Just out of interest, why do you want to know? How much physics have you done previously?Grunfeld said:I am more interested to the method to solve the problem I proposed, how to do so?
Are you allowed such help?Grunfeld said:I am doing a research project on the subject for a grade 11 physics course
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:I don't see why not. We can consult what ever resources available.
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,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?
w is angular velocity as is usually denoted by omega (\omega).Grunfeld said:Also, what's w in the equation v = rw
LHS stands for Left Hand Side, as in the left hand side of the equation.Grunfeld said:What is LHS?
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.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
Correct so,Grunfeld said:Ok i see, so the torque is equal to the Weight x static friction coefficient?
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.Grunfeld said:The angular velocity is v/r
so the time it takes for the ball to stop is (v/r)/(a)?
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: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.
It is the force of static friction which prevents the ball from slipping and allows the ball to accelerate if you like.Grunfeld said:Why is the static friction coefficient important here?
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.Grunfeld said:The ball is not considered to be a rigid body? What is a rigid body?