Kinetic Friction & Rotation: Questions & Answers

[Grunfeld]

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

 

[Hootenanny]

What is LHS?

LHS stands for Left Hand Side, as in the left hand side of the equation.

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?

 

[Grunfeld]

Ok i see, so the torque is equal to the Weight x static friction coefficient?

 

[Hootenanny]

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?

 

[Grunfeld]

The angular velocity is v/r
so the time it takes for the ball to stop is (v/r)/(a)?

 

[Hootenanny]

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.

 

[Grunfeld]

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?

 

[Hootenanny]

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.

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.

 

[Grunfeld]

The ball is not considered to be a rigid body? What is a rigid body?

 

[Hootenanny]

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.