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CoffeeJanitor
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(05-12-2011, 05:56 AM)
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Hey, GAF. I'm reviewing for a physics final tomorrow and I'm currently freaking out because I've forgotten how to do some of the most simple problems there are. I am currently stuck on this problem from around the beginning of the course (I would ask my classmates for help but I'm a commuter so I can't):

Point charges of -9 C, 3.0 C, and +9 C are located along the x axis at x = -1.0 cm, x = 0, and x = +1.0 cm, respectively.

Locate a point on the positive x axis where the magnitude of the electric field is zero.

I really, really need some help here, because if I can't remember how to do this tonight I don't think I'll be able to really study anything else. I would really appreciate it if someone could just help me with the general strategy here.

Thanks if you can help...
Last edited by CoffeeJanitor; 05-12-2011 at 06:01 AM.
Neuromancer
The Mayuh of f'n Bawston
(05-12-2011, 05:58 AM)
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Well there's always your coffee janitor career to fall back on, if this whole physicist thing doesn't pan out.
DragonKnight
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(05-12-2011, 06:01 AM)
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I think you need to look at a point on the x axis where the summation of the forces of all three charges equals zero because Fe=qE. It has been a while though. You need Newton's third law and the first relationship I gave you. You also need to set up three distance relationships to describe each particles proximity from the other.
CoffeeJanitor
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(05-12-2011, 06:10 AM)
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Originally Posted by DragonKnight

I think you need to look at a point on the x axis where the summation of the forces of all three charges equals zero because Fe=qE. It has been a while though. You need Newton's third law and the first relationship I gave you. You also need to set up three distance relationships to describe each particles proximity from the other.

That sounds right. I think I was trying to total out the electric fields from each point charge so that's probably why I've been getting it wrong. I'll post results when/if I get it. Thanks for the advice.
bangai-o
Banned
(05-12-2011, 06:11 AM)
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yahoo answers
IGotBillySoSpooked
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(05-12-2011, 06:13 AM)
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Originally Posted by bangai-o

yahoo answers

Yeah, that is a surprisingly great place for physics answers. None of that "show me your work" crap that you get on Physics Forum or whatever.

If you had asked a year ago, I would have been able to help you out. I don't recall how to do any of this right now, though.
RBH
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(05-12-2011, 06:13 AM)
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I believe you just have to use the electric field equation for each charge and then add up all three answers so that they add up to 0. I don't think that you need to use the electric force equation.


Magnitude of Electric Field ( E ) = k(Q)/r


Or maybe I'm just totally off on this...
DragonKnight
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(05-12-2011, 06:13 AM)
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Originally Posted by CoffeeJanitor

That sounds right. I think I was trying to total out the electric fields from each point charge so that's probably why I've been getting it wrong. I'll post results when/if I get it. Thanks for the advice.

This one isn't too bad. Because everything is one the x axis you don't need any vectors either
DanteFox
Meticulously designed by GodManPig to be a few sticks short of a teepee.
(05-12-2011, 06:17 AM)
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set the sum of the electric fields equal to zero, and solve for the distance r.


so...


E= -9 C/(4*pi*e*[x-(-1)]^2)+ 3 C/ (4*pi*e*[x]^2) + 9 C/ (4*pi*e*[x - 1 cm]^2) = 0

multiply both sides by 4*pi*epsilon and that simplifies it to

- 9 C / (x + 1)^2 + 3 C / (x^2) + 9 C / (x -1)^2 = 0

solve from there.
CoffeeJanitor
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(05-12-2011, 06:17 AM)
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^^^Damn that's what I've been doing. I think that maybe I had my signs messed up? Hmmmmm

Originally Posted by RBH

I believe you just have to use the electric field equation for each charge and then add up all three answers so that they add up to 0. I don't think that you need to use the electric force equation.


Magnitude of Electric Field ( E ) = k(Q)/r


Or maybe I'm just totally off on this...

The issue is that they're all at different points on the x axis. So you get

kq1/(r^2)+kq2/(r+.01)^2+kq3/(r+.02)^2=0

Or at least that's how I had it set up. That seems to be much much too difficult to solve, though, and plugging it into wolfram gave me the wrong answer

Thanks to everyone for the help!

