Proofing Moment of Inertia Solid Sphere

In summary: I'm sorry if this is confusing, I'm not familiar with the exact details.In summary, I don't think this problem can be solved using the pyramid method as proposed.
  • #1
chocophysuny
3
0

Homework Statement


Can i use "pyramid" method to derive the equation of Moment Inertia solid sphere?
The pyramid is such we slice a watermelon.
Sorry for my bad english.
Regards.


Homework Equations



2/5 MR^2

The Attempt at a Solution

 
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  • #2
Hello chocophysuny,
Welcome to Physics Forums,

chocophysuny said:
Can i use "pyramid" method to derive the equation of Moment Inertia solid sphere?


Have you tried it? You must show what you have tried to get help on PF (forum rules).

Sunil
 
  • #3
It doesn't seem to me that it would be a useful approach. It does not simplify anything. You need to slice it into elements that have an easily calculated MI, such as discs perpendicular to the axis of rotation.
 
  • #4
haruspex said:
It doesn't seem to me that it would be a useful approach. It does not simplify anything. You need to slice it into elements that have an easily calculated MI, such as discs perpendicular to the axis of rotation.

But my lecture say this method is easy enough to solve.
And, actually i confused to find the element of volume and element of MI.

So, in your opinion, this is not solvable?
 
  • #5
Sunil Simha said:
Hello chocophysuny,
Welcome to Physics Forums,




Have you tried it? You must show what you have tried to get help on PF (forum rules).

Sunil

Yes, I've tried it. But i didn't find the solution.
I'm confused with the element of volume and element of pyramid

"To find MI from the element, we need element of volume and element of MI, isn't it?"
 
  • #6
chocophysuny said:
"To find MI from the element, we need element of volume and element of MI, isn't it?"

I'm sorry, I didn't understand what you meant there.
 
  • #7
chocophysuny said:
But my lecture say this method is easy enough to solve.
And, actually i confused to find the element of volume and element of MI.

So, in your opinion, this is not solvable?

Maybe I misunderstand the proposed method. From your watermelon analogy, I assumed it involves slicing the sphere into thin wedges using planes through the axis of rotation. If so, I think you would then need to cut each wedge into laminae, either rectangular ones parallel to the axis or triangular ones orthogonal to the axis.
 

FAQ: Proofing Moment of Inertia Solid Sphere

What is the "Proofing Moment of Inertia Solid Sphere"?

The "Proofing Moment of Inertia Solid Sphere" is a physical property that describes how an object's mass is distributed around its rotational axis. It is a measure of an object's resistance to changes in its rotational motion.

What is the formula for calculating the "Proofing Moment of Inertia Solid Sphere"?

The formula for calculating the "Proofing Moment of Inertia Solid Sphere" is I = 2/5 * mr², where I is the moment of inertia, m is the mass of the object, and r is the radius of the object.

Why is the "Proofing Moment of Inertia Solid Sphere" important in physics?

The "Proofing Moment of Inertia Solid Sphere" is important in physics because it helps us understand and predict an object's rotational motion. It is also used in the design and engineering of various objects, such as wheels, gears, and flywheels.

How can the "Proofing Moment of Inertia Solid Sphere" be measured or determined experimentally?

The "Proofing Moment of Inertia Solid Sphere" can be measured or determined experimentally by using various methods such as the torsion pendulum, compound pendulum, or parallel-axis theorem. These methods involve measuring the object's period of oscillation or its mass and distance from the axis of rotation.

How does the "Proofing Moment of Inertia Solid Sphere" differ from the "Proofing Moment of Inertia Hollow Sphere"?

The "Proofing Moment of Inertia Solid Sphere" and the "Proofing Moment of Inertia Hollow Sphere" differ in that the solid sphere has a higher moment of inertia compared to the hollow sphere of the same mass and radius. This is because the mass is distributed farther from the axis of rotation in a solid sphere, resulting in a greater resistance to changes in rotational motion.

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