Physics help please n' thank you

[Deathslayer7]

Ok. So I'm still doing rotational inertia of solid bodies.

I understand how to do it for a rod, at any rotational point. Now i need help understanding it for a circle, but before i understand it for a circle, i need to understand it for a hoop.

So here are my thoughts:

So we have I = integral of r^2*dm.

using a hoop, we label the radius (R, so we can differentiate it from little r)

dm= p(rho)*dV

dV= dA/A=2*pi*r*dr/pi*R^2 (this is where i dont understand how they get this).

p(rho)=M/V=M/pi*R^2

Help? :smalleek:

[Deathslayer7]

forgot. rotational axis is through center.

[Deathslayer7]

correction i can do a hoop through the center, but not through any diameter.

[Deathslayer7]

Bueller Bueller Bueller.

Sorry. Got distracted waiting for someone.

So I take it no one can help me derive the formula for rotational inertia of a circle. :smallconfused:

[Jack Squat]

does this (http://library.thinkquest.org/16600/advanced/rotationalinertia.shtml) help?

I've been out of physics for a bit, so I can't say I remember enough to try and help someone with accuracy. Your best bet may infact be Google.

[Deathslayer7]

unfortuneately, not really. They show the proof for a rod. I need the proof for a circle/sphere. :smallsigh:

[Griever]

Attempting to figure this one out for you, harder than I expected. :smallredface:

I'm assuming "r" is being taken as the distance from the axis of rotation to the hoop, no? And "R" being the radius of the hoop/sphere (sphere is just a continuation on the hoop, anyway)

dV = dA/A is really throwing me for a loop (no pun intended)...I can't seem to figure out how a differential unit of volume * area = a differential unit of area... :smallannoyed:

[Deathslayer7]

I have the notes, and will type it up soon. Maybe that will help you put you on the right track. Expect it up soon. Within 5-10 minutes most likely.

[Deathslayer7]

I think this is hoop about any diameter. Not sure. Don't have it labeled. :smalltongue:

dm=(dA/A)*M=(area of hoop/total area)*Mass

dm=2*M*r*pi*dr/(R^2*pi)

pi's cancel giving you

dm= 2*M*r*dr/(R^2)

I= integral of r^2*2*M/(R^2)*dr*r

=2*M/(R^2)*the integral of r^3*dr

=2*M/(R^2)*(1/4)r^4 evaulated from 0 to R

=2*M/(R^2)*(1/4)*R^4

=1/2*M*R^2

Note:little r and big R are different. That is what confuses me.

Added in parenthesis so confusion is at a minimal.

[Griever]

I think this is hoop about any diameter. Not sure. Don't have it labeled. :smalltongue:

dm=(dA/A)*M=(area of hoop/total area)*Mass

dm=2*M*r*pi*dr/(R^2*pi)

pi's cancel giving you

dm= 2*M*r*dr/(R^2)

I= integral of r^2*2*M/(R^2)*dr*r

=2*M/(R^2)*the integral of r^3*dr

=2*M/(R^2)*(1/4)r^4 evaulated from 0 to R

=2*M/(R^2)*(1/4)*R^4

=1/2*M*R^2

Note:little r and big R are different. That is what confuses me.

Added in parenthesis so confusion is at a minimal.

I think I can shed some light on r vs. R, but not yet on *why* dm = dA/A * M... that part bugs me a lot.

I'm pretty sure "R" is a constant for the radius, which we simply integrate "r" over, as "r" is a changing variable that relates to the equation the distance the mass is from the rotational axis (which inertia is very dependent on).

dm = dA/A * M... grrrr, why don't you make any sense...

M is the mass of the hoop?
dA is the area of the hoop?
A is the area of the inside of the hoop?

[Deathslayer7]

dm = mass carried by the volume.

[Griever]

dm = dA/A * M... grrrr, why don't you make any sense...

M is the mass of the hoop?
dA is the area of the hoop?
A is the area of the inside of the hoop?

Wait... just thinking about this.

We are assuming the hoop to have a thickness of 0 (in theory), correct?

Hmmm... what if A is supposed to be the area of the hoop (2 * pi * r * h) and dA a small (read: tiny) piece of that area, which then sets up a ratio to get a differential piece of mass? Hmmm, that seems plausible.

[Deathslayer7]

honestly, i couldn't tell you, because for each shape, dm varies. Espcially if you have a rod and cut it in half.

