# Moments of Inertia Trouble

1. Nov 22, 2015

### brinstar

1. The problem statement, all variables and given/known data
Attached in a picture!

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

3. 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 j

II) 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! :-)

#### Attached Files:

• ###### Physics Problem 3.png
File size:
56.8 KB
Views:
33
2. Nov 22, 2015

### Staff: Mentor

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?

3. Nov 22, 2015

### brinstar

Oh okay, so I should be using 1/3*mr^2?

4. Nov 22, 2015

### Staff: Mentor

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?

5. Nov 22, 2015

### brinstar

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.

6. Nov 22, 2015

### Staff: Mentor

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.

7. Nov 22, 2015

### brinstar

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?

8. Nov 22, 2015

### Staff: Mentor

Yes. That's just what I would do.

9. Nov 22, 2015

### brinstar

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. Nov 22, 2015

### Staff: Mentor

Looks good. But I would first convert the distances to standard units (meters instead of cm).

11. Nov 22, 2015

### brinstar

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. Nov 22, 2015

### SteamKing

Staff Emeritus
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

13. Nov 22, 2015

### Staff: Mentor

Sure.

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.