# Homework Help: Angular Momentum - Train Placed on a Wheel

1. May 11, 2010

### Crusher8576

I've been trying this problem for a while now and can't seem to get the given answer. I know I'm probably making an elementary mistake somewhere but I can't find it.

1. The problem statement, all variables and given/known data
A track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis. A toy train of mass m is placed on the track and, with the system initially at rest, the train's electrical power is turned on. The train reaches a steady speed of 0.176 m/s with respect to the track. What is the angular speed of the wheel if its mass is 1.97m and its radius is 0.584 m? (Treat the wheel as a hoop, and neglect the mass of the spokes and hub.) (Answer 0.101 rad/s)

Ok so:
Vt= .176 m/s
mt= M
m2= 1.97M
r= 0.584m
I2=m2r2
Xf= ?
(X is angular velocity, subscript t for train and 2 for wheel)

2. Relevant equations

L=IX=rmv
Li=Lf
Lt+L2=0

3. The attempt at a solution

So I set it up with the initial equation
Lt = -L2
rmtvt = IX

Then expanded and substituted for corresponding masses
rMv = 1.97Mr2X

M cancels and so does an r, so
vt = 1.97rX

Then substitute the values
.176 = 1.97(.584)(X)

I get a final answer of

The answer is supposed to be .101 rad/s which I can't seem to get.

Also I used X for angular speed instead of omega because the text formatting wasn't working, and about 4 different symbols showed up instead of omega. Not sure why.

Last edited: May 11, 2010
2. May 11, 2010

### tiny-tim

Welcome to PF!

Hi Crusher8576! Welcome to PF!

(have an omega: ω )
hmm … I got the same answer as you until I read the question more carefully.

To find L, you need the actual speed of the train, which isn't 0.176, is it?

3. May 11, 2010

### Crusher8576

Aha I've got it. Thanks for the help.

If anyone was wondering how to do it:

The wheel is spinning in the opposite direction of the train, so the train may have a speed of .176, but that's in relation to the track. I had to find the actual speed of the train in relation to the ground.

So in the part where I have vt = 1.97rX, replace the Vt with what I called Va (for V actual), and also created Vw for (V wheel).

And Va = Vt - Vw = Vt - Xr

So then we're left with Vt - Xr = 1.97rX to solve for X.

Last edited: May 11, 2010