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Physics Help please!

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Seriously woman, these poor fools took the bait and all but did your hw for you. It would be wise for there to be some quality T&A in the morning for them....😀
 
Originally posted by: sciencewhiz
Originally posted by: element
Originally posted by: Afro000Dude
Originally posted by: element
Originally posted by: sciencewhiz
In some ways I was thinking about the problem wrong. However, element is still wrong. I got the same answer as fawhfe

I'd love to see element's equations. (PM if you don't want to give the solution away)
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I could be wrong, I did it off the top of my head without looking up anything.

first f=ma
2.5 * 15.8 (the 15.8 is from 3.0m/s in .19 seconds, thats 15.8 m/s^2 accel.)
nets you 39.5 Newtons of force

then 39.5 * 0.92 = 36.34 Nm

now how'd you do it?
Click to expand...

It's rev/s not m/s.
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yeah that's what I meant.
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It's not as simple as saying "yeah that's what I meant" You explicitly need to convert rev/s to m/s if you want to try to do it your way.
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yeah that's what I meant to do its been a long time since I did this stuff.
 
Originally posted by: Ready

3.0revs/.19 seconds = 1 rev / .0633333 sec,
So basically, you've covered the area of the circle in .063333333 seconds, thus you displaced the bar an area of Pi * R^2 meter (2.659 m^2) in .06333 seconds. Now suppose you don't rotate and instead push the center of the bar in a straight line in .063333 seconds with the same force. Since you expand the same amont of energy, you should in theory displace the bar the same amount of area right? Lets call X the distance you push the bar from the center. X times the length of the bar would be the total area you displace in .063333 seconds and should be = 2.659m^2. so X in meters = (2.659m^2./.92m)

This yield x = 2.89 meters. So pushing the Rod around in a circle once in .19 seconds is equal to pushing it down the center in a straight line of 2.89 meters in .19 seconds. But it's much easier to calculate things in a straight line. Our force is then merely the mass times acceleration..
Click to expand...
That was quite a complicated way to convert angular acceleration to linear acceleration, and while I don't exactly follow the area thing, the answer is right.


Mass = 2.5Kg
Acceleration = 2.89m/.19 s^2 = 15.2m/s ^2

Ok, now you have mass, you have acceleration, and you have distance 😀

Mine is a total cheap way of solving it. Instead of treating it as a rotating rod, convert it into something u push in a straight line right down the center.
Click to expand...

Can you explain this part a little more?
 
Originally posted by: Goosemaster
Seriously woman, these poor fools took the bait and all but did your hw for you. It would be wise for there to be some quality T&A in the morning for them....😀
Click to expand...

Come on man, this homework was pretty easy, and I like arguing physics. If being a woman could get someone to do my homework tonight, that'd be impressive, and I'd be sure to try it next week 🙂.

Btw, ready, you made 3 incorrect assumptions (thereby failing at teh physics)
1. 3 rev/s is only the final angular speed, not the speed as hes swinging the bat. You can't simply assume its traveling that fast for the area the bat swings out.
2. Energy has no correlation to area--if I were to go from A to B in a car at 200 mph, I would use more gas than at 50 mph (even w/o air resistance and friction it would take me more energy to accelerate)
3. You wrote and calculated acceleration in m/s when its a quantity in m/s^2
 
haha this thread is funny, if this was the typical guy saying "hey guys how do i write a while loop that gives the fibonacci sequence" everyone would give him a hard time and tell him to quit his major. but then again, no one has ever offered pictures.
 
Originally posted by: sciencewhiz
Originally posted by: Ready

3.0revs/.19 seconds = 1 rev / .0633333 sec,
So basically, you've covered the area of the circle in .063333333 seconds, thus you displaced the bar an area of Pi * R^2 meter (2.659 m^2) in .06333 seconds. Now suppose you don't rotate and instead push the center of the bar in a straight line in .063333 seconds with the same force. Since you expand the same amont of energy, you should in theory displace the bar the same amount of area right? Lets call X the distance you push the bar from the center. X times the length of the bar would be the total area you displace in .063333 seconds and should be = 2.659m^2. so X in meters = (2.659m^2./.92m)

This yield x = 2.89 meters. So pushing the Rod around in a circle once in .19 seconds is equal to pushing it down the center in a straight line of 2.89 meters in .19 seconds. But it's much easier to calculate things in a straight line. Our force is then merely the mass times acceleration..
Click to expand...
That was quite a complicated way to convert angular acceleration to linear acceleration, and while I don't exactly follow the area thing, the answer is right.


