# Moment of inertia/Parallel axis theorem (lots of work shown)

1. Nov 20, 2013

### BMcC

The outer ring of the wheel has a mass of 51 kg, the inner ring, 6 kg, and each spoke, 10 kg.

Find the mass moment of inertia of the wheel about an axis through A and perpendicular to the page. Assume r1 = 124 cm, r2 = 36 cm. Each spoke is 88 cm long.

DIAGRAM

What I've learned in class for these types of problems is to break the obscure shape into shapes we can work with, then add them all up.

The table in my textbook says the moment of inertia for a ring is I = mr2
It also says the moment of inertia for a rod is I = 1/3 mL2

Spokes

1/3 10kg*0.88m2 = 2.581 kg*m2

Since there are 5 spokes, 2.581*5 = 12.907 kg*m2

Outer big ring

51kg*1.24m2 = 78.42 kg*m2

Inner ring

6kg*0.36m2 = 0.778 kg*m2

Now that I have all of the moments of inertia, I've added them up.

12.907 kg*m2 + 78.42 kg*m2 + 0.778 kg*m2

= 92.105 kg*m2 = Ic

Since the question asks for the moment of inertia of the wheel about A, I've been using the parallel axis theorem to translate that inertia to the new axis.

This is where I'm not sure what to do. I've been taking my total inertia at C, Ic, and adding it to m*R2 to figure out the parallel axis inertia

92.105 kg*m2 + m*R2
92.105 kg*m2 + 107kg*1.24m2

= 256.63 kg*m2

Unfortunately this answer is incorrect. Can anybody help me? I'm sure it's just a stupid mistake somewhere...

2. Nov 20, 2013

### Tanya Sharma

ML2/3 is the MOI of a spoke about its end point .What is the MOI of a spoke about the center of wheel ?Use Parallel axis theorem.

3. Nov 20, 2013

### BMcC

M*l2/12?

4. Nov 20, 2013

### Tanya Sharma

ML2/12 + MR2 ,where 'R' is the distance between the axis passing through the center of the wheel and axis passing through the center of the rod.

5. Nov 20, 2013

### BMcC

Hmm, I'm still confused.

6. Nov 20, 2013

### SteamKing

Staff Emeritus
You can combine the inertia values of the inner and outer rings because both are concentric.

However, the inertia values for the spokes have to be treated differently. The formula you used for the inertia of each spoke gives the inertia value referenced about the end of the spoke. In order to use the Parallel Axis Theorem correctly, you must use the proper reference locations.

BTW, posting the same problem in multiple HW forums is NOT allowed.

7. Nov 20, 2013

### Tanya Sharma

8. Nov 21, 2013

### BMcC

Okay so I'd use this formula to get the moment of inertia for the 5 spokes. Then once I have that, is my parallel axis work at the bottom correct?

9. Nov 21, 2013

### Tanya Sharma

Yes...

10. Nov 21, 2013

### BMcC

So I used the wrong formula for the moment of inertia for the spokes because I chose the wrong axis

Spokes

1/12 10kg*0.88m2 = 0.6453 kg*m2

Since there are 5 spokes, 0.6453*5 = 3.227 kg*m2

My two rings haven't changed, so now it'd be:

3.227 kg*m2 + 78.42 kg*m2 + 0.778 kg*m2

= 82.42 kg*m2

So with the parallel axis theorem, I go:

82.42 kg*m2 + m*R2

You said R is the distance between the center of the wheel and the center of the spokes, right?

82.42 kg*m2 + (107kg)(0.44m + 0.36m)2

= 150.9 kg*m2

Is this what you meant?

11. Nov 21, 2013

### BMcC

Oh, nevermind. I see what you meant. I was supposed to add the m*r^2 to the original spoke moment of inertia. Thanks for your help!

12. Nov 21, 2013

### Tanya Sharma

You are welcome :)

Last edited: Nov 21, 2013