Register to reply

Please help - Rotational Kinetic Energy of Moving Wheel

by BlueSkyy
Tags: energy, kinetic, moving, rotational, wheel
Share this thread:
BlueSkyy
#1
Oct18-07, 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??
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
dvdqnoc
#2
Oct18-07, 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!
BlueSkyy
#3
Oct18-07, 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?

BlueSkyy
#4
Oct18-07, 10:13 PM
P: 36
Please help - Rotational Kinetic Energy of Moving Wheel

i would really appreciate it if anyone has any help...
dvdqnoc
#5
Oct18-07, 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!
BlueSkyy
#6
Oct18-07, 10:37 PM
P: 36
taking (4/15)/((4/15)+37) gives me an incorrect answer...


Register to reply

Related Discussions
Kinetic energy of a moving rotating rigid body Classical Physics 10
Rotational kinetic energy Introductory Physics Homework 1
Rotational Kinetic Energy Introductory Physics Homework 5
Calculate the rotational inertia of a wheel that has a kinetic energy of 20300 J when Introductory Physics Homework 1
Kinetic energy of a rotating wheel? Introductory Physics Homework 4