# Angular Momentum of Point Masses

1. Aug 17, 2013

### postfan

1. The problem statement, all variables and given/known data

A light rigid rod of length 3s where s = 1.5 m has small spheres of masses m = 0.24 kg, 2m, 3m and 4m attached as shown. The rod is spinning in a horizontal plane with angular speed w = 2.5 rad/s about a vertical axis.

Find the magnitude of angular momentum of the rod in each case. Enter your answers in [kg.m2.s-1].

Part A. The axis of rotation is passing through mass m.
Part B. The axis of rotation is passing through mass 2m.
Part C. The axis of rotation is passing through mass 3m.
Part D. The axis of rotation is passing through mass 4m.
Part E. The axis of rotation is passing through the center of mass of the configuration.
Part F. The axis of rotation is passing through the midpoint of the rod.
2. Relevant equations

L=I_cm*w

3. The attempt at a solution

Found the center of mass which is s/2

(Part A) Used the parallel axis theorem and got:
(m*9s^2/12+4s^2*m)*w=19/4*ms^2*w=4.75*.24*1.5^2*.25=6.4125, which is wrong. What am I doing wrong?

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Last edited: Aug 17, 2013
2. Aug 17, 2013

### voko

I do not think the C. o. M. is at s/2. Show your calculation.

3. Aug 17, 2013

### postfan

Taking the midpoint of the rod as 0:

m(-3s/2)+2m(-s/2)+3m(s/2)+4m(3s/2)

-3ms/2+3ms/2-ms+6ms=5ms/10m=s/2

Can't see what's wrong with it?

4. Aug 17, 2013

### D H

Staff Emeritus
Show how you obtained those two terms inside the parentheses (m*9s^2/12 and 4s^2*m).

5. Aug 17, 2013

### voko

If you define 0 as the midpoint, then your calculated result is the displacement from the midpoint. Where is that?

6. Aug 17, 2013

### postfan

voko-

The center of mass is at the 3m point.

D.M-

Center of mass around a point other that CM is = to ML^2/12+MD^2=M(3s)^2/12+M(2s)^2=
m*9s^2/12+4s^2*m

Last edited: Aug 17, 2013
7. Aug 17, 2013

### voko

Indeed, the 3m mass is at the C. o. M.

Now, how do you compute the moment of inertia about the C. o. M.?

8. Aug 17, 2013

### D H

Staff Emeritus
Neither one of those terms is correct.

ML2/12 is the moment of inertia of a thin rod of mass M and length L. You don't have a thin rod. You have four point masses attached to a light (i.e., essentially massless) rod. You can't use that formula for these point masses.

With regard to the second term, what is the total mass of the rod+ four point masses system?

9. Aug 17, 2013

### postfan

Ok so is the MoI just M(2s)^2? The total mass of the system is 10m.

10. Aug 17, 2013

### D H

Staff Emeritus
Don't just guess.

What's the moment of inertia of a single point mass m about some axis at some distance d from the axis?

What's the moment of inertia of a set of point masses mi about some axis, with point mass i at some distance di from the axis?

11. Aug 17, 2013

### postfan

1.MoI=md^2
2.MoI=m_i*d_i^2

Is that right?

12. Aug 17, 2013

### voko

Yes, these are correct. Show how you calculate the moment of inertia about the C. o. M. using these formulae.

13. Aug 17, 2013

### postfan

The MoI =m(2s)^2 right?

14. Aug 17, 2013

### voko

No, I do not think so. You have been asked to show how you compute that.

15. Aug 17, 2013

### postfan

The distance between the point particle and the CM is 2s, hence d=2s.

16. Aug 17, 2013

### voko

There are four particles, not one.

17. Aug 17, 2013

### postfan

OK so is the MoI = m(2s)^2+2m(s)^2+4m(s^2)/4?

18. Aug 17, 2013

### D H

Staff Emeritus
What's that "/4" at the end?

19. Aug 17, 2013

### postfan

The average of the four masses.

20. Aug 17, 2013

### D H

Staff Emeritus
Why are you averaging? Once again,