Physics car tire throwing competition

In summary: The parallel axis theorem states that the axis of rotation of a body is perpendicular to the axis of symmetry of the body.
  • #1
Elliot
4
0
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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  • #2
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
 
  • #3
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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  • #4
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.
 
  • #5
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.
 
  • #6
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
 
  • #7
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.
 

1. What is the purpose of a physics car tire throwing competition?

The purpose of a physics car tire throwing competition is to apply principles of physics, such as velocity, force, and trajectory, to throw a car tire as far as possible. It is a fun and challenging way to demonstrate understanding of physics concepts.

2. What factors affect the distance a car tire can be thrown?

The main factors that affect the distance a car tire can be thrown include the initial velocity of the throw, the angle at which the tire is thrown, the weight of the tire, and air resistance. Other factors such as surface friction and wind conditions may also play a role.

3. How can I increase the distance of my car tire throw?

To increase the distance of your car tire throw, you can try adjusting the angle and force of your throw to achieve the optimal trajectory. You can also use a lighter tire or reduce air resistance by throwing in a windless environment. Additionally, you can use techniques such as spin or adding weight to the tire to maximize its momentum.

4. How does the shape of the tire affect its throw distance?

The shape of the tire can affect its throw distance in several ways. A larger surface area can increase air resistance and slow down the tire, while a smaller surface area can reduce air resistance and increase the distance. The weight distribution and center of mass of the tire can also impact its trajectory and distance.

5. What safety precautions should be taken during a physics car tire throwing competition?

It is important to prioritize safety during a physics car tire throwing competition. Participants should wear protective gear, such as gloves and safety glasses, and ensure that the throwing area is clear of any obstacles or bystanders. It is also recommended to use a designated area for the competition and have a designated person to supervise the event.

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