- #1

- 40

- 1

Alright, I absolutly do not understand this concept. I missed the lecture and the textbook does not seem to explain it very well.

Here is a sample question:

Two particles, each with mass m = 1.10 kg, are fastened to each other, and to a rotation axis at O, by two thin rods, each with length d = 0.670 m and mass M = 0.252 kg. The combination rotates around the rotation axis with angular speed w = 0.417 rad/s. Measured about O, what are the combination's (a) rotational inertia and (b) kinetic energy?

http://www.intercomrealestate.com/FileSystem/Fig10_35.gif [Broken]

I know how to get [tex]E_k[/tex] from the rotational intertia, pretty much just plug and chug, but as far as finding the rotational intertia...

When I opened my textbook I just saw a lot of integrals and paniced. Well, actually they give you the common integals for all the shapes, the one for a rod is [tex]\frac{1}{12}ML^2[/tex]. But I'm not sure what I'm supposed to do with that.

I have tried to do the question, but with no avail. But I will attempt to give some working that I've done.

[tex]I = \sum m_i*r_i^2[/tex]

[tex]I = \frac{1}{12}ML^2[/tex] (For rod around central axis)

[tex]I = I_com + Mh^2[/tex] (Parallel axis theorm)

Applying all that I got:

[tex]md^2 + m(2d)^2 + \frac{1}{12}M(2d)^2 + Md^2[/tex]

Its not right, and I'm not 100% sure why.

Edit: Oh wait, there are two rods. Ah, well, I'll try to figure it out from there. Any help is still appreciated

Edit2: [tex]md^2 + m(2d)^2 + \frac{1}{12}(2M)(2d)^2 + (2M)d^2[/tex]

Heh; figured it out. Wish I could delete this thread.

Here is a sample question:

Two particles, each with mass m = 1.10 kg, are fastened to each other, and to a rotation axis at O, by two thin rods, each with length d = 0.670 m and mass M = 0.252 kg. The combination rotates around the rotation axis with angular speed w = 0.417 rad/s. Measured about O, what are the combination's (a) rotational inertia and (b) kinetic energy?

http://www.intercomrealestate.com/FileSystem/Fig10_35.gif [Broken]

I know how to get [tex]E_k[/tex] from the rotational intertia, pretty much just plug and chug, but as far as finding the rotational intertia...

When I opened my textbook I just saw a lot of integrals and paniced. Well, actually they give you the common integals for all the shapes, the one for a rod is [tex]\frac{1}{12}ML^2[/tex]. But I'm not sure what I'm supposed to do with that.

I have tried to do the question, but with no avail. But I will attempt to give some working that I've done.

[tex]I = \sum m_i*r_i^2[/tex]

[tex]I = \frac{1}{12}ML^2[/tex] (For rod around central axis)

[tex]I = I_com + Mh^2[/tex] (Parallel axis theorm)

Applying all that I got:

[tex]md^2 + m(2d)^2 + \frac{1}{12}M(2d)^2 + Md^2[/tex]

Its not right, and I'm not 100% sure why.

Edit: Oh wait, there are two rods. Ah, well, I'll try to figure it out from there. Any help is still appreciated

Edit2: [tex]md^2 + m(2d)^2 + \frac{1}{12}(2M)(2d)^2 + (2M)d^2[/tex]

Heh; figured it out. Wish I could delete this thread.

Last edited by a moderator: