Simple conceptual question - rolling object question:

In summary, the conversation discusses the formulas for torque and moment of inertia, as well as how to apply them to a disk rolling on a level ground. It is concluded that the moment of inertia can be calculated using the parallel axis theorem and that the direction of the frictional force should be in the opposite direction of the applied force for the disk to slow down. The specific equations and variables used in the problem are also mentioned.
  • #1
frasifrasi
276
0
Ok,

First of all, for the formulas torque = r X F

If the object is being rotated around an axis parallel to the axis going throught the center of mass, what is the "r" in this case? is it the distance from the axis of rotation to the center of mass? And in what direction would the force point in this case (is it coming from the edge of the object -- say it's is cylinder -- or from the center of mass?


--> now for my question, for a disk rolling on a level ground, what would its inertia be?
if I set the fixed point where the disk is touching the ground in the initial position, would the inertial simply be 1/2MR^2 or do I have to add anything to it?


thank you
 
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  • #2
Anyone? please...
 
  • #3
Frankly, I have no idea which r to use. But What I would do is I would try and derive it.

F = ma
Torque = Moment of inertia x Angular Acceleration

Now moment of inertia can be calculated using parallel axis theorem. After which you can find Torque. Angular acceleration can be calculated from linear acceleration at any point multiplied by distance to axis of rotation.

To answer your question about the moment of inertia of a disk rolling on the ground, if you were to take the axis of rotation to be the spot it hits the ground, you would have to use parallel axis theorem to calculate the moment of inertia which would be 1/2MR^2 + MD^2 D being the distance from the centre of mass to the axis of rotation.
 
  • #4
r is the distance from the axis to the point of application of the force. It is nothing to do with the center of mass.
When the disc is rolling on a level ground all the perticlas of the disc are rotating symmetrically about the axis of the cylinder. Therefore you have to use 1/2MR^2. If the whole disc rotates around a fixed point then you have to use parallel axis theorem to find moment of inertia.
 
  • #5
I see, the book used 1/2Mr^2 + mr^2 , so I was just wondering why they did that. I guess they were taking into consideration both the cylinder/disk as a whole and the individual particles...Any ideas? this is prob 95 (6th) or 85(5th) on fundamentals of physics.
 
  • #6
this is prob 95 (6th) or 85(5th) on fundamentals of physics.
Unless I read the problem I can't say anything.
 
  • #7
Here is the question:
http://img131.imageshack.us/img131/2309/screenhunter001gb1.jpg [Broken]

why did they use 1/2mr^2 + mr^2 in getting the solution (based on solutions manual)

for inertia instead of just 1/2mr^2 ?

Thank you...
 
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  • #8
Maybe it is because they are looking at the cylinder itself as a cylinder and a point mass?
 
  • #9
frasifrasi said:
Here is the question:
http://img131.imageshack.us/img131/2309/screenhunter001gb1.jpg [Broken]

why did they use 1/2mr^2 + mr^2 in getting the solution (based on solutions manual)
for inertia instead of just 1/2mr^2 ?

They have taken MI about the point of contact, which will be the MI of the cylinder about its own axis plus mr^2, since r is the dist of the CM of cylinder from the pt of contact.
 
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  • #10
It’s an easy problem. I’ll just use symbols here. Let P be the force, i.e., 12 N.
F=frictional force, w=omega. F is acting at the pt of contact in the opp dirn.

1. P-F=ma ( force = ma)
2. v=rw => a=rdw/dt
3. P*r + F*r= Idw/dt ( torque = MI around axis*angular accn)

You don’t need to take the MI about the pt of contact. Can you understand these eqns? Then solve them.
 
  • #11
3. P*r + F*r= Idw/dt ( torque = MI around axis*angular accn)
The torque due to the applied force is in the clockwise direction. Therefore torque due to frictional force sould be anticlockwise direction so that the body will be slowed down. In fact the frictional force acts in the direction of the velocity of COM. Therefore Idw/dt = P*r - F*r.
 
  • #12
It doesn't matter, because F will turn out be -ve from the equations I have given.

But for beginners, it's better to do it your way. In that case, eqn 1 also should be changed to P+F=ma.
 

1. What is a rolling object?

A rolling object is an object that is in motion and moving along a surface without sliding or slipping. Examples of rolling objects include a ball, a wheel, or a cylinder.

2. What is the difference between rolling and sliding?

The main difference between rolling and sliding is the presence of friction. In rolling, the object is in contact with the surface and is able to rotate, reducing the amount of friction and allowing the object to move smoothly. In sliding, there is no rotational motion and the object slides along the surface, creating more friction and making it harder to move.

3. Why do rolling objects continue to move without an external force?

Rolling objects continue to move because of their moment of inertia and angular momentum. The moment of inertia is a measure of an object's resistance to changes in rotational motion, and the angular momentum is the product of the moment of inertia and the angular velocity. As long as these values remain constant, the object will continue to roll without an external force.

4. How does the shape of an object affect its rolling motion?

The shape of an object can greatly affect its rolling motion. Objects with a larger radius, such as a larger wheel, will have a higher moment of inertia and will require more force to start rolling. Objects with a smaller radius, such as a smaller wheel, will have a lower moment of inertia and will require less force to start rolling. Additionally, objects with a more streamlined shape will experience less air resistance and will roll further and faster.

5. Can a rolling object ever come to a complete stop?

In theory, a rolling object can come to a complete stop if all of its energy is dissipated through friction. However, in most cases, there will always be some small amount of friction present, causing the object to continue rolling for a longer period of time. Additionally, other factors such as air resistance and the object's surface may also affect its ability to come to a complete stop.

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