EDIT: GOT IT. VICTORY
Last edited by CoffeeJanitor; 05-12-2011 at 06:25 AM.
RBH
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(05-12-2011, 06:30 AM)
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Originally Posted by CoffeeJanitor

EDIT: GOT IT. VICTORY

Nice. How did you get it?
Plasmid
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(05-12-2011, 06:31 AM)
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Gravity makes you go down.

boom physics.
DanteFox
Meticulously designed by GodManPig to be a few sticks short of a teepee.
(05-12-2011, 06:33 AM)
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Originally Posted by CoffeeJanitor

^^^Damn that's what I've been doing. I think that maybe I had my signs messed up? Hmmmmm

The issue is that they're all at different points on the x axis. So you get

kq1/(r^2)+kq2/(r+.01)^2+kq3/(r+.02)^2=0

Or at least that's how I had it set up. That seems to be much much too difficult to solve, though, and plugging it into wolfram gave me the wrong answer

Thanks to everyone for the help!

EDIT: GOT IT. VICTORY

why do you have it as r+.02?
CoffeeJanitor
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(05-12-2011, 06:36 AM)
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Originally Posted by DanteFox

why do you have it as r+.02?

Typo. I just made a force diagram and that made everything much clearer.
DragonKnight
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(05-12-2011, 06:37 AM)
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Originally Posted by CoffeeJanitor

Typo. I just made a force diagram and that made everything much clearer.

It really does
DanteFox
Meticulously designed by GodManPig to be a few sticks short of a teepee.
(05-12-2011, 06:39 AM)
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Originally Posted by CoffeeJanitor

Typo. I just made a force diagram and that made everything much clearer.

did you get what wolfram alpha gave you or was it different? I actually had trouble getting hte answer by hand lol. Post the answer man!
CoffeeJanitor
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(05-12-2011, 06:46 AM)
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Originally Posted by RBH

Nice. How did you get it?

I'll try to make a diagram. Arrows represent what direction the E-field is pointed (negative charges have efield pointed toward them, postive point away). q1 = -9, q2 = 3, q3 = 9


-------->q1<------- <-------------q2------------> <----------q3------->

The problem states the e-field is in the positive x axis, which is located to the right of q2. so, if we look at that we can see that the e field due to q1 and q3 is pointed left (negative) and the e field due to q2 is positive (the zero point is in between q2 and q3 just by observation).

the equation thus becomes

-9/(x + .01)^2 +3/(x^2) - 9/(x -.01)^2 = 0

And then I just solved from there (and by solving from there I mean I used Wolfram to do it for me. No way I'm computing that).

Hope that made sense.

Originally Posted by DanteFox

did you get what wolfram alpha gave you or was it different? I actually had trouble getting hte answer by hand lol. Post the answer man!

Answer is .341 cm = .00341 meters. I'm cramming right now so I skipped the algebra. Looked really ugly lol
Last edited by CoffeeJanitor; 05-12-2011 at 06:52 AM.
RBH
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(05-12-2011, 06:53 AM)
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Originally Posted by CoffeeJanitor

I'll try to make a diagram. Arrows represent what direction the E-field is pointed. q1 = -9, q2 = 3, q3 = 9


-------->q1<------- <-------------q2------------> <----------q3------->

The problem states the e-field is in the positive x axis, which is located to the right of q2. so, if we look at that we can see that the e field due to q1 and q3 is pointed left (negative) and the e field due to q2 is positive (the zero point is in between q2 and q3 just by observation).

the equation thus becomes

-9/(x + .01)^2 +3/(x^2) - 9/(x -.01)^2 = 0

And then I just solved from there (and by solving from there I mean I used Wolfram to do it for me. No way I'm computing that).

Hope that made sense.


Answer is .341 cm = .00341 meters. I'm cramming right now so I skipped the algebra. Looked really ugly lol

Ah, now I see. Thanks for the explanation and best of luck on your final tomorrow.
CoffeeJanitor
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(05-12-2011, 07:01 AM)
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Originally Posted by RBH

Ah, now I see. Thanks for the explanation and best of luck on your final tomorrow.

