Question
A bicycle has wheels of radius 0.3 m. Each wheel has a rotational inertia of 0.088 kg× m2 about its axle. The total mass of the bicycle including the wheels and the rider is 79 kg. When coasting at constant speed, what fraction of the total kinetic energy of the bicycle (including rider) is the rotational kinetic energy of the wheels?
Answers
The total kinetic energy is the sum of
Translational KE = (1/2)MV^2 = 39.5 V^2, and
Rotational KE = 2*(1/2)*I*w^2
= I*V^2/R^2 = 0.978 V^2
I is the moment of inertia of a single wheel.
The fraction of the energy that is rotational is
0.978V^2/(0.978+39.5)V^2 = 0.0243
Note that the V cancels out.
Translational KE = (1/2)MV^2 = 39.5 V^2, and
Rotational KE = 2*(1/2)*I*w^2
= I*V^2/R^2 = 0.978 V^2
I is the moment of inertia of a single wheel.
The fraction of the energy that is rotational is
0.978V^2/(0.978+39.5)V^2 = 0.0243
Note that the V cancels out.
ans is right!