Kinetic Friction & Rotation: Questions & Answers

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Discussion Overview

The discussion revolves around the relationship between kinetic friction and rotation, specifically how these concepts apply to a ball rolling on a flat surface. Participants explore the equations governing frictional forces, deceleration, and the conditions under which a ball rolls without slipping. The scope includes theoretical considerations, mathematical reasoning, and practical implications in real-world scenarios.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions how the deceleration of a rotating ball relates to the frictional coefficient of the surface.
  • Another participant states that the force of kinetic friction can be expressed as F = μR, where R is the normal reaction force.
  • There is a discussion about the difference between sliding and rolling motion, with some suggesting that if a ball is rolling without slipping, it will not accelerate further.
  • Participants discuss how to calculate the distance a ball travels given certain parameters, indicating that Newton's second law and its rotational analogue can be used.
  • One participant raises a concern about the initial premise of no forces acting on the ball, suggesting that this may not apply in real-world scenarios.
  • There is mention of rolling resistance and its effect on a ball's motion, with a formula provided for calculating this force.
  • Some participants express uncertainty about the relationship between angular and linear deceleration and how they relate to the time it takes for the ball to stop.
  • Questions arise regarding the specifics of motion when a ball rolls without slipping and the implications of static friction in this context.

Areas of Agreement / Disagreement

Participants express multiple competing views regarding the effects of friction and the conditions under which a ball will stop rolling. There is no consensus on how to calculate the time it takes for the ball to stop or the implications of rolling resistance in practical scenarios.

Contextual Notes

Participants note limitations in the discussion, such as assumptions about the rigidity of the ball and surface, the nature of forces acting on the ball, and the conditions under which static friction applies. There is also mention of the need to consider real-world factors like rolling resistance.

Who May Find This Useful

This discussion may be useful for students and enthusiasts of physics, particularly those interested in mechanics, rotational motion, and the dynamics of friction. It may also benefit individuals exploring practical applications of these concepts in sports or engineering contexts.

  • #31
Grunfeld said:
I am more interested to the method to solve the problem I proposed, how to do so?
Just out of interest, why do you want to know? How much physics have you done previously?
 
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  • #32
I am doing a research project on the subject for a grade 11 physics course
 
  • #33
Grunfeld said:
I am doing a research project on the subject for a grade 11 physics course
Are you allowed such help?
 
  • #34
I don't see why not. We can consult what ever resources available.
 
  • #35
Grunfeld said:
I don't see why not. We can consult what ever resources available.
In that case I will continue, but first I will make it clear that I am not aware that I am doing anything contrary to your school or examining board policies.
Grunfeld said:
Ok, i get the general idea of it, but I still am unsure of the specifics.

Say on a flat surface, a ball with mass 1 kg and radius 5 cm initially starts rolling with a velocity of 1 m/s. The ball is not slipping. The the coefficient of rolling resistance is 0.01. How far will the ball travel?
Now we have that over we must make some assumptions. Firstly, we assume that the ball is uniform. And secondly we ignore any change in radius of the ball due to deformation. Now, the moment of inertia of a solid sphere is,

I = \frac{2}{3}mr^2

And the rotational analogue of Newton's second Law,

\sum \tau = I\alpha

where the LHS is the net torque. What is the net torque in this case?
Grunfeld said:
Also, what's w in the equation v = rw
w is angular velocity as is usually denoted by omega (\omega).
 
  • #36
What is LHS?

I am guessing that the torque is the the weight times the friction coefficient divided by r,
which is 9.8x1x0.01/0.05 = 1.96
 
  • #37
Grunfeld said:
What is LHS?
LHS stands for Left Hand Side, as in the left hand side of the equation.
Grunfeld said:
I am guessing that the torque is the the weight times the friction coefficient divided by r,
which is 9.8x1x0.01/0.05 = 1.96
There are two frictional forces acting on the ball. One is the static frictional force, which acts tangential to the ball collinear with the surface. The second is the rolling resistance which is usually considered to act through the centre of mass (COM) of the ball. Since the rolling resistance acts through the COM, it doesn't create a torque. Therefore, the only torque on the ball is that of the static friction.

Do you follow?
 
  • #38
Ok i see, so the torque is equal to the Weight x static friction coefficient?
 
  • #39
Grunfeld said:
Ok i see, so the torque is equal to the Weight x static friction coefficient?
Correct so,

\mu_s\not{M}g\not{R} = \frac{2}{3}\not{M}R^{\not{2}}\alpha

\Rightarrow \mu_sg = \frac{2}{3}R\alpha

Hence the angular acceleration is,

\alpha = \frac{3}{2}\frac{\mu_s g}{R}

Can you take the next step?
 
  • #40
The angular velocity is v/r
so the time it takes for the ball to stop is (v/r)/(a)?
 
  • #41
Grunfeld said:
The angular velocity is v/r
so the time it takes for the ball to stop is (v/r)/(a)?
Correct, of course we assumed no slipping so we have a condition of the minimum value of the coefficient of static friction, which can be determined using linear mechanics. Furthermore, this is rather a crude model, the actual form of the net torque is a function of the normal reaction forces acting on the tyre as is depicted in the link I provided earlier. We also assume that the ball is on the verge of slipping (i.e. the frictional force is maximal), which is generally the case for bodies undergoing deformation.
 
Last edited:
  • #42
I thought you said the normal friction forces does not matter in rotating motion (it will roll forever), and the rotation friction is the force that slows the ball down. Why is the static friction coefficient important here?
 
  • #43
Grunfeld said:
I thought you said the normal friction forces does not matter in rotating motion (it will roll forever), and the rotation friction is the force that slows the ball down.
That was for rigid bodies only, the normal force now acts to create a torque on the ball. See my previous posts and the link I provided.
Grunfeld said:
Why is the static friction coefficient important here?
It is the force of static friction which prevents the ball from slipping and allows the ball to accelerate if you like.
 
  • #44
The ball is not considered to be a rigid body? What is a rigid body?
 
  • #45
Grunfeld said:
The ball is not considered to be a rigid body? What is a rigid body?
You said previously that you could use any available resource, well google is an excellent resource. If you have any specific questions I'll be more than happy to answer them here, but for more general questions, it is often better to do one's own research.
 

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