Now density is no longer constant. :smallsigh: Then you can have the rotational axis anywhere you want. :smallsigh::smallsigh:

Life sucks. :smallsigh::smallsigh::smallsigh:

[Deathslayer7]

On another note, can you help me with a problem?

Problem:

A small lead sphere of mass 25g is attached to the origin by a thin rod of length 74 cm and negilible mass. The rod pivots about the z-axis in the xy plane. A constant force of 22 N in the y direction acts on the sphere.

a) Considering the sphere to be a particle, what is the rotational inertia about the origin.

b) if the rod makes an angle of 40 degrees with the positive x axis, find the angular acceleration.

For part a, i got 0.14, but in the back of the book, they got 0.014, which i think is a typo.

for part b, i am confused.

We have our I, we found it in part a.

Torque =I*alpha

Torque=F*r*sin(theta)

so, F*r*sin(theta)/I= alpha

=74.7 degrees per sec^2,

but the answer they have is 91 rad/s^2.

When i try to convert the degrees to radians, i get the wrong answer.

[Griever]

This post to be editted with my answer for that question. Just posting to show that I am, in fact, paying attention.

for a, real quick, though.

I = mr^2

m = 25 g
r = .074 m

I get I = .1369

[Deathslayer7]

sig digits, rounds to .14, which is what i got. :smalltongue: Sorry. I mistyped the answers.

I got .14, while the back of the book got .014. :smallredface:

[Griever]

Wait, 74 cm = .74 m, right?

[Deathslayer7]

yes indeed it does. :smallsmile:

[Griever]

yes indeed it does. :smallsmile:

Then... .74 ^ 2 = .5476 then multiplied by 25 = 13.69 for the I... right? :smallconfused:

[Deathslayer7]

multiplied by .025.

The mass they give you is in grams, you want to change it to kg.

and arent you supposed to be helping me? :smallconfused::smalltongue:

wait. that means i did have it right.

0.014 is the answer i get for I :smallsigh:

[Griever]

Found your mistake, you are using the wrong angle.

The angle made between F and F (perp.) is 40 degrees, which should then go into a cos, not a sin.

Make sense?

[Deathslayer7]

no. why is it cos and not sin? :smallconfused:

[Griever]

Draw out the force diagram and you'll get a right triangle on one of its non-90 corners with the 90 degree corner on the left balancing on top of the sphere.

You can figure out using geometry that the angle between F and F (perp.) are 40 degrees, and since you want the non-hypotenuse side that is next to that corner, you use cos (adj/hyp)

[Deathslayer7]

I think i got it. I'll draw a quick picture in paint to make sure.

no. I'm still confused. Will post the pic, but not sure what is what.

[Deathslayer7]

http://i198.photobucket.com/albums/aa125/death_slayer7/Physics1.jpg


yay nay? :smallconfused:

[Griever]

Negative y direction? Was assuming, since you didn't say otherwise, it was in the positive y.

[Deathslayer7]

it doesnt specify. so i assume it would go up then.

[Deathslayer7]

ok. can you draw a quick pic, and show me? :smallconfused:

[Griever]

http://i287.photobucket.com/albums/ll125/Griever__/ForceDiagram2.jpg

That is... rather small.... one moment.

EDIT: That legible yet?

[Deathslayer7]

kinda. :smalltongue: I'll live with it though. Got it drawn bigger on my paper. :smallsmile:

[Deathslayer7]

ok. So on the upper angle, how do you know which one is theta? :smallconfused:

[Griever]

http://i287.photobucket.com/albums/ll125/Griever__/ForceDiagram2-1.jpg

I can take either angle to be theta. If I use 50, I'll go with sin, if I use 40, cos.

[Deathslayer7]

ok. i understand where i messed up. I forgot that the distance is r perpendicular. I got it. Thanks.

What would i do without you and averagejoe, I dont know, and is too horrible to contemplate. :smalleek:

[Deathslayer7]

I thank you for staying up so late though to help me. You are about two hours ahead of me, and it is almost 11 pm here.

So thanks. :smallsmile:

[Griever]

I thank you for staying up so late though to help me. You are about two hours ahead of me, and it is almost 11 pm here.

So thanks. :smallsmile:

Don't worry about it. :smallwink:

Calculus 3 tomorrow, best get alittle sleep. Night.