You can use energy conservation in rotation problems just like you did in the energy conservation lessons for linear motion. Note that rotating objects also have kinetic energy of rotation and this must be included when accounting for energy conservation.
∑Einitial=∑EfinalThe total energy does not change in closed systems
KErot=12Iω2
Example
A hoop of mass 1 kg and radius .75 m, begins at rest at the top of a ramp, 3m above the ground. What is it's rotational velocity at the bottom of the ramp if the hoop rolls down the ramp without slipping?
To solve this problem, we'll apply energy conservation to the hoop. The hoop only has gravitational potential energy at the top of the ramp, and it has both rotational and linear kinetic energy at the bottom of the ramp.
Ei=EfPE=KE+KErotstart with conservation of energymgh=12mv2+12Iω2substitute the proper value for each energy termmgh=12mv2+12mr2ω2substitute in the moment of inertia of the hoopmgh=12m(ωr)2+12mr2ω2we can put v in terms of omega because the hoop is rolling without sliding2gh=(ωr)2+r2ω2now we will multiply by 2 and divide by m to simplify the equation2gh=2ω2r2ω=√ghrsolving for ωω=√9.8m/s2∗3m.75mplug in the known valuesω=7.2rad/s
Interactive Simulation
Review
- Your bike brakes went out! You put your feet on the wheel to slow it down. The rotational kinetic energy of the wheel begins to decrease. Where is this energy going?
- The upper pulley has mass of 2kg and a radius of 30cm. The lower pulley has mass of 4kg and a radius of 1.0m. The 12 kg mass is 5m off the ground.
- With what velocity will the 12kg mass hit the ground if μk=0?
- With what velocity will the 12kg mass hit the ground if μk=0.2?
Review (Answers)
1. The energy goes into the heat generated by the friction between your foot and the tire.
2. a. 5.87m/s b. 4.79 m/s