Rolling Without Slipping on an Inclined Plane

[Heisenberg7]

I have asked this question last year (on discord; IPhO server) and I believe I wasn't satisfied by the answer at that time, but I let it go. Today, as I was going through some physics videos on YouTube a video about it popped up. So, I would like to address this issue now.

Let's imagine an object on an inline, in our case, let it be a ball (look at the picture). For the first case, let's assume that the coefficient of friction between the ball and the ground is 0. Would the ball start to roll or would it just slide down the incline without any rotation? Personally, to me, the second case seems a bit counterintuitive, but that's the answer I got from the physics community above. The reason why it seems a bit counterintuitive is this: If we put the ball on the incline, the center of mass of the ball would lose support thus it will cause torque about the lowest point (due to gravity) which would cause the whole ball to rotate. Now, I'm not sure if this is the case, this is what I supposed would happen by pure intuition.

For the second case let's assume that there is friction. If my supposition is correct, how would the torques relate? Now, this is probably a bit of an undefined question. Torque due to friction about the lowest point would be 0 since $r = 0$. And if my supposition is not correct (also why), what would be the work done by torque (assuming there is friction)?

 

[kuruman]

The reason why it seems a bit counterintuitive is this: If we put the ball on the incline, the center of mass of the ball would lose support thus it will cause torque about the lowest point (due to gravity) which would cause the whole ball to rotate.

In rolling motion without slipping, the point of contact is instantaneously at rest while the rest of the ball rotates about that point. Rolling motion is a combination of translational motion of the center of mass and rotational motion about the center of mass. You need a non-zero torque about the center of mass for the latter to happen. This is not the case in the absence of friction.

You may wish to take a look at this article. It might clarify your thinking.

For the second case let's assume that there is friction. If my supposition is correct, how would the torques relate?

When you draw force diagrams in which torques are involved, it is important to draw the arrows representing the forces with their tail at the point of application of each force. Thus, in the diagram on the left (rough plane), N should have its tail at the point of contact, and so should the force of friction. Gravity acts at the center of mass and it's OK to draw its components separately.
Here is how they are related. There are two points about which it makes sense to calculate torques

About the center of mass:
Friction generates a clockwise torque $fR$.
Gravity generates zero torque because it acts at the center of mass.
The normal force generates zero torque because its lever arm is zero.
According to Newton's second law, the net torque is equal to the moment of inertia about the center of mass times the angular acceleration about the CM.

About the point of contact P:
Friction generates zero torque because it acts at the point of contact.
Gravity generates clockwise torque $mgR\sin\theta.$
The normal force generates zero torque because it acts at P.
According to Newton's second law, the net torque is equal to the moment of inertia about the point of contact times the angular acceleration about P.

Note: For rolling without slipping the angular acceleration about the CM is equal to the angular acceleration about P.

 

[jbriggs444]

I have asked this question last year (on discord; IPhO server) and I believe I wasn't satisfied by the answer at that time, but I let it go. Today, as I was going through some physics videos on YouTube a video about it popped up. So, I would like to address this issue now.

Let's imagine an object on an inline, in our case, let it be a ball (look at the picture). For the first case, let's assume that the coefficient of friction between the ball and the ground is 0. Would the ball start to roll or would it just slide down the incline without any rotation? Personally, to me, the second case seems a bit counterintuitive, but that's the answer I got from the physics community above. The reason why it seems a bit counterintuitive is this: If we put the ball on the incline, the center of mass of the ball would lose support thus it will cause torque about the lowest point (due to gravity) which would cause the whole ball to rotate. Now, I'm not sure if this is the case, this is what I supposed would happen by pure intuition.

Your intuition is simply incorrect here. In the absence of friction there is no torque anywhere.

The normal force of slope on ball provides no torque about the center of mass.
Gravity provides no torque about the center of mass.

For the second case let's assume that there is friction. If my supposition is correct, how would the torques relate? Now, this is probably a bit of an undefined question. Torque due to friction about the lowest point would be 0 since $r = 0$. And if my supposition is not correct (also why), what would be the work done by torque (assuming there is friction)?

So now you are taking the reference axis at the point of contact of cylinder on slope, yes? And you are assuming a coefficient of friction adequate so that there is rolling without slipping, yes?

You have the normal force which provides zero torque.
You have the frictional force which provides zero torque.
You have gravity which provides non-zero torque.

That torque can be easily calculated. You know the force and you can easily calculate the [perpendicular] offset of the force from reference axis.

Since you have selected a reference axis that matches the instantaneous center of rotation, the work done by torque will match the work done by the force providing that torque. The rate at which that work is provided will match torque times instantaneous rotation rate. It will also match force times instantaneous velocity of the center of mass.