Hey thanks. And once again I appreciate the help.
The Lamp
Banned
(11-16-2011, 03:50 PM)
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I wanted to bump this thread because I have a physics question as well.

We just got to angular momentum which I hear is a non-intuitive concept to grasp at first, and I'm having trouble.

We've got a problem that says,

A man, mass M, stands on a massless rod which is free to rotate about the center in the horizontal plane. The man has a gun (massless) with one bullet, mass m. He shoots the bullet with velocity Vb, horizontally. Find the angular velocity of the man as a function of the angle, theta, which the bullet's velocity vector makes with the rod.

So you've got a guy standing on one end of a rotating rod, and he shoots a bullet from an angle and starts moving.

My prof told me I have to determine first that the angular momentum in the direction we are moving is conserved. To do this, I must show that external torque in that direction is 0.

Since it's frictionless, the only two external forces acting on the man are gravity and normal force. When I use right-hand-rule and cross product of those two forces with the r displacement vector from the origin of pivot to the man, I get that their torques are in the horizontal axis, which has nothing to do with the direction I'm interested in.

L = r x p, so to determine the direction of angular momentum for the man, I do r x mv which she said the velocity vector was pointed into the board/paper



So that r x p = angular momentum in the upwards z-direction.

I don't understand why the velocity vector goes into the board? If the man shoots the bullet in that direction, he's going to move outwards from shooting the bullet inwards.

If anybody could clear up this one conceptual misunderstanding for me (which I've been looking all over the internet to try to understand) I would greatly appreciate it.
Fantastical
Death Prophet
(02-06-2012, 03:18 AM)
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Hello GAF, I have a simple question and I didn't want to start a new thread.

Basically someone is throwing a ball straight up on the moon (the acceleration due to gravity 1/6 that on Earth). All I'm given is the time of flight (27 seconds). The homework is asking for the maximum height.

How am I supposed to answer this? I feel like I need more information, but I'm really bad at Physics.
The Technomancer
card-carrying scientician
(02-06-2012, 03:19 AM)
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They don't give you the mass of the ball? Even as an arbitrary constant?
riceckr
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(02-06-2012, 03:23 AM)
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wouldn't you use your kinematic equations

d = v(f)t -1/2at^2

velocity final would be 0

d = -1/2(acceleration due to gravity) (time)^2

would it be half the time since we are measuring up and down
The Technomancer
card-carrying scientician
(02-06-2012, 03:24 AM)
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Yeah, I'm dumb, you don't need mass for this. There's an annoying two year hole in my Mechanical Engineering education where they don't have us do anything kinematic at all anymore, and I'm just emerging from the other side.
Fantastical
Death Prophet
(02-06-2012, 03:25 AM)
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Originally Posted by riceckr

wouldn't you use your kinematic equations

d = v(f)t -1/2at^2

But I don't know v(f) or d, right?
riceckr
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(02-06-2012, 03:26 AM)
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Originally Posted by Fantastical

But I don't know v(f) or d, right?

d would be what you are trying to find, distance. v(f) would be 0 at the top of the arc

like I said earlier, I think you may need to divide the time of flight by half
Fantastical
Death Prophet
(02-06-2012, 03:29 AM)
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Originally Posted by riceckr

d would be what you are trying to find, distance. v(f) would be 0 at the top of the arc

So you are saying I half the time to get the time it takes to get to the top of the arc? I thought about that, but it's someone throwing it (so there is an initial height from which it is thrown) and the time of flight is when it goes back to the surface, so I can't half it can I since it goes beyond the point it was thrown at, right?
Orayn
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(02-06-2012, 03:30 AM)
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The ball's position can be calculated by the following equation.

y = V*t - (1/2)*(1/6)*9.8*t^2

Where 'y' is the ball's height, 't' is time in seconds, and 'V' is the initial velocity of the ball when it's thrown upwards. Since they tell you that the total flight time is 27 seconds, you plug in 27 for 't' and solve to find the initial velocity.

