Simple conceptual question - rolling object question:

[frasifrasi]

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

 

[frasifrasi]

Anyone? please...

 

[nova-ex]

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.

 

[rl.bhat]

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.

 

[frasifrasi]

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.

 

[rl.bhat]

this is prob 95 (6th) or 85(5th) on fundamentals of physics.
Unless I read the problem I can't say anything.

 

[frasifrasi]

Here is the question:
http://img131.imageshack.us/img131/2309/screenhunter001gb1.jpg

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...

 

[frasifrasi]

Maybe it is because they are looking at the cylinder itself as a cylinder and a point mass?

 

[Shooting Star]

Here is the question:
http://img131.imageshack.us/img131/2309/screenhunter001gb1.jpg

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.

 

[Shooting Star]

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.

 

[rl.bhat]

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.

 

[Shooting Star]

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.