Moment of inertia and Parallel axis theorem

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SUMMARY

The moment of inertia for a solid, uniform disk of mass M and radius R oscillating about an axis through point P, perpendicular to the plane of the disk, is calculated using the Parallel Axis Theorem. The formula applied is I = I_{cm} + Mb^2, where I_{cm} = 0.5 MR^2 is the moment of inertia about the center of mass. Given the distance b from the center C to point P, the final expression for the moment of inertia is I = 1.5 MR^2. The discussion clarifies the application of Steiner's formula in this context.

PREREQUISITES
  • Understanding of moment of inertia and angular momentum
  • Familiarity with the Parallel Axis Theorem
  • Knowledge of rotational dynamics
  • Basic calculus for deriving formulas
NEXT STEPS
  • Study the derivation of the Parallel Axis Theorem in detail
  • Learn about the implications of angular momentum in rotational systems
  • Explore advanced applications of moment of inertia in engineering contexts
  • Investigate the differences between moment of inertia for various shapes and axes of rotation
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Physics students, mechanical engineers, and anyone studying rotational dynamics and mechanics will benefit from this discussion.

soopo
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1. Thehttp://nft01.nuk.edu.tw/lib/exam/97/master/97ap-master.pdf" statement, all variables and given/known data
A solid, uniform disk of mass M and radius R is oscillating about an axis through P. The axis is perpendicular to the plane of the disk. Suppose the friction at P can be ignored. The distance from P to the center, C, of the disk is b (see figure 1). The gravitational acceleration is g.

What is the moment of inertia for rotation about the axis through P?

The Attempt at a Solution



I know the formula

L = Iw

Moment of inertia is I, while rotation per second is w.
The question asks to calculate the moment of inertia for rotation.
This seems to be apparently "L".

However, I am not sure, since this problem should be apparently solved with Parallel Axis theorem i.e.
I = I_{cm} + MR^2
where likely I_{cm} = .5 MR^2 is the moment of inertia though the center of mass.
However, I amn not sure, since this would imply

I = 1.5 MR^2How find the moment of inertia for rotation about the axis though P?
 
Last edited by a moderator:
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soopo said:
1. Thehttp://nft01.nuk.edu.tw/lib/exam/97/master/97ap-master.pdf" statement, all variables and given/known data
A solid, uniform disk of mass M and radius R is oscillating about an axis through P. The axis is perpendicular to the plane of the disk. Suppose the friction at P can be ignored. The distance from P to the center, C, of the disk is b (see figure 1). The gravitational acceleration is g.

What is the moment of inertia for rotation about the axis through P?

The Attempt at a Solution



I know the formula

L = Iw

Moment of inertia is I, while rotation per second is w.
The question asks to calculate the moment of inertia for rotation.
This seems to be apparently "L".
No. L is the angular momentum of the object when rotating at angular speed \omega.

However, I am not sure, since this problem should be apparently solved with Parallel Axis theorem i.e.
I = I_{cm} + MR^2
where likely I_{cm} = .5 MR^2 is the moment of inertia though the center of mass.
However, I amn not sure, since this would imply

I = 1.5 MR^2

How find the moment of inertia for rotation about the axis though P?
You have found it. Why do you think it is wrong?

AM
 
Last edited by a moderator:
Andrew Mason said:
No. L is the angular momentum of the object when rotating at angular speed \omega.

You have found it. Why do you think it is wrong?

AM

I did not know that the same Steiner's formula applies for moment of inertia of rotating objects.
There is no "w" in the Steiner's formula which made me to think that there is something wrong in the result.

Thank you for your answer!
 

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