Moment of inertia of a non uniform disc

In summary, the centre of gravity for a non uniform disc is located a distance x from the centre and has a moment of inertia around the axis perpendicular to the surface passing through the centre of gravity.
  • #1
Rnm
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1

Homework Statement


A non uniform disc of radius R has a mass of M. Its centre of gravity is located at a distance x from the centre. Find the moment of inertia of mass (moi) around the axis perpendicular to the surface passinf through the centre of gravity.

Homework Equations


Parallel axis theorem
Ioo = Igg + m[r^2]

Ioo = moi around axis required
Igg = moi around axis through centre of gravity
m = mass of the object
r = distance between the 2 axises(?dont know the plural)

Moi around the centre of a uniform disc
1/2m[r^2]

The Attempt at a Solution

[/B]
I assumed that since Ioo = 1/2M[R^2]
Then using parallel axis theorem i took
Igg =1/2M[R^2] - M[x^2]
The problem is you can get a negative answer for Igg for certain x(>R/root2). So i think I am wrong some where in my assumptions. Can someone please clarify
 
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  • #2
Rnm said:
I assumed that since Ioo = 1/2M[R^2]
Why would you assume that? (That's the MOI of a uniform disk through its center.)

Have you posted the full problem? I don't see sufficient information to solve for anything.
 
  • #3
+1

Looks like something is missing.
 
  • #4
Doc Al said:
Why would you assume that? (That's the MOI of a uniform disk through its center.)

Have you posted the full problem? I don't see sufficient information to solve for anything.

This is the first part of the question.
The 2nd part involves placing this disk on a horizontal table at rest and letting go. The disk starts to roll (with no slipping) along the table . I need to use the answers i obtained in the first part to find the vertical and horizontal components of the reaction force on the disk by the table

I wrote 3 equations for this with Newtons 2nd law in the vertical and horizontal directions and the 2nd law of rotation in the direction of the angular acceleration (a)
I can't figure out how to write the last equation without using Igg. (Since it is stated to use the previous result).
 
  • #5
Are you sure the problem says it's a NONuniform disc not a uniform disc?

Consider the case when x=0. There are lots of ways that such a disc can be "nonuniform" and the moment of inertia will be different in each case. For example the disc could be thicker in the middle or thicker at the edge. That would change the moment of inertia while keeping x unchanged. You couldn't calculate the moment of inertia without knowing the mass distribution.
 
  • #6
CWatters said:
Are you sure the problem says it's a NONuniform disc not a uniform disc?
Rnm said:
Its centre of gravity is located at a distance x from the centre
 
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  • #7
Rnm said:
A non uniform disc of radius R has a mass of M. Its centre of gravity is located at a distance x from the centre. Find the moment of inertia of mass (moi) around the axis perpendicular to the surface passinf through the centre of gravity.
As others have posted, this is not nearly enough information. As an example, consider a uniform disc mass m with a point mass M-m stuck on at distance y from the disc's centre. The moment of inertia about the mass centre of the combination is mr2/2+Mmx2/(M-m).

Edit: No, that can't be right... something nasty anyway.
 
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  • #8
No other data is given in the question. I double checked. I suppose the question is incomplete then? But if so what would be missing in the question?
 
  • #9
Rnm said:
what would be missing in the question?
I cannot think of one simple extra piece of information that would allow you to find the moment.
I suggest you just write it as J and proceed to the second part of the question.

Edit: I found an old question exactly the same except that it did not expect you to determine the moment of inertia. It just said it is I, and asked about the oscillation. This suggests to me that asking for the moment of inertia was just a mistake.
 
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  • #10
haruspex said:
I suggest you just write it as J and proceed to the second part of the question.
Will do. Ill try to find the missing information about the question. Thanks for helping me out.
 

Related to Moment of inertia of a non uniform disc

1. What is moment of inertia of a non-uniform disc?

The moment of inertia of a non-uniform disc is a measure of its resistance to rotational motion. It depends on the mass distribution of the disc, with more mass located farther from the axis of rotation resulting in a larger moment of inertia.

2. How is moment of inertia of a non-uniform disc calculated?

The moment of inertia of a non-uniform disc can be calculated by breaking the disc into small segments and using the formula I = ∫r^2 dm, where r is the distance from the axis of rotation to each segment and dm is the mass of the segment. The integral is taken over the entire disc's mass distribution.

3. How does the moment of inertia of a non-uniform disc compare to that of a uniform disc?

The moment of inertia of a non-uniform disc is generally larger than that of a uniform disc with the same mass and dimensions. This is because the mass is concentrated at the center of a uniform disc, resulting in a smaller moment of inertia compared to a non-uniform disc with the same mass distribution.

4. What factors affect the moment of inertia of a non-uniform disc?

The moment of inertia of a non-uniform disc is affected by the mass distribution, shape, and size of the disc. A larger mass concentrated farther from the axis of rotation will result in a larger moment of inertia, while a smaller mass concentrated closer to the axis will result in a smaller moment of inertia.

5. Why is the moment of inertia of a non-uniform disc important?

The moment of inertia of a non-uniform disc is important in understanding and predicting the rotational behavior of the disc. It is used in many applications, such as calculating the torque required to rotate the disc and determining its angular acceleration. It is also important in engineering and design, as it affects the stability and performance of rotating systems.

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