What Is the Instantaneous Speed of the Point of Contact on a Rolling Wheel?

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Homework Help Overview

The discussion revolves around the instantaneous speed of the point of contact on a rolling wheel, specifically a wheel with a diameter of 30 cm rolling at 120 rpm. Participants explore the implications of the wheel's motion and the definitions of instantaneous speed in this context.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Participants discuss the relationship between the wheel's rotation and the speed of the point of contact, with some suggesting that the point of contact has zero instantaneous speed when rolling without slipping. Others question their understanding of the point of contact and its motion relative to the ground.

Discussion Status

The discussion is active, with participants providing insights into the geometry of the problem and the physics of rolling motion. Some guidance has been offered regarding the definitions and implications of the wheel's motion, while multiple interpretations of the point of contact's behavior are being explored.

Contextual Notes

There is some confusion regarding the definitions and assumptions about the point of contact and its instantaneous speed, as well as the relevance of certain equations like the moment of inertia in this context.

newton9point8
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Homework Statement


A 30cm diameter wheel rolls without slipping at 120rpm. The point of contact with the ground has an instantaneous speed of:

a-5.2mm/s
b-1.04cm/s
c-2.6 m/s
d-zero


Homework Equations


I'm not really sure about my final answer but i think is d-zero because i have a diagram of a wheels that has 0m/s on the bottom of the wheel.



RotationalKinematics44.gif


RotationalKinematics46.gif



The Attempt at a Solution



If the correct answer is not zero i think i will need the following formula

I=MR^2

but I'm only given diameter and rpm's so i don't know if there's anohter formula I'm missing.
 
Last edited:
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for each rotation the wheel goes around, how many cm of wheel touch the ground?

I=MR^2 is inertia; you shouldn't need that for this, you're already given the speed of rotation and the geometry of the wheel

This starts as a geometry problem.
 
okay... so i might need to multiply it by 2*3.1426 or something right?
 
Newton9point8 said:
okay... so i might need to multiply it by 2*3.1426 or something right?

the equation for the perimeter of a circle is P = 2*pi*r

where r is the radius. You're given the diameter.
 
Pythagorean said:
for each rotation the wheel goes around, how many cm of wheel touch the ground?
QUOTE]

30cm right?
 
no, 30 cm is the diameter. The diameter is 2 times the radius.

The radius relates to the perimeter as stated in the above equation.
 
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Newton9, your first answer was corret. "Rolling withing slipping" means the point of contact has zero instantaneous velocity. The object would be slipping if the contact point had a non-zero velocity.
 
thanks guys :smile:
 
hrm, I must have a bad definition of 'point of contact', because I'd think if a wheel is rolling along, the position where it touches the floor is the point of contact, and if it's not slipping, then both the wheel and the floor are touching in different places at ever time step, and that would imply the point of contact moves.

Can someone help me with my misconception here?
 
  • #10
If you look at the third diagram given in the first post you can see that the centre of the wheel has a constant speed v (translational velocity). the top of the wheel has a speed of 2v which is the tangent velocity of the point on the rim of the wheel and the translational velocity. Thus the point at the bottom of the wheel (the point of contact) has a tangental velocity that is opposite to the translational velocity and the two cancel.
 
  • #11
Kurdt said:
If you look at the third diagram given in the first post you can see that the centre of the wheel has a constant speed v (translational velocity). the top of the wheel has a speed of 2v which is the tangent velocity of the point on the rim of the wheel and the translational velocity. Thus the point at the bottom of the wheel (the point of contact) has a tangental velocity that is opposite to the translational velocity and the two cancel.

wow... that's kind of unintuitive. It almost seems as if you're in a frame with a velocity equal to the translational velocity of the wheel.
 
  • #12
There's a famous photo of a wheel with its centre as a green light and a point on the rim as a red light and it illustrates it beautifully but I'm having trouble finding it on the net. If I do find one i'll give you the link and perhaps it will shed more light on it for you.

EDIT: this helps me anyway.

http://img242.imageshack.us/img242/8971/rollingmotionoi9.th.jpg

Apologies for the crude photo but I gave up on the internet.
 
Last edited by a moderator:
  • #13
Much appreciated, thank you!
 

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