Inertia Translation Rotation Moments And Couples, car on accelerating Treadmill

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I have been asked to come up with a formula by a rather persistent person who is Addiment that a wheel on an accelerating treadmill attached to a mass (such as a car) would not produce enough translational force to move the Car.

The Question as posed to that persons;

Please forgive any and ALL incorrect terminologies (I am not quite that educated, as you will probably guess )

We have a car with the bearings degreased and completely free of any imperfections such as bumps or imperfect roller bearings etc etc.

We have disconnected any drive train portions. (The cars engine is not running, it is not needed for this test, there is no drive being produced and delivered to the wheels by the cars engine, the wheels are free from any mechanical resistance friction from the diff etc etc.)

The wheels are Perfectly balanced.

To eliminate the rolling frictions from the question, try to imagine the wheels are as hard as steel. (Please also remove any friction from the bearings as a reason the car may move, IE Imagine frictionless)

For the purpose of the Example we will assume one more thing;

The Tyre and Treadmill traction is flawless.

NOW

What we are going to do is Start the belt and Accelerate the belt from 0mph to 200mph within 3 seconds while the car is sitting freely on top. (Belt direction is backwards car face forward, speedometer reads +mph not -mph ) after three seconds the Experiment is over and all data collected.

Here is the Question.

What will happen to the car?
(will it move during those three seconds? )
<relative to the Ground. Not relative to the treadmill surface>.
 
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Answers and Replies

  • #2
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You gotta be joking me, what kind of nonsense is this?

Thread Locked.
 
  • #3
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Are you serious ? or just kidding ?

What? Have I got the wrong kind of sub forum here ?

Or is the question not making sence?

Bryce.
 
  • #4
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I don't even know what you are asking. Stuff like this:

all up these items are so perfect that a small drop of water on one side of Top dead Centre of any one wheel would cause the wheel to rotate back and forth for at least 100 repetitions as if that drop was on a swing (When the wheel bearing axel system was suspended off the ground).

Please go back and rewrite your question in a way that people can understand it.

I have been asked to come up with a formula by a rather persistent person who is Addiment that a wheel on a treadmill attached to a mass (such as a car) would not produce enough translational force to move the Car.

Why wouldn't a wheel on a treadmill be able to produce enough force to move a car? Go google a video of a car on a dyno and you can see how they have to strap it down so it doesnt jump off the rig and fly into a wall: it's the same setup you are asking.

https://www.youtube.com/watch?v=0yMd2n66uiQ
 
  • #5
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Edited.

The car is not running , it is not going to move of its own accord. I am looking at "free to spin" wheels on the "accelerating treadmill". only the treadmill will be producing any force to the wheel.

Sorry I didn't make that clear. my boo

Bryce.
 
  • #6
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I don't understand your question.
 
  • #7
Danger
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The wheels should just spin like crazy, with the inertia of the car holding it where it is.
 
  • #8
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I don't understand your question.

the question is two fold.

first fold is the Poll (least importance)

Second fold is "can some one please provide me with a formula that would calculate the force on the axle of the car as the wheel is being spun up by the treadmill"

The wheel is being spun by the treadmill.

the wheel is not connected to any drive train through the diff.

if it makes it easy to understand. lets just say that all 4 wheels are just like the front wheel, in that they have no means of being spun by the cars engine.

The treadmill will accelerate up to 200mph (the speedometer on the car will read + mph)
The treadmill will accelerate at a constant rate.

We will have a camera next to the treadmill (not on the treadmill) that will monitor the car for any movement. if the car dose not move then the camera will show the car in the exact same spot on screen with the wheels being spun up.
If the car moves then the car will move on screen.

How do I calculate the Amount of movement the car will experience during the three seconds that the treadmill is Accelerating to 200mph?

Bryce.

PS
If you do not understand the question with this help in this post. Please provide more details as to what you do understand and what seems Nonsense. For now, "I do not know what part you do not understand" so I am lost as to where to start.
 
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  • #9
Dale
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The car will move.

Even if the wheels are perfectly balanced and the bearings are perfectly frictionless the wheels still have some mass and some moment of inertia. This will require some specific torque in order to achieve the angular acceleration corresponding to 0-200 mph in 3 s. If you draw a free-body diagram of the wheels you will see that the weight of the car cannot produce any torque, so all of the torque will be provided by the horizontal force of the treadmill. Since that force must therefore be non-zero there will be a net horizontal force on the vehicle which, by Newton's 2nd law, will cause the vehicle to accelerate.
 
