# Simple Moment of Inertia problem

1. Aug 19, 2009

### controlfreaks

1. The problem statement, all variables and given/known data
A dunmbell is made up of a solid rod of mass 0.5Kg and two solid spheres each of mass 2Kg. The rods length is 1.6m and the spheres are both of radius 0.2m.
1. I need to find the moment of inertia of the dumbell about an axis through its centre of mass and perpendicular to the rod.
2. Also i need to find the moment of inertia of the dumbell about an axis through the centre of one of the spheres and perpendicular to the rod

2. Relevant equations
Moment of interia of rod about it's centre is I = (M x L^2)/12
Moment of interia of solid sphere about it's centre is I = (2 x M x R^2)/5

3. The attempt at a solution
I did the porblem using these equations but i'm gettin complety different answers to the right ones. Can anybody point me in the right direction with this? Is there another equation i'm missing here?

2. Aug 19, 2009

### kuruman

For the moment of inertia of a sphere you probably used
$$I = \frac{2}{5}m R^{2}$$

about an axis that goes through the centre of the sphere. That's not what you want here. You want the moment of inertia about an axis that goes through the centre of mass of the dumbell. To do this, use the parallel axes theorem. Same for part (2).

For future reference: Please show what you did, even if you know it is not the right answer. It is easier to help you if we know where you went wrong instead of guessing.

3. Aug 19, 2009

### controlfreaks

so will there be 2 axes? both in the Z direction through the spheres on the dumbell? is that why the Parallel Axis Theorem is used. I've looked up about it but what i need to do here is still unclear to me.

4. Aug 19, 2009

### kuruman

The parallel axes theorem says that the moment of inertia of an object about an axis that is parallel to an axis going through the object's centre of mass (CM) at distance d is given by

$$I =I _{CM} + md^{2}$$

where m is the mass of the object. So if you want to find the moment of inertia of one of the spheres about the centre of mass of the dumbbell, you will need

(a) The distance between the centre of the sphere and the centre of mass of the dumbbell (that's d)
(b) The moment of inertia of the sphere about its centre of mass (you know that already).

You need to do the same for the other sphere and then the connecting rod and add the three terms.

5. Aug 19, 2009

### controlfreaks

ok here's what i got

i've used the old formula (2mr^2)/5 to work out Icm of one of the spheres and got an answer of 32

but when i plug that into the formula with I = Icm + md^2 i'm not getting the right answer.

the answers for the two problems on this are

a) 4.171
b) 8.671

I think i'm going wrong getting Icm!

6. Aug 19, 2009

### kuruman

Posting just numbers does not help me figure out where you went wrong. I need to know exactly how you got these numbers in order to help you. What did you plug in, where and how did you get it in the first place?

7. Aug 19, 2009

### controlfreaks

sorry, here's what i did

m=2x10^3
r=0.2

Icm = (2mr^2)/5
Icm = [2(2x10^3)(0.2)(0.2)]/5 = 32

8. Aug 19, 2009

### kuruman

In the SI system (which is what you should be using here) the mass is expressed in kg. Since the mass of the sphere is already given in kilograms, multiplying by 1000 is incorrect. Please get into the habit of showing the units every time you put down a number. It is an unbelievably good habit.