Physics car tire throwing competition

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Homework Help Overview

The discussion revolves around a physics problem involving a tire-throwing competition where a 23.5 kg tire is swung and released. The problem includes concepts of angular motion, torque, and moment of inertia, with specific parameters such as the tire's mass, radius, and the torque applied during the throw.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the appropriate equations for angular motion and question the original poster's use of torque and radius in their calculations. There is uncertainty about the correct moment of inertia for the tire and how to apply the parallel axis theorem. Some participants suggest considering the tire as a hollow cylinder or a ring.

Discussion Status

The discussion is active, with participants providing guidance on relevant equations and clarifying concepts related to moment of inertia. There is an ongoing exploration of how to correctly apply the parallel axis theorem and relate it to the problem's parameters.

Contextual Notes

Participants note that the original poster's calculations may be affected by the choice of radius and the interpretation of the tire's moment of inertia. There is a mention of the need for clarity regarding the dimensions and relationships involved in the equations being used.

Elliot
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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 acceleration3. T=Inertia * angular acceleration
T=Inertia * (linear acceleration/radius)
T=Inertia * (linear velocity^2/radius^2)
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 I am supposed to use. the answer is supposed to be 7.68s, but I am close with 7.15. can someone tell me what I am doing wrong?
 
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Hello Elliot, welcome to PF :)

You can leave the template headings into 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
 
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
 
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Elliot said:
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

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.
 
BvU said:
the tyre can't be considered as a point mass
That's not the error. With that error, the OP calculation would have been m*1.12.
BvU said:
You need the parallel axis theorem
Yes, that's the error. Elliot, do you know that theorem and how to apply it?
Elliot said:
is it going to be that of a hollow cylinder
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.
 
haruspex said:
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.
Yea, I know the theorem. I am just trying to figure out how to combine it with the orbital radius. I've tried subtracting 1.1 from 0.35, and many other variables, but I can't get the answer
 
Elliot said:
Yea, I know the theorem. I am just trying to figure out how to combine it with the orbital radius. I've tried subtracting 1.1 from 0.35, and many other variables, but I can't get the answer
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.
 

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