  • #10
Dale
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PS if a is the tangential acceleration of the wheels (200 mph/3 s) and assuming that the wheels are each uniform cylinders of mass m and that the mass of the car (including the wheels) is M then the force, f, generated by the treadmill is
f = m a/2

and if there are n wheels (usually 4) the acceleration, A, of the car (including the n wheels) is
A = n f/M
 
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  • #11
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PS if a is the tangential acceleration of the wheels (200 mph/3 s) and assuming that the wheels are uniform cylinders of mass m and that the mass of the car (including the wheels) is M then the force, f, generated by the treadmill is
f = m a/2

and the acceleration, A, of the car (including the wheels) is
A = f/M

So a one tonne car with 40kgs being held in the 4 10kg wheels (30cm radius)

So that’s 321.86kph/3 s = roughly 107

So f = 40 107/2 = 2140 (is this a meaningful number or is this in Newtons or ?? Clueless here ;))

So now the car acceleration;

Acceleration = 2140/1000kg = 2.14 (Again what is that in Ie is that KG/kgs

HELP ! <I need to lay down for a while>
 
  • #12
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http://gizmodo.com/assets/resources/2007/07/nascartreadmill.jpg [Broken]
http://www.gizmodo.com.au/2007/07/180mph_car_treadmill_great_for.html [Broken]
 
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  • #13
vanesch
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I fully subscribe to Dale's explanation. The car has to move because the spin of the wheels has to increase, and they have a finite moment of inertia, so a torque is needed to speed them up, and that will result in a net force on the axle of the wheel.

Now, if the wheels were also massless (or better, had 0 moment of inertia for their axis), then the car wouldn't move.

That said, it is a small effect.
 
  • #14
Danger
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Now, if the wheels were also massless (or better, had 0 moment of inertia for their axis), then the car wouldn't move.

Wasn't that part of the original question, or did I misread it? :confused:
 
  • #15
Dale
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So that’s 321.86kph/3 s = roughly 107
The units there would be km/h/s which is a terrible unit to use. Convert your speed into m/s so that your acceleration is in m/s^2.

Also, f is the force per wheel. So if your vehicle has 4 wheels then you need to multiply by 4 to get A, I will go back and clarify my PS above.
 
  • #16
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Wasn't that part of the original question, or did I misread it? :confused:
Don't worry Danger, your not the first to assume the Wheels were 0 mass.

http://www.rcgroups.com/forums/showthread.php?t=1007606

Bryce.

Dale;

So its f = m a/2
Wheels in my case were a little light. They were 10kg each so I just used 40kg as a one solid wheel example. (It makes no difference if I do it this way ?) just being conservative with the weight or the wheels ;)

f = 40 29/2 = 580

and

A = f/M

A = 580/1000 = .58

What dose that mean. .58 (Help Help :) )

Thanks for taking the time to work with me, I am a Complete (almost) Novice when it comes to physics.

Bryce.

Let me guess. No Laughing (To loud ;)) Is that .58m/s^2 there for, at the end of the third second the car would be moving 1.74m/s
 
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  • #17
Dale
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Let me guess. No Laughing (To loud ;)) Is that .58m/s^2 there for, at the end of the third second the car would be moving 1.74m/s
Yes, that is correct.
 
  • #18
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Yes, that is correct.

Thanks again for your help :D

I have one last question.

When you say Cylinder. do you mean that all the mass is at the outer points of the wheel? or that the mass is distributed throughout the whole radius?

IE a steel rod v a Steel pipe of same mass and same outer Diameter. Wouldn't the pipe have more rotational inertia ?


Hehe. I thought we were going to need 3.14159 a couple of times in the equation.

Bryce.
 
  • #19
Dale
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When you say Cylinder. do you mean that all the mass is at the outer points of the wheel? or that the mass is distributed throughout the whole radius?

IE a steel rod v a Steel pipe of same mass and same outer Diameter. Wouldn't the pipe have more rotational inertia ?
I meant steel rod, not steel pipe. The upper left figure in the http://hyperphysics.phy-astr.gsu.edu/HBASE/mi.html#cmi" page.
 
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  • #20
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Can you help me understand this Formula;
http://hyperphysics.phy-astr.gsu.edu/HBASE/mi.html#mix

I = kmr^2

What is k ?

http://www.colorado.edu/physics/phys2010/phys2010_fa04/lab/lab5.pdf" [Broken] just says it is a constant, between 0 and 1, depending on the shape. I can understand how the Solid disk would have the inner parts of mass, and the lesser amount of, don't have as much acceleration due to the reduce Radius.


Also, that link you gave me http://hyperphysics.phy-astr.gsu.edu/HBASE/mi.html#cmi the top left pic. has a second formula for angular Inertia. what is that ? looks like
T = Ia

Thanks again
Bryce.
 
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