PDA

View Full Version : Physics help (Rotational Inertia)

Deathslayer7
11-02-2008, 08:32 PM
see post 24.
:smallsmile:

averagejoe
11-02-2008, 08:37 PM
I know I=md^2, but not sure what it is asking for.

This is what it's asking for; rotational inertia and moment of inertia are the same thing. For multiple particles, I is equal to the sum of the I's of the individual particles (or the integral of d^2dm if it's a continuous distribution, which it isn't for this problem.)

Deathslayer7
11-02-2008, 08:39 PM
that comes later :smallwink:

first i just want to understand the beginning.

averagejoe
11-02-2008, 08:40 PM
I'm not sure what you mean; I was talking about part a).

Deathslayer7
11-02-2008, 08:44 PM
From what i understand, you have to find I for the three points at the particles. When you find the points, the rotational axis is always on one of those points, so that moment inertia is 0.

so rotational axis is at m1.

we get

I(1)= m2*d2^2 +m3*d3^2 ??? :smallconfused:

averagejoe
11-02-2008, 08:46 PM
Yes, exactly.

Deathslayer7
11-02-2008, 08:47 PM
but you also have to do that through all 3 points right?

averagejoe
11-02-2008, 08:52 PM
Yes; it's kind of like a three part question. It's like,

a)

i) What is the moment of the inertia if you rotate the object through m1, perpendicular to the xy plane?

ii) What is the moment of the inertia if you rotate the object through m2, perpendicular to the xy plane?

iii) What is the moment of the inertia if you rotate the object through m3, perpendicular to the xy plane?

Deathslayer7
11-02-2008, 08:55 PM
that is what i did, and i got 53, 58, and 117, which is what they got in the book. :smallsmile:

then part b. I'm not sure how to even approach that.

I know F= I*alpha, but somehow i dont think it is that simple.

averagejoe
11-02-2008, 09:08 PM
that is what i did, and i got 53, 58, and 117, which is what they got in the book. :smallsmile:

then part b. I'm not sure how to even approach that.

I know F= I*alpha, but somehow i dont think it is that simple.

Well, it's sort of that simple, except it's torque that equals I*alpha. The only other thing is now you have to take into account the angle of force, splitting it up into components. I don't have a clear picture of this, though, since I don't know where m1, m2, or m3 are with respect to the coordinates.

You do know polar coordinates, right?

Deathslayer7
11-02-2008, 09:12 PM
m1 is at (0,0)
m2 is at (0,3)
m3 is at (4,0)

that should help.

Deathslayer7
11-02-2008, 09:17 PM
------> Force
/
m2
| \
| \
| \
m1 ---------m3

looks something like that. hope that helps.

if you quote it, it looks better.

Deathslayer7
11-02-2008, 09:34 PM
i just had an epihany (sp).

I was wondering how they got T(z)= r(perpendicular)*F

but drawing out the force vector to m3, the axis of rotation, r is the distance between the vector that is perpendicular to the rotational motion.

averagejoe
11-02-2008, 09:36 PM
Okay, well, the first thing you want to do is change the coordinates so that (0,0) is at m3, because that's what's rotating and these problems are easier like that. Note that the problem doesn't change; you still get the same moment of inertia, this just makes the math easier. Also, your force vector is still the same vector; you can always transform vectors in space, and all that will change is where the vector is applied.

So now you want to change your system to polar coordinates (since that's generally easier with circular motion. Stuff like this is intuition you just have to build up.) by using the rules x/y=tan(t) and r^2=x^2+y^2, where I used t instead of theta. Now you have to decide how much of the force is in the t direction as opposed to the r, since force in the r direction won't provide any rotation. What you do is you attach the force to m2. Now think of two new lines, one following the rod between m2 and m3 (that is, in the "r" direction) and the other crossing "r", that is, going tangent to an imaginary circle centered at m3 with a radius of 5. Now you have your force of 4.5 newtons and 30 degrees from the horizontal, and you have to find out how much is in the r direction and in the theta direction. You can do this with sines and cosines, using the force vector as your hypotenuse. However, to do this you'll also want to find the angle between the line between m2 and m3 and the force vector, but that should be fairly straightforward geometry. (The reason why this can be more complex is r and theta change, whereas x and y are always the same. However, since you're only applying the force once this isn't an issue for this problem.)

Edit: Double ninja'ed; T(z) is just torque, I think. The definition of it is kinda weird.

