Angular Momentum of Point Masses

[postfan]

Homework Statement

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.

Homework Equations

L=I_cm*w

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?

 

[voko]

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

 

[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?

 

[D H]

(Part A) Used the parallel axis theorem and got:
(m*9s^2/12+4s^2*m)*w

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

 

[voko]

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

 

[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

 

[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.?

 

[D H]

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

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?

 

[postfan]

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

 

[D H]

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 m_i about some axis, with point mass i at some distance d_i from the axis?

 

[postfan]

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

Is that right?

 

[voko]

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

 

[postfan]

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

 

[voko]

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

 

[postfan]

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

 

[voko]

There are four particles, not one.

 

[postfan]

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

 

[D H]

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

 

[postfan]

The average of the four masses.

 

[D H]

Why are you averaging? Once again,

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

 

[postfan]

Oh ok, so the MoI is m(2s)^2+2m(s)^2+4m(s^2), right?

 

[D H]

Correct. Now what about that second term, the one due to the parallel axis theorem?

 

[postfan]

The second term is m(2s)^2, right?

 

[D H]

Why m? What is the mass of the rid + four point masses system?

 

[postfan]

Oh the second term is 10m(2s^2), right?

 

[D H]

Yes.

 

[postfan]

OK, since I have got the correct moment of inertias and I know the angular velocity I have managed to get the right answers to the problems. Thanks for all your help!