Mechanism of a cylinder rolling down a ramp?

[kendro]

Hi guys... Can anyone hlp me with the mechanism of a cylinder rolling down a ramp? Here's my understanding about the problem. Can anybody corrected me?
If the CM is located at point O (the center of the cylinder), then the torque of for the rotational mechanism is provided by the friction force right? Then, the lever arm will be R (the radius of the cylinder). Then, the Torque will always be I (the moments of inertia) x alpha (angular acceleration)...

I got very confused when the CM is not located at the center of the cylinder, such as a distance d from point O. The weight of the CM will not in-line with the point where the cylinder is in contact with the ramp. Therefore, what is the normal force? Is it still W x cos(theta)? But then, if the CM is not located at the center of the cylinder, the weight of it will provide another torque other than the torque provided by the friction, with the lever arm of d x cos (beta)? But, then this torque will not be the same as the cylinder rolls down the ramp because the CM will change its position relative to the ramp, right? I am not sure about this...

Can anybody help to show me the correct force diagram for the attached diagram? Thank you very much guys... I really appreciate any kind of help...

 

[quasar987]

Here's my humble take on it.

There's a torque provided by gravity, another provided by friction and another by normal force. All with lever d because of the theorem "when many forces act on a solid body of mass M, the motion of the CM is that of a particle of mass M on which all the forces on the body are exerted"

 

[thepatientmental]

The mechanism of a cylinder rolling down a ramp involves several forces and torques acting on the cylinder. First, let's define some variables:

- CM: Center of mass of the cylinder
- O: Contact point between the cylinder and the ramp
- R: Radius of the cylinder
- d: Distance between the CM and point O
- W: Weight of the cylinder
- theta: Angle between the ramp and the horizontal
- alpha: Angular acceleration
- I: Moment of inertia

When the CM is located at point O, the torque for the rotational mechanism is provided by the friction force, which acts at point O and has a lever arm of R. This means that the torque is equal to R x Ff, where Ff is the magnitude of the friction force.

However, when the CM is not located at point O, the weight of the cylinder will also contribute to the torque. The weight of the cylinder acts at the CM, which is a distance d from point O. This creates a torque of d x W x cos(theta), as you correctly stated.

The normal force in this case will still be W x cos(theta), as it is always perpendicular to the contact surface between the cylinder and the ramp. However, the normal force will also have a lever arm of d x cos(beta), where beta is the angle between the line connecting the CM and point O and the horizontal. This normal force will provide a counter-torque to the weight of the cylinder, balancing out the torque created by the weight.

As the cylinder rolls down the ramp, the CM will change its position relative to the ramp, causing the torque created by the weight to change. However, the normal force will also adjust accordingly, keeping the cylinder in rotational equilibrium.

To accurately depict this situation, you can draw a free-body diagram of the cylinder, showing all the forces acting on it and their respective lever arms. This will help you better understand the forces and torques involved in the mechanism of a cylinder rolling down a ramp.

I hope this helps clarify the problem for you. Let me know if you have any further questions. Good luck!