Deathslayer7
11-02-2008, 09:42 PM
t(z) they have is torque parallel to the z-axis, which does kind of lead into the parallel axis theorem, but not here.

what they have:

theta = arcsin (3/5)
theta= 30+37 degrees = 67 degrees.

r(perpendicular)=5.0*sin(67)=4.6m

then they get T(Z)=20.7 N*m

Honestly, I might possibly understand your way, but i would need to see a drawing. Too hard in words. :smalltongue:

Deathslayer7
11-02-2008, 09:56 PM
I got it. Thanks for the help. :smallbiggrin:

After the epihany, it all came naturally.

Sad part was i just kept staring at the problem just trying to figure out what they did. :smalltongue:

averagejoe
11-02-2008, 10:02 PM
I got it. Thanks for the help. :smallbiggrin:

After the epihany, it all came naturally.

Sad part was i just kept staring at the problem just trying to figure out what they did. :smalltongue:

Unfortunately, that's a large part of physics. But, yeah, you're actually not bad at this, a lot of the time it seems like you already have the answer, you just need to realize you have it. However, I'm going to post the picture I made anyways. Note: not to scale. :smalltongue:

http://i171.photobucket.com/albums/u282/teddifunken/physics.jpg

The goal is to find the force in the direction of the line theta.

Deathslayer7
11-02-2008, 10:04 PM
and this is when you have m3 at the origin right?

Deathslayer7
11-02-2008, 10:07 PM
but isnt the force supposed to be at m2 not m3? :smallconfused:

averagejoe
11-02-2008, 10:08 PM
Well, it's valid for any coordinates, moving m3 to the origin was to make the calculation for torque easier. For this all you basically have to do is find the angle between F and theta (or F and r) and use cosine (sine) to get theta. All of this is completely independent of what coordinates you choose. I may have even jumped the gun a bit when advising you to change coordinates when I did, it's just one of those things that's become automatic for me.

Edit: D'oh, yes it is supposed to be at m2, I just became confused. In that case r would be sticking straight up from m2 and theta would be sticking horizontally to the left.

Deathslayer7
11-02-2008, 10:13 PM
possibly some other time i might learn your method. for now, i will stick with the xy coordinate system. :smalltongue:

averagejoe
11-02-2008, 10:15 PM
Oh, sorry, I thought you did know polar coordinates. Let me know when I do something like that; I don't always remember what people already know at the lower levels.

Deathslayer7
11-02-2008, 10:17 PM
i do know polar, but i haven't used them in forever. If my memory was refreshed, I'm pretty sure i could do them.

Like from calc 2.

start a t1, you have theta 1.
end at t2, have theta 2. The change in theta depends on how much of the curve you travelled, or some such thing.

Deathslayer7
11-02-2008, 11:01 PM
ok. so i'm look at sample problem 9-3, and they use the same figure as above, but this time they ask for a rotational mass from the center of mass of the figure.

i don't understand how they did that.

x(cm) = summation of m(n)*x(n)/summation of m(n)

then y(cm) summation of m(n)8y(n)/summation of m(n)

averagejoe
11-02-2008, 11:17 PM
Well, first you have to find the center of mass, which is just a weighted average of all the masses. It's basically vector addition, with longer vectors corresponding to particles with more mass. You have:

[m1(0,0)+m2(0,3)+m3(4,0)]/[m1+m2+m3]

Note that I'm using a more general definition of center of mass than you; however, they are equivalent since you have to add up the x and y entries separately. So the x coordinate of the center of mass is:

xc=[0(m1)+0(m2)+4(m3)]/(m1+m2+m3)

and the y coordinate is:

yc=[0(m1)+3(m2)+0(m3)]/(m1+m2+m3)

This is just the definition of the center of mass.

So, now you just get the rotational mass the same way. The calculations are a bit harder, but the concepts are the same. You still have I=sum(md^2), but the d's are harder to calculate since you have to calculate the distance between the center of mass and m1, m2, and m3.

Deathslayer7
11-02-2008, 11:22 PM
ok. I just wasn't sure how they found the center of mass, but that makes sense. Thanks. :smallsmile:

they then just used right triangles to find the distance.

ex:

r^2 = x(cm)^2 + y(cm)^2

averagejoe
11-02-2008, 11:28 PM
ok. I just wasn't sure how they found the center of mass, but that makes sense. Thanks. :smallsmile:

they then just used right triangles to find the distance.

ex:

r^2 = x(cm)^2 + y(cm)^2

Yes, that's right. They found the center of mass by experiment, if I'm not mistaken. I've never seen a mathematical derivation of it, but that doesn't mean they're not out there.