# Physics car tire throwing competition

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1. Nov 3, 2014

### Elliot

1. in a tire-throwing competition, a man holding a 23.5 kg car tire quickly swings the tire through three full turns and releases it, much like a discus thrower. the tire starts from rest and is then accelerated in a circular path. the orbital radius "r" for the tire's center of mass is 1.10 m, and the path is horizontal to the ground. the man applies a constant torque of 20.0 N m to accelerate the tire at a constant angular acceleration. Assume that all of the tire's mass is at a radius R = 0.35 m from its center.
what is the time, t, required for the tire to complete three full revolutions?

2. T=[2(pi)(r)]/v
Torque = Inertia * angular acceleration

3. T=Inertia * angular acceleration
20=[(23.5*.35^2)]*(linear velocity^2/.75^2)
linear velocity = 2.9 m/s
T=[2(pi)(1.1)]/2.9 = 2.38
2.38 * 3 revolutions = 7.15

I think the problem is my radius, I don't know which ones im supposed to use. the answer is supposed to be 7.68s, but im close with 7.15. can someone tell me what im doing wrong?

Last edited by a moderator: Nov 4, 2014
2. Nov 4, 2014

### BvU

Hello Elliot, welcome to PF :)

You can leave the template headings in to remind yourself of what is what. And use normal font for the paragraphs, reads a lot easier.

Under 2) I expect some relevant equations for angular motion with constant angular acceleration.
$\tau = {2\pi r\over v}$ makes no sense. That has the dimension of time.

Then, under 3) I expected some legible rendering of your attempt thus far. Not a repetition of $\tau = I\alpha$ and a new relationship $\tau = {a\over r}$ which again doesn't make sense: it has the dimension of (time)-2
I certainly don't understand how you can jump from that to $\tau = {v^2\over r^2}$ !? That righthand side has yet another dimension.

The r = 0.75 also surprises me. Why ?

Look up some sensible equations and see how you can use them . e.g. here or here

Oh, and you also have a problem working out $I$ : mr2 is not correct, because the tyre can't be considered as a point mass. You need the parallel axis theorem

3. Nov 4, 2014

### Elliot

for the moment of inertia of the tire, is it going to be that of a hollow cylinder? I know the first step is to calculate the moment of inertia, but I don't know which shape to use. if we use hollow cylinder, we have to find two radii, which I have no clue how to do

Last edited: Nov 4, 2014
4. Nov 4, 2014

### collinsmark

A hollow cylinder (or a circular hoop) plays a role in the effective moment of inertia. But that's not the whole story since the tire is held at an orbital radius of 1.10 m. As BvU pointed at (alluded to), much of this problem involves the parallel axis theorem.

5. Nov 4, 2014

### haruspex

That's not the error. With that error, the OP calculation would have been m*1.12.
Yes, that's the error. Elliot, do you know that theorem and how to apply it?
Yes, the axis of rotation being parallel to the axis of the cylinder. But why make it 3D? You can treat it as a ring.

6. Nov 5, 2014

### Elliot

Yea, I know the theorem. Im just trying to figure out how to combine it with the orbital radius. ive tried subtracting 1.1 from 0.35, and many other variables, but I cant get the answer

7. Nov 6, 2014

### haruspex

So quote the theorem (it now belongs in the Relevant Equations list) and post your thoughts on how to relate the problem data to the variables in the theorem.