Kinetic energy physics problem?

Question

A bicycle has wheels of radius 0.24 m. Each wheel has a rotational inertia of 0.087 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'm just really confused about this question

any help would be greatly appreciated

Answer 1

Rotational kinetic energy = 0.5 I ω^2.
For two wheels of same I, total energy Er = I ω^2

Translational kinetic energy = 0.5 m v^2
Since v = r ω
Translational kinetic energy = 0.5 m r^2 ω ^2

Total of all energies:
ET =0.5 m r^2 ω ^2 + I ω^2
= [0.5 m r^2 + I] ω^2

Er / ET = I / [0.5 m r^2 + I]
= 0.087 / [0.5 *74 *0.24^2 + 0.087]
=0.80 = 80%

Answer 2

Use energy conservation. The potential energy of the limb due to gravity is: mgh (mass*9.8*height). This will turn into kinetic energy as the limb falls. Kinetic energy is given by: 1/2 mv^2 (mass times velocity squared, divided by 2). equate these two expressions and solve algebraically for the variable you need: mgh=1/2(mv^2) For this problem, remember that you want kinetic energy at h=6m, so your h is actually (h-6). Good luck!

Answer 3

Pick a speed....... Any speed.
Calculate the angular velocity of one wheel.
Calculate the rotational kinetic of the wheel.
Double it (for both wheels)
Calculate the kinetic energy of the bike and rider (including the mass of the wheels)
Add the rotational kinetic energy of the wheels to the kinetic energy of the bike and rider. This is the total kinetic energy of the system.
Divide the rotational kinetic energy of the wheels by the total kinetic energy and multiply by 100 to get the percentage.

Doug

Answer 4

I wish I could answer your question, but you have my sympathy because I'm now also doing a physics lab and I'm having a lot of trouble to .................