Acceleration of a rotating system

In summary, the author is trying to find the acceleration of a system where a disk is attached to an axle that is placed on an incline. The author is not sure how to find total inertia, but believes that they can find the inertia of a cylinder and a disk and subtract the two. Then they can add the inertia of a cylinder to the inertia of a disk to get the total inertia of the system.
  • #1
ahuebel
11
0
I have this problem where I need to find the acceleration of a system of a disk attached to an axle that is placed on an incline made of two parallel bars. I think I basically know how to do the problem, I just need to know how to figure out total inertia. I know the inertia of a cylinder (the axle) and a disk is (1/2)MR^2 but I am not sure how to find total inertia since the axle basically goes through the disk.

TIA for any help.
 
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  • #2
Until we see the details of what you did, it's hard to say where you went wrong. Are the masses of cylinder and disk equal? I assume there are two different "R"s involved: that of the disk and that of the axle.
 
  • #3
ahuebel said:
I also know that friction acts at both contact points (the axle on each bar) so basically 2mgsin(theta) - 2f = ma.
If M is the total mass of the "disk + cylinder" and f is the total friction, then: Mgsin(theta) - f = Ma.
 
  • #4
I edited my post after you responded. Sorry about that.
You are right about your last message and the masses are not equal.

The mass of the axle is .1 kg (not including the part inside the disk), and has a radius of .02m. The mass of the disk is .8kg and its radius is .12cm. This seems wrong since those radii would make the disk smaller than the axle. The book must have meant .12m.

So it seems I can basically find the inertia of a disk of radius .02m and subtract that from the inertia of a disk of .12m and add that inertia to a cylinder of radius .02m to get total inertia, does this seem right? This assumes the same density for both disk and axle, I guess I can assume that...

Isnt the inertia of a disk the same as the inertia of a cylinder (1/2 MR^2)?
 
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  • #5
ahuebel said:
So it seems I can basically find the inertia of a disk of radius .02m and subtract that from the inertia of a disk of .12m and add that inertia to a cylinder of radius .02m to get total inertia, does this seem right?
Why are you subtracting anything? Find the inertia of the disk; find the inertia of the axle (outside of the disk, which is what you are given). Add them.

Isnt the inertia of a disk the same as the inertia of a cylinder (1/2 MR^2)?
Right.
 
  • #6
I guess I am confused because the inertia of the disk includes a portion of the axle (that portion inside the disk) which leaves basically two cylinders on the sides of the disk if you can visualize that. We can just assume a continuous object, find that inertia and go on with it? That just doesn't seem intuitive to me...
 
  • #7
Why don't you state EXACTLY what information is given in the problem? Then we can figure it out.
 

Related to Acceleration of a rotating system

1. What is the acceleration of a rotating system?

The acceleration of a rotating system refers to the rate of change of its rotational velocity. It is a vector quantity and is measured in radians per second squared (rad/s²) or revolutions per second squared (rev/s²).

2. How is the acceleration of a rotating system calculated?

The acceleration of a rotating system can be calculated using the formula a = ω²r, where a is the acceleration, ω is the angular velocity, and r is the radius of rotation. Alternatively, it can also be calculated using the formula a = αr, where α is the angular acceleration.

3. What factors affect the acceleration of a rotating system?

The acceleration of a rotating system is affected by the radius of rotation, the angular velocity, and the angular acceleration. Additionally, the mass and shape of the rotating object can also affect its acceleration.

4. How does the direction of rotation affect the acceleration of a rotating system?

The direction of rotation does not affect the magnitude of acceleration of a rotating system. However, it does affect the direction of the acceleration vector, which is always perpendicular to the plane of rotation.

5. Can the acceleration of a rotating system be negative?

Yes, the acceleration of a rotating system can be negative. This indicates that the rotational velocity is decreasing, or the direction of rotation is changing. It is important to note that the negative sign only denotes the direction of acceleration and not its magnitude.

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