Total moment of inertia of a two particle system

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Homework Help Overview

The discussion revolves around calculating the total moment of inertia of a two-particle system with respect to the y-axis. Participants are exploring the definitions and calculations related to moment of inertia, specifically for particles positioned at varying distances from the axes of rotation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the distances of particles from the axes and the corresponding moment of inertia calculations. There are questions about the assumptions regarding mass and the need for different axes of rotation. Some participants express confusion about the total moment of inertia and its relation to the y-axis.

Discussion Status

The discussion is active with participants sharing their calculations and questioning the correctness of their results. Some guidance has been offered regarding the interpretation of the problem and the calculations involved, but there is no clear consensus on the correct total moment of inertia.

Contextual Notes

There is a lack of explicit information about the masses of the particles and the setup of the problem, leading to assumptions being made. Participants are also referencing a diagram that is not provided in the thread.

Linus Pauling
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1. Find the moment of inertia I_x of particle a with respect to the x-axis (that is, if the x-axis is the axis of rotation), the moment of inertia I_y of particle a with respect to the y axis, and the moment of inertia I_z of particle a with respect to the z axis (the axis that passes through the origin perpendicular to both the x and y axes).

Particle a is located 3r from the y axis, and particle b is located r away.



2. I = mr^2
I = SUMM(m_i*r_i^2)



3. I_a = 9mr^2
I_b = mr^2

Total I = 10mr^2

No idea what I am doing wrong
 
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Linus Pauling said:
Particle a is located 3r from the y axis, and particle b is located r away.
Give the complete coordinates and the mass of each particle.
 
Particle a:

distance from y-axis = 3r
distance from x-axis = r

Particle b:

distance from y-axis = r
distance from x-axis = -4r

It doesn't say anything explicitly about mass, and so I think we assume each particle has mass m.
 
Linus Pauling said:
3. I_a = 9mr^2
I_b = mr^2

Total I = 10mr^2

No idea what I am doing wrong
Looks like you've found I_y. What about I_x?
 
I_x would be m(r)^2 + m(-4r)^2 = mr^2 + 16mr^2 = 17mr^2.

Am I simply adding that to I_y to get I_total? It asks for total I with respect to the y axis, though, why would I need I_x?
 
Shoot, just realized I posted to wrong question. Here's what I am trying to answer:

Find the total moment of inertia I of the system of two particles shown in the diagram with respect to the y axis
 
Linus Pauling said:
Here's what I am trying to answer:

Find the total moment of inertia I of the system of two particles shown in the diagram with respect to the y axis
Assuming the diagram matches your description, your answer seems correct to me. (Post the diagram, if you can.)
 
134726A.jpg
 
Looks good to me. I say your answer is correct.
 
  • #10
Ok, I requested the correct answer: 11mr^2

WTF?!
 
  • #11
Linus Pauling said:
Ok, I requested the correct answer: 11mr^2
Sounds bogus to me. You might want to mention this to your instructor.
 
  • #12
Ok, now:

Using the formula for kinetic energy of a moving particle K=\frac{_1}{^2}mv^2, find the kinetic energy K_a of particle a and the kinetic energy K_b of particle b.
Express your answers in terms of m, omega, and r separated by a comma.

I know I need to calculate linear speed, how do I do that? I know it's v=omega*r but I don't see how to put this together...
 
  • #13
Linus Pauling said:
I know I need to calculate linear speed, how do I do that?
What's the relationship between tangential speed and angular speed for something going in a circle?
 
  • #14
Just edited before you replied: v=omega*r

But what are omega and r here?!
 
  • #15
Particle A:

Ka = 0.5m(3*omega*r)^2

Kb = 0.5m(omega*r)^2

Apparently Kb is wrong but I did it the same way as Ka....
 
  • #16
Ok I got Kb = m(omega*r)^2

What is total kinteic eneregy?
 
  • #17
Linus Pauling said:
Ka = 0.5m(3*omega*r)^2

Kb = 0.5m(omega*r)^2

Apparently Kb is wrong but I did it the same way as Ka....
These both look OK to me.

Linus Pauling said:
Ok I got Kb = m(omega*r)^2
How did you get that? (Are you posting this problem exactly as given, word for word?)

What is total kinetic energy?
Just add up the kinetic energy of each mass.
 

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