Then, you use the equation for the ball's velocity to find out when the ball starts falling. To do that, take this equation for the ball's velocity, plug in V, and solve it for 't' when the velocity is equal to zero.

dy/dt = V - (1/6)*9.8*t

Where 'dy/dt' is the rate of change in the ball's position, also known as the velocity. This gives you the time when the ball reaches its maximum height. Put that time back into the first equation and you have your answer.

EDIT: This all assumes the ball was launched right from the surface and lands back on the surface. I did everything in as generic a way as I could. It's more work, but you'll find it useful in other situations.
Last edited by Orayn; 02-06-2012 at 03:39 AM.
riceckr
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(02-06-2012, 03:33 AM)
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^^

what he said, sorry it's been years
areal
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(02-06-2012, 03:38 AM)
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Originally Posted by riceckr

what he said, sorry it's been years

What you wrote was fine.
Fantastical
Death Prophet
(02-06-2012, 03:42 AM)
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Originally Posted by Orayn

The ball's position can be calculated by the following equation.

y = V*t - (1/2)*(1/6)*9.8*t^2

Where 'y' is the ball's height, 't' is time in seconds, and 'V' is the initial velocity of the ball when it's thrown upwards. Since they tell you that the total flight time is 27 seconds, you plug in 27 for 't' and solve to find the initial velocity.

Then, you use the equation for the ball's velocity to find out when the ball starts falling. To do that, take this equation for the ball's velocity, plug in V, and solve it for 't' when the velocity is equal to zero.

dy/dt = V - (1/6)*9.8*t

Where 'dy/dt' is the rate of change in the ball's position, also known as the velocity. This gives you the time when the ball reaches its maximum height. Put that time back into the first equation and you have your answer.

Okay, but in the first equation I don't know y or V, do I? The picture shows a ball thrown by a person with an initial height above the surface, and then coming back down to the surface and all they give me is the time of flight.

Sorry if I'm doing something wrong, but the problem I've been having is that they don't give you the height from which it is thrown I guess.
Orayn
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(02-06-2012, 03:44 AM)
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Originally Posted by Fantastical

Okay, but in the first equation I don't know y or V, do I? The picture shows a ball thrown by a person with an initial height above the surface, and then coming back down to the surface and all they give me is the time of flight.

Sorry if I'm doing something wrong, but the problem I've been having is that they don't give you the height from which it is thrown I guess.

You don't know V, that's what you're solving for. y is zero because you're plugging in "t=27," which is when the ball hits the ground.
KarmaCow
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(02-06-2012, 03:47 AM)
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Usually with these types of questions, you can assume 'in the air' is just higher than the starting point. I really doubt they want something more complicated that requires relating the time in the air to the initial height.
Fantastical
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(02-06-2012, 03:48 AM)
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Originally Posted by Orayn

You don't know V, that's what you're solving for. y is zero because you're plugging in "t=27," which is when the ball hits the ground.

Oh, thank you!

Also thanks to riceckr, areal, and KarmaCow. :D
Vermillion
Member Formerly Known as JokerOfSpades
(02-08-2012, 02:41 AM)
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Help?

So here's the thing - I think I know 1 and 2 (I presume I just subtract continually). With the alpha particles, I'd subtract 2 from the Z and 4 from the A each time, correct?

So what is a beta particle, then? What do I do for that?

And I know nothing of 4 - 6. I'm not a Physics major - this is half a history class and half science. And the problems are rather difficult for me (I suck at science, you see).

I'm not asking for the answers - I just want to know what to do.

EDIT: Explain it to me as simply as possible, please. I don't even know what the Comprehensive Test Ban Treaty is...
Last edited by Vermillion; 02-16-2012 at 08:03 AM.
Cyan
Purple Drazi
(02-08-2012, 02:45 AM)
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Originally Posted by JokerOfSpades

So what is a beta particle, then? What do I do for that?

Beta decay turns a neutron into a proton, and an electron is emitted.

