- #1
kmarinas86
- 979
- 1
Say I have three wheels. Their characteristics are the following:
WHEEL ONE
20 kg mass
10 cm radius
Accelerated from 0 to 10 m/s in one second
WHEEL TWO
10 kg mass
20 cm radius
Accelerated from 0 to 10 m/s in one second
WHEEL THREE
10 kg mass
10 cm radius
Accelerated from 0 to 20 m/s in one second
We assume all wheels are close enough to uniform mass density and that their moment of inertia is calculated according to (mass*radius^2), the case for a cylinder. The values above were specifically chosen so that the final angular momentum for all three wheels are the same. We now determine the rotational kinetic energy for all three wheels.
WHEEL ONE
20 kg mass
10 cm radius
Accelerated from 0 to 10 m/s in one second
Angular Momentum = 20 kg*m/s^2
Final Rotational Kinetic Energy = (1/2) * (20 kg * (10 cm)^2) * ((10 m/s)/(10 cm))^2 = 1000 Joules
WHEEL TWO
10 kg mass
20 cm radius
Accelerated from 0 to 10 m/s in one second
Angular Momentum = 20 kg*m/s^2
Final Rotational Kinetic Energy = (1/2) * (10 kg * (20 cm)^2) * ((10 m/s)/(20 cm))^2 = 500 Joules
WHEEL THREE
10 kg mass
10 cm radius
Accelerated from 0 to 20 m/s in one second
Angular Momentum = 20 kg*m/s^2
Final Rotational Kinetic Energy = (1/2) * (10 kg * (10 cm)^2) * ((20 m/s)/(10 cm))^2 = 2000 Joules
Consider that angular momentum was transferred from one wheel to the other. Apparently what I said suggests that the rotational kinetic energy does not follow a one-to-one relationship with angular momentum. What other forms of energy should be present in this scenario?
WHEEL ONE
20 kg mass
10 cm radius
Accelerated from 0 to 10 m/s in one second
WHEEL TWO
10 kg mass
20 cm radius
Accelerated from 0 to 10 m/s in one second
WHEEL THREE
10 kg mass
10 cm radius
Accelerated from 0 to 20 m/s in one second
We assume all wheels are close enough to uniform mass density and that their moment of inertia is calculated according to (mass*radius^2), the case for a cylinder. The values above were specifically chosen so that the final angular momentum for all three wheels are the same. We now determine the rotational kinetic energy for all three wheels.
WHEEL ONE
20 kg mass
10 cm radius
Accelerated from 0 to 10 m/s in one second
Angular Momentum = 20 kg*m/s^2
Final Rotational Kinetic Energy = (1/2) * (20 kg * (10 cm)^2) * ((10 m/s)/(10 cm))^2 = 1000 Joules
WHEEL TWO
10 kg mass
20 cm radius
Accelerated from 0 to 10 m/s in one second
Angular Momentum = 20 kg*m/s^2
Final Rotational Kinetic Energy = (1/2) * (10 kg * (20 cm)^2) * ((10 m/s)/(20 cm))^2 = 500 Joules
WHEEL THREE
10 kg mass
10 cm radius
Accelerated from 0 to 20 m/s in one second
Angular Momentum = 20 kg*m/s^2
Final Rotational Kinetic Energy = (1/2) * (10 kg * (10 cm)^2) * ((20 m/s)/(10 cm))^2 = 2000 Joules
Consider that angular momentum was transferred from one wheel to the other. Apparently what I said suggests that the rotational kinetic energy does not follow a one-to-one relationship with angular momentum. What other forms of energy should be present in this scenario?