Mass = 2.5Kg
Acceleration = 2.89m/.19 s^2 = 15.2m/s ^2

Ok, now you have mass, you have acceleration, and you have distance 😀

Mine is a total cheap way of solving it. Instead of treating it as a rotating rod, convert it into something u push in a straight line right down the center.
Click to expand...

Can you explain this part a little more?
Click to expand...



Darn it, I coverted it from angular to linear so it's easier to understand. The idea is that energy is energy. If I expend x amount of energy to push the rod in a circle, I would displace a certain amount of area. If I once again expand the same amount of energy to push it down the center in a straight line, I should still displace the same amount of area. I know this only works for a uniform mass.
 
Originally posted by: fawhfe
Originally posted by: Goosemaster
Seriously woman, these poor fools took the bait and all but did your hw for you. It would be wise for there to be some quality T&A in the morning for them....😀
Click to expand...

Come on man, this homework was pretty easy, and I like arguing physics. If being a woman could get someone to do my homework tonight, that'd be impressive, and I'd be sure to try it next week 🙂.
Click to expand...

😀..I love that stuff
 
Don't you love solving differential equations? 😀

Oh crap, that reminds me. I have E&M homework to do.
 
Originally posted by: fawhfe
Come on man, this homework was pretty easy, and I like arguing physics. If being a woman could get someone to do my homework tonight, that'd be impressive, and I'd be sure to try it next week 🙂.
Click to expand...

Not even a women could get me to do that. We're just starting waveguides, and only briefly touched on propogation modes. I don't like E&M.
 
isn't torque a b sin theta, where a and b are the magnitudes of the radius and force applied to the "wrench"?
 
Originally posted by: sciencewhiz
Originally posted by: Darien
I can't believe this is at ~50 posts. I haven't seen anyone work out #2 though.
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I worked out #2, I'm just not posting it 😉

It's even easier then #1.
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Yeah, at least with #1, if you aren't given the moment of inertia you'd have to do an integral to calculate it. a simple one, but at least it's not pure algebra.
 
Originally posted by: EmperorIQ
haha this thread is funny, if this was the typical guy saying "hey guys how do i write a while loop that gives the fibonacci sequence" everyone would give him a hard time and tell him to quit his major. but then again, no one has ever offered pictures.
Click to expand...

The AT double standard in full swing 😀
 
Originally posted by: sciencewhiz
Originally posted by: fawhfe
Come on man, this homework was pretty easy, and I like arguing physics. If being a woman could get someone to do my homework tonight, that'd be impressive, and I'd be sure to try it next week 🙂.
Click to expand...

Not even a women could get me to do that. We're just starting waveguides, and only briefly touched on propogation modes. I don't like E&M.
Click to expand...

i still don't understand those cr@p
 
Originally posted by: Darien
Originally posted by: EmperorIQ
haha this thread is funny, if this was the typical guy saying "hey guys how do i write a while loop that gives the fibonacci sequence" everyone would give him a hard time and tell him to quit his major. but then again, no one has ever offered pictures.
Click to expand...

The AT double standard in full swing 😀
Click to expand...

Hey, I help out in any physics thread I see, and honestly, I've never had to argue in one before, so its no wonder this one has gotten so long.
 
Originally posted by: fawhfe
Originally posted by: Darien
Originally posted by: EmperorIQ
haha this thread is funny, if this was the typical guy saying "hey guys how do i write a while loop that gives the fibonacci sequence" everyone would give him a hard time and tell him to quit his major. but then again, no one has ever offered pictures.
Click to expand...

The AT double standard in full swing 😀
Click to expand...

Hey, I help out in any physics thread I see, and honestly, I've never had to argue in one before, so its no wonder this one has gotten so long.
Click to expand...

hehe that was my fault. i was being neftastic. i only knew enough to be dangerous, to complicate matters.

now, about those transverse electromagnetic waves...
 
Originally posted by: fawhfe
Originally posted by: Darien
Originally posted by: EmperorIQ
haha this thread is funny, if this was the typical guy saying "hey guys how do i write a while loop that gives the fibonacci sequence" everyone would give him a hard time and tell him to quit his major. but then again, no one has ever offered pictures.
Click to expand...

The AT double standard in full swing 😀
Click to expand...

Hey, I help out in any physics thread I see, and honestly, I've never had to argue in one before, so its no wonder this one has gotten so long.
Click to expand...

OK, so generalizations don't work all the time 😛
 
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