But how did you get given these problems without knowing anything about them? Did you miss classes, or not read assigned textbook chapters or something? Seems weird.
Vermillion
Member Formerly Known as JokerOfSpades
(02-08-2012, 02:46 AM)
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So in mathematical terms... would I be subtracting a neutron and adding a proton?
Vermillion
Member Formerly Known as JokerOfSpades
(02-08-2012, 02:49 AM)
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Originally Posted by Cyan

Beta decay turns a neutron into a proton, and an electron is emitted.

But how did you get given these problems without knowing anything about them? Did you miss classes, or not read assigned textbook chapters or something? Seems weird.

Teacher is a science professor. The class was just made, and he's not yet sure how to teach it. We're supposed to let him know if the problems are too difficult, but he's moving at a pretty quick pace. And we don't have a science textbook - ours are about history.

Manhattan Project course.
Cyan
Purple Drazi
(02-08-2012, 02:49 AM)
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Originally Posted by JokerOfSpades

So in mathematical terms... would I be subtracting a neutron and adding a proton?

Yeah, the atomic number should go up one.

For the rest of the problems, try working through the steps they give you. I only glanced through, but it looks like they gave you all the information you need at the beginning. More plug-in math than physics.
Vermillion
Member Formerly Known as JokerOfSpades
(02-08-2012, 02:50 AM)
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Originally Posted by Cyan

Yeah, the atomic number should go up one.

For the rest of the problems, try working through the steps they give you. I only glanced through, but it looks like they gave you all the information you need at the beginning. More plug-in math than physics.

Alright then - thanks!
mysticwhip
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(02-08-2012, 02:56 AM)
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thread reminded me of my physics exam i borked yesterday
Fantastical
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(02-12-2012, 06:10 AM)
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Hey guys I'm back and doing homework on a Saturday out of necessity. I actually have so many questions but I'll ask one where I actually think I may be thinking correctly.

"NASA uses a plane often called the "vomit comet" to simulate the weightlessness of space in astronaut training. The plane travels in circular arcs at a velocity of V = 222 m/s."

Basically what I did is V=sqrt(gr) -> V^2/g=r. Is my logic correct here? I knew it had something to do with centripetal force, but I wasn't exactly sure how to use the equation there but I googled it and used the equation from here: http://hyperphysics.phy-astr.gsu.edu...nics/hump.html.
krameriffic
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(02-12-2012, 06:33 AM)
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Originally Posted by Fantastical

Hey guys I'm back and doing homework on a Saturday out of necessity. I actually have so many questions but I'll ask one where I actually think I may be thinking correctly.

"NASA uses a plane often called the "vomit comet" to simulate the weightlessness of space in astronaut training. The plane travels in circular arcs at a velocity of V = 222 m/s."

Basically what I did is V=sqrt(gr) -> V^2/g=r. Is my logic correct here? I knew it had something to do with centripetal force, but I wasn't exactly sure how to use the equation there but I googled it and used the equation from here: http://hyperphysics.phy-astr.gsu.edu...nics/hump.html.

You didn't state what you are looking for, but I assume it is the radius of a circle in which you would experience weightlessness at a velocity of 222 m/s. What you did is what I would call equationally correct, but it lacks any understanding of why one would experience weightlessness in that situation. It's just a rearrangement of a_centripetal=v^2/r with g plugged in for the centripetal acceleration.

The most important thing in introductory Newtonian mechanics courses is the free-body diagram. If they didn't teach you how to properly draw these, look it up and learn it. It will help you understand all of these subtle dynamical phenomena that they might test you on. It's common in high school level physics classes to rely heavily on the equations in lieu of the conceptual connection between them and the actual physics, so knowing how to manipulate formulas might be good enough for that.
Last edited by krameriffic; 02-12-2012 at 06:36 AM.
Al-ibn Kermit
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(02-12-2012, 06:34 AM)
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Yeah, that sounds like the correct method. A=v^2/r so you swap the radius with acceleration to find what the radius should be.

And what kramer says about free-body diagrams is good advice to follow. In my experience, physics questions are always based on the homework but applied to situations that are very different so you can't just copy a formula and plug-and-chug. You have to be able to conceptualize all the forces.

