM AND SOLUTIONS

Problem 1 :
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:

 E f = E o mgh = Iσ 2 h = = = .102 meters

Problem 2 :
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 forσ , 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:

 E f = E o Mv 2 + I = mgh (4)v 2 + (1.6) = (4g)(10) 2v 2 + .8v 2 = 40g v = = 11.8 m/s