2grhodes_at_work:

quote:

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This equation is wrong, by the way. In 2D, it would be:
M = I * dw/dt, where dw/dt is the time derivative of w, or the angular acceleration.

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He ask about rotation about axis ( on plane )
M = FR = mar
where a center orientaited acceleration ( i mean tan acceleration = 0 and w = const )
a = w*w*r
M = m*r*r*w*w , get I = mr*r
I think this will be all , what he need.
And where you see here "wrong" ?

If it come from plane to 3D space, certainly all you formulas will be true , but if it didn''t know moment at all , i don''t think it understandant your post

Since the original poster has answers to his questions, I''d like to resolve some final issues and then close this thread. The sub-topic regarding race cars spinning out (another message) should be continued in a new thread if there is interest.

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Original post by Bandures
He ask about rotation about axis ( on plane )
M = FR = mar
where a center orientaited acceleration ( i mean tan acceleration = 0 and w = const )

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I see a number of possible problems with your analysis here, but before I reply I''d like to clarify my understanding of your variables. It will help me better understand what you''re trying to do.

Question 1) Do you mean for R = r = the distance away from the center of mass where the new force (F) is applied?

Question 2) When you say a is the "center orientaited acceleration," do you mean for a to be the centripetal acceleration of the point on the object surface where the force F is applied *OR* do you mean for a to be the translational acceleration of the center of mass of the object?

I understand that dw/dt = 0 (and tangential acceleration also = 0) when w = const.

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Original post by Bandures
a = w*w*r

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This equation implies a particular answer to question #2.

quote:

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Original post by Bandures
M = m*r*r*w*w , get I = mr*r

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Question 3) In your equation I = mr*r, do you mean for "r" to be the same as the "R = r" from question #1?

Please answer my questions, and then we''ll be able to settle our differences!


Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

Original Post by WarMage.

flame_warrior, I''m not sure I''m convinced. How would you calculate the point at which to break the tires loose, and the angular momentum associated with it? If it works, that sounds cool (I did post the somewhat longwinded one above), but could you give a little more background to your solution?


Hi,
Its tough for me to explain here. It cannot be explained without the use of drawings(by me anyway). I can point you to some books where you can find this info though. Try to find some books on Theory of machines which has Gyroscopic Effects and Precessional Motions. Any Mechanical Engineering(studying in 5th or 6th sem) Student should have it.

I can give you the equations for overturning though if you want it that is.

1) yes , r - radius of inertion ( all mass on this radius )
2) "the centripetal acceleration of the point on the object surface where the force F is applied"
Ofcouse F = F applied only if we have fixed center.

quote:

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Original post by Bandures
1) yes , r - radius of inertion ( all mass on this radius )
2) "the centripetal acceleration of the point on the object surface where the force F is applied"
Ofcouse F = F applied only if we have fixed center.

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Okay, excellent. This clarifies your assumptions very well. There may still be some missing information, and a picture would immensely help but I can clarify a few things with what we have. I''m afraid I still see inconsistencies in your approach. First of all, the correct standard terminology for your "r" is "radius of gyration", not "radius of interia/inertion."

Lets go back to your previous post:

quote:

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Original post by Bandures
M = FR = mar
where a center orientaited acceleration ( i mean tan acceleration = 0 and w = const )

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Two things are inconsistent in this part of your post. First, if you allow M to be nonzero, then the tangential (and rotational) accelerations would also be *nonzero*, leading to w being *not* constant----in conflict with your stated assumption. This is because M *causes* rotational/tangential acceleration. Thus, given your assumption that tan acceleration = 0 and w = const, it would be impossible to have a nonzero M. Second, you have stated that a is the centripetal acceleration. By writing M = mar you are relating moment, which causes rotational acceleration, to centripetal acceleration. Centripetal acceleration is radial and is orthogonal to rotational acceleration and cannot produce a nonzero M. Thus, M = mar is incorrect. (Your force, F = ma, is centripetal force.)

Lets look at one further part of your previous post:

quote:

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Original post by Bandures
a = w*w*r

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Well, of course this is an equation for centripetal acceleration, and as described above it cannot be related here to the calculation of M, and so your expanded equation,

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M = m*r*r*w*w

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is also incorrect.

It appears to me you have attempted to change equations for forces and accelerations related to an object moving in a perfect circle at constant speed into equations that can represent an object that may undergo arbitrary forces and moments about its center of mass. uncutno''s original question did deal with forces applied both radially through the center of mass causing translational acceleration and tangentially causing rotational acceleration. Simple constant centripetal acceleration isn''t applicable to that problem.

Does this clear up some misunderstandings?


Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

i think thier overcomplicating the answers to the question
in the second post he wanted to how to calculate when the car will slide in a turn

so i would say because the front wheels begin the turn and causes rotation as far as the center mass of the car is concered the driver turns the wheel to compensate and increase friction on the front tires the back tires do not turn as a result a micro vibrational effect occurs that decreases the grip friction of the back tires and the back of the car begins to slide in part from momentum and in part from rotation i have no idea how to put it into math but their it is

to add this is the crucial part of the calculation because if the center mass of the cars momentum overcomes the (friction grip) of the tires it will slide into the wall
if the friction overcomes the gravity of the cars mass or the momentum the car may roll or may violenty reverse rotation

u could would also have to consider the fact that the ground affects friction water on the road the type of road the speed of the car the condition of the tires the shocks windflow other factors affect each tire in a turn obviously the inside tires have less force on them as the outside ones

u also have to take into consideration the fact that the higher the cars center mass is the more unevenly distributed the (force and weight) on the tires will be a car very low to the ground car will have a very evenly distributed amount of friction to each tire ect...

the turning of the cars front wheels generates the
rotation relative to the back wheels of the car the amount of
friction of the front tires to the back tires determines increaseing or decreaseing rotation once the the turn has begun