Understanding Moments of Inertia for Solving Homework Problems

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Homework Help Overview

The discussion revolves around understanding moments of inertia, particularly in the context of homework problems involving the calculation of rotational inertia for various shapes, including rods and plates. The original poster expresses difficulty with the concept and application of moment of inertia in their homework tasks.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the moment of inertia for point masses versus extended objects, questioning the application of formulas. There are discussions about using specific formulas for rods and plates, and whether to split the problem into simpler components. The original poster also questions the relationship between center of mass and axis of rotation.

Discussion Status

Participants are engaging in a productive dialogue, with some providing hints and clarifications regarding the formulas and concepts involved. There is an acknowledgment of the need to convert units and an exploration of the differences in rotational inertia for various objects. While there is no explicit consensus, guidance has been offered on how to approach the problem.

Contextual Notes

The original poster has shared specific calculations and assumptions, including the total mass of the object and its dimensions. There is also mention of homework constraints, such as the requirement to use standard units.

brinstar
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Homework Statement


Attached in a picture!

Homework Equations


I = mr^2
m = (x*y)
cm = (x/2, y/2)

The Attempt at a Solution


I was able to solve Part 1 fairly fast, but Part II has drawn a blank for me overall. I was thinking of that rotational thing from Calculus where you spin a 2D figure around the axis to make a 3D shape, but I'm really just not sure. To be honest, I'm having a very difficult time even understanding the concept of moment of inertia >.< So even if I somehow guessed how to do part II right, I really don't know what I'm doing in that section.

I) mass 1 = (20 cm)(10 cm) = 200 cm^2
mass 2 = (10 cm)(20 cm) = 200 cm^2
center of mass 1 = (20 cm i, 5 cm j)
center of mass 2 = (5 cm i, 10 cm j)

Center of Mass = (summation of (mass)(coordinates) / summation of masses) = [density of object*(200*(20 i + 5 j))+(200*(5 i + 10 j)] / density of object*(200 + 200)
= (4000 i + 1000 j + 1000 i + 2000 j) / 400
= 12.5 cm i + 7.5 cm jII) total mass of object = 4 kg
I = summation of mass*r^2
= (1 kg*100 cm^2)+(3 kg*300 cm^2)
=(100 kgcm^2)+(900 kgcm^2) = 1,000 kgcm^2

Thank you for any help and have a nice day! :-)
 

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brinstar said:
I = mr^2
This is the moment of inertia of a point mass, not an extended object.

Hint: Can you find the moment of inertia of a rod rotating about one end?

Read this: Moment of Inertia
 
Doc Al said:
This is the moment of inertia of a point mass, not an extended object.

Hint: Can you find the moment of inertia of a rod rotating about one end?

Read this: Moment of Inertia

Oh okay, so I should be using 1/3*mr^2?
 
brinstar said:
Oh okay, so I should be using 1/3*mr^2?
Sure, if you use it correctly. Do you understand that the MOI of a plate about one edge is the same as that of a rod about one end?
 
Doc Al said:
Sure, if you use it correctly. Do you understand that the MOI of a plate about one edge is the same as that of a rod about one end?

I didn't know that, but I would suppose so since the area-of-force on the object is going to be farther from the center of mass for both a plate and a rod. It'd be higher than if the point of force was closer to the center of mass.
 
Moving mass parallel to the axis of rotation will not affect the moment of inertia about that axis. So you can squeeze the mass of the plate into a rod and get the same MOI.
 
Doc Al said:
Moving mass parallel to the axis of rotation will not affect the moment of inertia about that axis. So you can squeeze the mass of the plate into a rod and get the same MOI.

Oh okay. I think I confused center of mass and axis of rotation there.

So for this problem, would I be splitting the equation into two rods spinning about the y axis, and then add it all up, since the top square would be parallel to the bottom rectangle?
 
brinstar said:
So for this problem, would I be splitting the equation into two rods spinning about the y axis, and then add it all up, since the top square would be parallel to the bottom rectangle?
Yes. That's just what I would do.
 
Doc Al said:
Yes. That's just what I would do.

ah so would this be right?

II) total mass of object = 4 kg
I = summation of (1/3) mass*r^2
= (1/3)(1 kg*100 cm^2) + (1/3) (3 kg*900 cm^2)
=(100/3 kgcm^2)+(900 kgcm^2) = 933.333333 kgcm^2 = 933 kgcm^2
 
  • #10
brinstar said:
ah so would this be right?

II) total mass of object = 4 kg
I = summation of (1/3) mass*r^2
= (1/3)(1 kg*100 cm^2) + (1/3) (3 kg*900 cm^2)
=(100/3 kgcm^2)+(900 kgcm^2) = 933.333333 kgcm^2 = 933 kgcm^2
Looks good. But I would first convert the distances to standard units (meters instead of cm).
 
  • #11
Doc Al said:
Looks good. But I would first convert the distances to standard units (meters instead of cm).

ah okay, 0.093 kgm^2

and rotational inertia is different for different objects, right? Should I just memorize those formulas?

thank you so much!
 
  • #12
brinstar said:
ah okay, 0.093 kgm^2

and rotational inertia is different for different objects, right? Should I just memorize those formulas?

thank you so much!
There are formulas for the MOI for different objects. These formulas can be derived using calculus. Most people keep a table of these formulas handy:

https://en.wikipedia.org/wiki/List_of_moments_of_inertia
 
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  • #13
brinstar said:
and rotational inertia is different for different objects, right?
Sure.

brinstar said:
Should I just memorize those formulas?
It's a good idea to have a few basic ones memorized... just in case. If you have a bit of calculus under your belt, be sure to understand how they are derived.
 
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