Physics Help Please!?

Question

A bicycle has wheels of radius 0.29 m. Each wheel has a rotational inertia of 0.078 kg* m2 about its axle. The total mass of the bicycle including the wheels and the rider is 74 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?

I just don't know what they are asking for?

And thanks in advance! :)

Answer 1

What you will find is that the KE associated with the wheels turning is directly related to the bikes velocity, as is the KE of the bike and rider.

Kinetic Energy of Wheel
=1/2 * I *w^2
I = rotational inertia
w= angular velocity=radians per sec
w=bike velocity/radius
w=Vb/.29m
KE=1/2*0.078kg*m^2*(Vb/.29m)^2
Since there are two wheels, the KE of both wheels, combined, would be:
= 0.078kg*m^2 *(Vb)^2/(.29m)^2
=0.078kg*m^2 *(Vb)^2/.841m^2
=9.27kg *(Vb)^2

The KE of the Rider plus bike is
=1/2 m*Vb^2
=1/2*74kg*Vb^2
=37kg*Vb^2

Total KE = KE bike + KE wheels
=(37+9.27)kg *Vb^2
=46.27kg*Vb^2

Ratio KE Wheels/ Total KE

9.27kg*Vb^2/46.27kg*Vb^2

=19.8%

Note: In making the ratio, I assumed that the question meant that the total KE of the bicycle included the bike, the rider and the wheels. The question is actually ambiguous on this.