Total moment of inertia of a two particle system

[Linus Pauling]

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

 

[Doc Al]

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.

 

[Linus Pauling]

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.

 

[Doc Al]

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?

 

[Linus Pauling]

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?

 

[Linus Pauling]

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

 

[Doc Al]

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.)

 

[Linus Pauling]

 

[Doc Al]

Looks good to me. I say your answer is correct.

 

[Linus Pauling]

Ok, I requested the correct answer: 11mr^2

WTF?!

 

[Doc Al]

Ok, I requested the correct answer: 11mr^2

Sounds bogus to me. You might want to mention this to your instructor.

 

[Linus Pauling]

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...

 

[Doc Al]

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?

 

[Linus Pauling]

Just edited before you replied: v=omega*r

But what are omega and r here?!

 

[Linus Pauling]

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....

 

[Linus Pauling]

Ok I got Kb = m(omega*r)^2

What is total kinteic eneregy?

 

[Doc Al]

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.

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.