Trajectory of a rigid body with two wheel with different speed

  • Thread starter Rossi88
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  • #1
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Hi,

first sorry for my bad English, I am more skilled to read than to write in English
I don't know if this is the right section for this topic.

The problem is described in the title of the topic.
My guess is that each of the two wheels goes on with circular trajectory around a same center. There isn't skidding of the wheels.
If the distance between the wheel is D, the speed of the slowest wheel is v1 and the speed of the other wheel is v2 (v1 and v2 is the linear speed), then:

R1=v1*d/(v2-v1)

is the distance between the centre and the slowest wheel, and of course the distance between the centre and the other wheel is

R2=v1*D/(v2-v1)+D=v2*D/(v2-v1)


is my guess correct?
My guess is based on time that each wheel needs to do the circle, and this time needs to be equal for the two wheel.
It seems to be correct, if one wheel is stopped (for example v1=0) the other does a circle around it, infact for the above formulas:
R1=0
R2=D
if the speed of the two wheel are the same
R1->inf
R2->inf

There is someone that knows an other demonstration more rigorous of this?
I attended a mechanical course some year ago, with euler's formula, lagrange and other but I don't remember much and I don't know how to determine the trajectory of a rigid body with two point with different speed.

A special thank to anyone that will help me.
 
Last edited:

Answers and Replies

  • #2
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Your equations look good to me, even letting v1=-v2, means R1 and R2 = D/2, which looks good.
 
  • #3
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Your equations look good to me, even letting v1=-v2, means R1 and R2 = D/2, which looks good.

your observation is interesting, and is a further proof of the goodness of the equations.
Thank you.

if someone other have a demonstration I will be grateful to him
 
  • #4
406
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I think if you took a toy car and put two different size wheels on either side, It would trace the outside of a circle depending on the initial velocity you give it. As long as the wheels are on the same axle that is.
 

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