Anyone good at physics?

[Kali the Flygon]

Hmm, okay. So if a rod about the middle was two independent rods rotating about the end, their lengths would be half... so (1/3)m(.5l)² = (1/12)ml². But wouldn't you have two of those rods? How come it isn't (1/6)ml²?

While you have two, which doubles the mass, you also have to add the two separate inertias together, which doubles the 1/6 to a 1/3.

I was thinking it was possibly a matter of units, but if you're saying it still doesn't work, I'm getting confused. Perhaps you can send me your step by step work in a PM, and I can check it myself.

 

[opaltiger]

I can contribute nothing to the physics, but if you've checked your work multiple times and keep getting the same result, the problem probably isn't in your end. Textbooks get solutions wrong all the time, and I have even less faith in online quizzes.

 

[Sandstone-Shadow]

So it turns out the teacher thought I was doing it right too and just gave me credit for the problem... hrr.

Hmm, okay. So if a rod about the middle was two independent rods rotating about the end, their lengths would be half... so (1/3)m(.5l)² = (1/12)ml². But wouldn't you have two of those rods? How come it isn't (1/6)ml²?

While you have two, which doubles the mass, you also have to add the two separate inertias together, which doubles the 1/6 to a 1/3.

But isn't the inertia for each of them (1/12)ml²? Or wait, it would actually be (1/12)(1/2)ml² since the mass would be half too, which would equal (1/24)ml², so then if you have two of those, you'd get (1/12)ml². Ahh, I understand that now. Thanks! =D

Alright, we started something new and the teacher hardly explained it at all, and basically shot me down when I tried to ask about it. How would I do a problem like this?

Given M = 2 i + 2 j - 6 k and N = i - 6j - k, calculate the vector product M multiplied by N.

We learned how to calculate cross products, but only for vectors with just i and j components, not i, j, and k. And the answer is looking for a vector with i, j, and k components. I'm confused, not only because I don't know how to take the cross product of something with three terms, but I'm also confused because when taking the cross product of vectors with just i and j components, you end up with just k components. Now apparently the i and j stay in there too?

 

[opaltiger]

First make a 3x3 matrix where the top row consists of the unit vectors and the other two rows of the coefficients in both vectors (so second row is 2, 2, -6 and third is 1, -6, -1).

Then calculate the determinant of that matrix. It's as easy as that. If you don't know how to do that, I'll try and explain, though it's a little difficult in words.

Cover the first column (the i column). Then find the determinant of the 2x2 matrix which is left (if you ignore the j and k elements). Multiply this by i. Next cover the j column and find the determinant of the new 2x2 matrix formed from the numerical elements. Multiply this by j and subtract it from the first result. Finally, cover the third column, calculate the 2x2 determinant, multiply by k, and add to the first two.

Full disclosure: instead of subtracting the j result, you can reverse the order of the two columns and add it instead. But I find this way easier.

 

[Kali the Flygon]

To say the same thing opaltiger said in a similar, but slightly different way, if you know how to do cross products of 2-component vectors, you can do them for 3-component vectors in the following way:

The cross product of two 3-component vectors is another 3-component vector, X.

Remove the i components of the two vectors, and do the cross product between just the j and k components. The answer will be the i component of X. Then, remove the j components of the two initial vectors, and, importantly... switch the order of the k and i components, so that k comes before i. Then do the cross product with just the k and i components, and write the answer as the j component of X. Finally, remove the k components of the two initial vectors and perform the cross product between the i and j components, and write the answer as the k component of X.

In purely mathematical speak, if:
M = ai + bj + ck,
N = di + ej + fk, then

X = MxN = (b*f - c*e)i + (c*d - a*f)j + (b*d - a*e)k.

Hope this helps.

 

[Sandstone-Shadow]

Ah, okay... that makes sense, thanks guys! =)

Ugh, I'm stuck again. We're studying angular momentum, and I've tried reading over my notes from class, looking at the book, and looking at various other webpages. What I'm figuring out is that angular momentum (for, I'm assuming, a point) is r (the position vector) crossed with p (the linear momentum) so you end up with L = r X mv.

And I also found that for a system of particles, angular momentum is L = Iω.

First of all, is what I posted correct?

And second of all, how do I use those things to solve something like:

The position vector of a particle of mass 1.70 kg is given as a function of time by r = (6.00 i + 5.10 t j) m. Determine the angular momentum of the particle about the origin, as a function of time.

For this one, L = (6i + 5.1tj) X 1.7v. (Right?) But it doesn't give you v at all; this seems like a really basic concept to me but I can't figure it out.

I'm running into similar issues in other problems, but the other problems are all more complicated than the one that I posted, argh.

 

[Kali the Flygon]

Yes, those equations look correct, though it's I x ω, as opposed to other forms of products. I assume that's what you meant.

The velocity is related to the change in the radius with respect to time, in particular v = dr/dt.

 

[Sandstone-Shadow]

Yes, those equations look correct, though it's I x ω, as opposed to other forms of products. I assume that's what you meant.

Hmm, okay. Our teacher told us it was just Iω, because for some "pure rotation" thing the angle between the position vector r and the velocity v is always 90 degrees? I really wasn't sure what she meant though, and she never explained it or explained what "pure rotation" is. Is there any truth to what she's saying?

The velocity is related to the change in the radius with respect to time, in particular v = dr/dt.

Ah... awesome, I finally got one right! =D Thank you so much for your help so far, I really appreciate it.

 

[Kali the Flygon]

Well, I suppose you're right about Iω, considering the direction of the inertia is just the axis of rotation, and the angular velocity measures in the perpendicular direction to that axis always.

Glad I could help... even if it's just to get you to write things out and spur your own thoughts... *references visitor message*