Please help - Rotational Kinetic Energy of Moving Wheel

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Homework Help Overview

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's wheels have a specified radius and rotational inertia, while the total mass of the bicycle and rider is provided.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between total kinetic energy, rotational kinetic energy, and translational kinetic energy. There are attempts to derive equations based on the given parameters, and questions arise regarding the necessity of theoretical values for velocity and how to properly incorporate them into the calculations.

Discussion Status

Some participants have provided guidance on how to approach the problem without introducing theoretical values for velocity, suggesting that these values may cancel out in the final calculations. There is an ongoing exploration of the correct formulation of kinetic energy components, but no consensus has been reached on the final steps or calculations.

Contextual Notes

Participants are navigating the constraints of the problem, including the lack of a specified velocity and the rules against providing complete solutions. This has led to some confusion regarding the correct approach to the calculations.

BlueSkyy
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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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