Please help  Rotational Kinetic Energy of Moving Wheelby BlueSkyy Tags: energy, kinetic, moving, rotational, wheel 

#1
Oct1807, 09:14 PM

P: 36

1. The problem statement, all variables and given/known data
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? 2. Relevant equations K=1/2 I w^2 I = m r^2 < (for a hoop) 3. 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?? 



#2
Oct1807, 09:21 PM

P: 23

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! 



#3
Oct1807, 09:51 PM

P: 36

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? 



#4
Oct1807, 10:13 PM

P: 36

Please help  Rotational Kinetic Energy of Moving Wheel
i would really appreciate it if anyone has any help...




#5
Oct1807, 10:18 PM

P: 23

First of all, you dont 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 im going with this? TOTAL = (4/15)v^2 + (37)V^2 2KEr = (4/15)v^2 I cant tell you the whole solution because its against the rules. Hopefully you'll be able to figure it out. Good luck! 



#6
Oct1807, 10:37 PM

P: 36

taking (4/15)/((4/15)+37) gives me an incorrect answer...



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