Edit: just noticed that you're not the OP.
Last edited by Al-ibn Kermit; 02-12-2012 at 06:38 AM.
Exuro
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(02-16-2012, 02:18 AM)
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Just started getting into electricity/magnetism in physics and it's making my brain hurt. Could anyone give me some pointers on what to do here? I've only messed with point charges so far so multitude of them in shapes is a little scary.


A semicircle of radius 'a' is in the first and second quadrants, with the center of curvature at the origin. Positive charge +Q is distributed uniformly around the left half of the semicircle, and negative charge -Q is distributed uniformly around the right half of the semicircle in the following figure.What is the magnitude of the net electric field at the origin produced by this distribution of charge?

We haven't done much with integration in the class yet(homework-wise) and this looks like I'll need to use it somehow. I know that the direction of the electric field will be going in the -x direction but I'm not sure how I would calculate the magnitude.
Orayn
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(02-16-2012, 02:35 AM)
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Originally Posted by Exuro

Just started getting into electricity/magnetism in physics and it's making my brain hurt. Could anyone give me some pointers on what to do here? I've only messed with point charges so far so multitude of them in shapes is a little scary.



We haven't done much with integration in the class yet(homework-wise) and this looks like I'll need to use it somehow. I know that the direction of the electric field will be going in the -x direction but I'm not sure how I would calculate the magnitude.

Close. I always think of distribute electric charges as a bunch of little electric charges dQ or -dQ, so integrating is really just adding up what they do to a positive test charge.



The way you sum up all those small charges this would be to integrate in polar coordinates from 0 to pi/2 for the distributed -Q, and from pi/2 to pi for the +Q part. For a positive test charge at the origin, the total magnitude is 2Q in the +x direction, since all the Y components cancel out.
Exuro
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(02-16-2012, 03:16 AM)
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Thanks for the response. I made a quick image of the entire field showing what I had.



So I can see at the origin it's going to the right. Guess I was looking at the field as a whole where it's going to the -x direction. So then the magnitude of this wouldn't really be any different than if it were positive point charge and negative point charge anywhere on the x axis? So anything with a positive charge of some shape and a mirrored shape/negative charge would do the same thing?

I was looking in the book for how to do some integration(of which the examples have been super simple) stuff and was getting completely confused with how to do that with 2 quartercircles of opposite charges.
Last edited by Exuro; 02-16-2012 at 03:18 AM.
The Lamp
Banned
(02-16-2012, 03:31 AM)
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Originally Posted by Exuro

Just started getting into electricity/magnetism in physics and it's making my brain hurt. Could anyone give me some pointers on what to do here? I've only messed with point charges so far so multitude of them in shapes is a little scary.



We haven't done much with integration in the class yet(homework-wise) and this looks like I'll need to use it somehow. I know that the direction of the electric field will be going in the -x direction but I'm not sure how I would calculate the magnitude.

I just took an electricity/magnetism exam last night and made a 100. I love these types of problems. Let me know if this is not the right answer because I might have missed something in there (it's been a long day), but this is the general thought process.

By the way, where did you get that maplet/application for electric field? :o

maxxpower
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(02-16-2012, 03:36 AM)
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Quick question: I'm supposed to use s-shifting to find the Laplace transform of sinhxcosx but there's no e^at so I don't know how exactly to use s-shifting in this case.
Exuro
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(02-16-2012, 03:40 AM)
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Originally Posted by The Lamp

I just took an electricity/magnetism exam last night and made a 100. I love these types of problems. Let me know if this is not the right answer because I might have missed something in there (it's been a long day), but this is the general thought process.

By the way, where did you get that maplet/application for electric field? :o

It's an applet for my masteringphysics course. Here's a link. http://phet.colorado.edu/en/simulati...ges-and-fields

Okay so I'm looking at your work. Shouldn't the y components cancel out? As the force goes away from the positive side it also comes back down, thus they should cancel out no? Maybe I'm just reading it wrong or something.

EDIT: Also found this which looks like nearly the same question. http://www.physicsforums.com/showthread.php?t=334566
Last edited by Exuro; 02-16-2012 at 04:01 AM.

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