# Moment of inertia from diffrent axis

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1. Jun 6, 2015

### **Mariam**

1. The problem statement, all variables and given/known data
Two particles of mass m1 = 14 kg and m2 = 24 kg are connected by a massless rod of length 1.4 m. Find the moment of inertia of this system for rotations about the following pivot points. Assume the rotation axis is perpendicular to the rod.

A) center of rod
B) end of m1
C) end of m2
D) at center of mass

2. Relevant equations
I=1/12 m l^2
I=1/3 m l^2
I'=Icm+md

3. The attempt at a solution
first I calculate the center of mass :
Xcm= ((x1m1)*(x2m2))/(m1+m2)
= (1.4*24)/(14+24)=
Xcm= 0.88 m from m1

for A) and D)
(i am confused which to use for the formula of I=1/12 m l^2 and which to use parallel axis theorem.)

for B)
I=1/3(24)(1.4)^2
I=15.68 kgm^2

(here i am a bit confused as to whether or not to add the 14 kg mass. So is it correct or not?)

for C)
I=1/3(14)(1.4)^2
I=9.15 kgm^2
(same problem as B)

for D)
first I calculate the center of mass :
xcm= ((x1m1)*(x2m2))/(m1+m2)
= (1.4*24)/(14+24)=0.88 m from m1

Extra question: What if the rod had a mass? How will moment of inertia be then?

Sorry, if my questions are basic but I have a physics test tomorrow and my teacher didn't explain moment of inertia a lot.

2. Jun 6, 2015

### ehild

Your relevant equations refer to the moment of inertia of a rod, and m means the mass of the rod. It is zero now.
What is the moment of inertia of a point mass m if it is at distance r from the axis of rotation?

3. Jun 6, 2015

### **Mariam**

Hello, yes I see my mistake...
So I need to use I=mr^2 instead
But how about my other questions is the way I solved it right?

4. Jun 6, 2015

### SammyS

Staff Emeritus

B) & C) are incorrect. What is the (1/3) doing there?

D) Your formula for C.M. should have + in the numerator, not * .

xcm= ((x1m1)+(x2m2))/(m1+m2)

5. Jun 7, 2015

### erisedk

I have a related question, sorry if it's not exactly relevant to the OP.
For part (D), considering the two masses to be m1 and m2, and distance between them l, instead of figuring out the centre of mass and then taking the necessary distances, using reduced mass simplifies the math, i.e. Icom= (m1m2)/(m1+m2) . r^2.

6. Jun 7, 2015

### **Mariam**

Ok here is my solution:
A) I= 14(0.7)^2 + 24*(0.7)^2= 18.6 kgm^2

Here I used the length to each mass as half the total length of rod
B) I= 24*(1.4)^2= 47.04 kgm^2

C) I=14*(1.4)^2= 27.44 kgm^2

D) xcm=0.88 m
So I = 14*(0.88)^2+24*0.52^2= 17.33 kgm^2

Is this correct? For my extra question about mass having a rod and the question mentioned by erisedk, I don't have time to try them now because I am going to my exam now :/
Anyway thanks a lot for your help :)

7. Jun 7, 2015

### SammyS

Staff Emeritus
Yes. Those look good.

8. Jun 7, 2015

### **Mariam**

First of all, I just finished my exam, and a very similar question came but with three masses, so I think I did well :)

Next, I'm here again for the extra question about having a rod with mass.
Will the calculations be like this...
Let's say the rod had mass 50 kg...
A) I=14*0.7^2+ 24*0.7^2 + (1/12*50*1.4^2 + 50*(0.88-0.7))= 35.8 kgm^2
Here I used parallel axis theorem because it's not center of mass is this right??

B) I=24*0.7^2 + (1/3* 50* 1.4^2)= 44.4 kgm^2
C) I= 14*0.7^2 + (1/3* 50* 1.4^2)= 39.5 kgm^2

D) I= 14*0.7^2+ 24*0.7^2 + (1/12*50*1.4^2) = 26.8 kgm^2

9. Jun 7, 2015

### ehild

Question A) asks the angular momentum with respect to the centre of the rod. It is the sum of the momenta all with respect that centre. The momentum of the rod with respect its centre is 1/12 mL2. There is no sense of the fourth term.

The 0.7 is wrong, m2 or m1 are at distance of 1.4 m from the axis of rotation.

That is when you have to use the parallel axis theorem. The question asks the moment of inertia about the CM. As the rod has mass, you have to take it into account when determining the CM. Then use the distances of the individual masses from the new CM to calculate the contribution to the moment of inertia, and Parallel axis theorem to get the contribution from the rod (the distance of the centre of the rod from the CM, and the momentum with respect to the middle of the rod) .

Last edited: Jun 7, 2015
10. Jun 7, 2015

### **Mariam**

Thanks I understand my mistakes now.

So basically the 1/12 ml^2 is for center of rod and not center mass

And for B and C I just copied and pasted it without changing the 0.7. Silly mistake

11. Jun 7, 2015

### ehild

1/12 mL2 is for the center of mass of a single homogeneous rod, which is at the middle of the rod.

12. Jun 7, 2015

### erisedk

13. Jun 7, 2015

### ehild

It is simple Math. m1 is at distance $d_1=\frac{m_2L}{m_1+m_2}$ from the CM and m2 is at distance $d_2=\frac{m_1L}{m_1+m_2}$ The moment of inertia is $I=m_1d_1^2+m_2d_2^2=m_1\left(\frac{m_2L}{m_1+m_2}\right)^2+m_2\left(\frac{m_1L}{m_1+m_2}\right)^2$. You can factor out $\frac{m_1 m_2}{m_1+m_2}L^2$,
$I=\frac{m_1 m_2}{m_1+m_2}L^2\left(\frac{m_2}{m_1+m_2}+\frac{m_1}{m_1+m_2}\right)$.
The last factor is equal to ... ?
Two-body problems usually solve with the method of reduced mass. We place the CM in the origin of the frame of reference, and we substitute the two particles with a single one with the reduced mass, and we use the distance between the two bodies as coordinate.

Last edited: Jun 7, 2015
14. Jun 7, 2015

Ok thanks!