Please help - Rotational Kinetic Energy of Moving Wheel

AI Thread Summary
The discussion revolves around calculating the fraction of the total kinetic energy of a bicycle that is attributed to the rotational kinetic energy of its wheels. The bicycle has a total mass of 74 kg, with wheels having a radius of 0.3 m and a rotational inertia of 0.08 kg*m². Participants clarify that the total kinetic energy consists of both rotational and translational components, and emphasize that a theoretical velocity is unnecessary since it cancels out in the final calculations. The correct approach involves substituting angular velocity in terms of linear velocity and using known values for mass and inertia. The final step requires careful calculation to find the correct fraction of rotational kinetic energy relative to total kinetic energy.
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Homework Statement



A bicycle has wheels of radius 0.3 m. Each wheel has a rotational inertia of 0.08 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?

Homework Equations



K=1/2 I w^2
I = m r^2 <--- (for a hoop)

The Attempt at a Solution



I don't even know where to start...
Should I use the radius and the mass for the Inertia equation? Should I divide the mass by two since there are two wheels??
 
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Total KE in this case is 2KEr + KEt, where KEr is rotational kinetic energy and KEt is translational/linear kinetic energy. Because the bike itself as a whole is moving linearly, it has KEt. And since there are two wheels, there is KEr - but 2KEr because of the two wheels.

So the total ikinetic energy is 2KEr + KEt and the rotational kinetic energy of the wheels is 2KEr. I'm sure you know where to go from there. Good luck!
 
okay...i got it wrong.

here's what i did:
(2)(1/2 I w^2) + (1/2 m v^2)

that's the 2KEr + KEt

i made up a theoretical velocity (5) since they didn't give me one...and found w using that velocity and the radius of the wheels

so...

(.08)(16.667^2) + (1/2)(74)(5^2)

then took (.08)(16.667^2) over the total...which is wrong. what step do i need to correct?
 
i would really appreciate it if anyone has any help...
 
First of all, you don't need to make a theoretical velocity because they cancel out in the end. Now...

(2)(1/2 I w^2) + (1/2 m v^2)
(2)(1/2) is 1. You know that w=v/r, so plug in v/r for w, and you get
I(v/r)^2 + (1/2 m v^2)
You know I, r, and m, so you get
((.08)V^2)/.3 + (1/2)(74)V^2
Do you see where I am going with this?
TOTAL = (4/15)v^2 + (37)V^2
2KEr = (4/15)v^2
I can't tell you the whole solution because its against the rules. Hopefully you'll be able to figure it out. Good luck!
 
taking (4/15)/((4/15)+37) gives me an incorrect answer...
 
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