# Torque/Intertia: Balancing a Tire

1. Mar 10, 2009

### Wobble

1. The problem statement, all variables and given/known data

2. Relevant equations

$$\Sigma$$$$\tau$$= r x F = I*$$\alpha$$

Angular Momentum
L= r x p = r x mv = r x mr$$\omega$$ = mr2 * $$\omega$$2

editediteditediteditediteditediteditediteditediteditediteditedit
This formatting looks weird. It should be r^2 *(omega)^2

and I*(alpha for the next part)

editeditediteditediteditediteditediteditediteditedit

3. The attempt at a solution

Right now I'm working on part a

I was thinking the I$$\alpha$$ of each wheel would give me the center of mass
Torque = I \alpha + I\alpha
but I don't have an acceleration value, and that wont give me where the center of mass. I think I need to find the center of mass first, and then figure out how adding the mass to one wheel will change the axis of rotation.

2. Mar 10, 2009

### Staff: Mentor

Yes, figure out the center of mass first. Find the combined center of mass of the "perfect tire" of mass M and the extra mass m. (Where's the center of mass of the "perfect tire" by itself?)

3. Mar 10, 2009

### Wobble

Well for the center of mass of the tire, I don't really have a radius for it?

The mass m is 1 ft away from the center of the tire, but how far away is mass M from the center of the tire? Do I use 1 ft for it a well?

But then I would get (M+m)/(M+m) which is just 1...

($$\Sigma$$mr)/($$\Sigma$$m)

is the center of mass equation, with just a general m and r in there

Last edited: Mar 10, 2009
4. Mar 10, 2009

### Wobble

Well, I could say that the perfect tire's R for the center of mass equation is 0, and that would just give me 0 as the center of mass. Then the center of mass for the two objects would be:

(M*0+m*R)/(M+m)
and that would give me my new center of mass, which would be the distance between the center of mass and the axis of rotation

I'm not sure how to continue with part B though.

5. Mar 11, 2009

### Staff: Mentor

Good.

The center of mass moves in a circle, so what's its acceleration?

6. Mar 11, 2009

### Wobble

Torque= R cross F= I times alpha

Alpha is then a/R. But I need a velocity in there somehow to solve part C. 50mph isn't an angular velocity, nor does a/R contain velocity in it.

How do I account for the change in the center of mass? Its going to involve my answer to A somehow, which I got to be a really small number. Is it just 1 + this number?

edit: I need a way to relate a/R to linear velocity

Last edited: Mar 11, 2009
7. Mar 11, 2009

### Staff: Mentor

Parts A, B, and C have nothing to do with torque.

To solve B, think of all the mass being located at the center of mass. Thus the mass is rotating in a circle of known radius (from part A). What force must be exerted to make it go in a circle? (Hint: Centripetal.)

8. Mar 11, 2009

### Wobble

Looking a wikipedia

1. (2piR)/V=T

2. (2piV)/a=T

3. a=(v^2)/R

and for it in radians per second
4. omega=2pi/T

So do I just want equation 3, substitute it into F=ma,

It says "frequency of the vibration" though, which doesn't sound like a force answer

edit:
Well now I see omega=2pi/T=V/R <--- is that all I need? I feel like I need to take this and substitute it into something else

edit2:
http://www.phy.cmich.edu/people/andy/Physics110/Book/Chapters/Chapter6_files/image054.gif

I'm thinkin that I want this.

I'm not sure how to use R though. Is it just simply my answer for part A, or is it 1 +/- my answer in part A? (Which I got to be 3/803)

Last edited by a moderator: Apr 24, 2017
9. Mar 11, 2009

### Staff: Mentor

Yes, where v is the tangential speed of the center of mass and R is the distance from the axis to the center of mass.

Part C asks for both force and frequency. What's the frequency of rotation of the tire?