But that's not the radius that the mass rotates, which is what counts.
EDIT: It's not a 2.5 kg mass at the end of a weightless rod, which is what you are calculating, but a uniform 2.5 kg rod.
But that's not the radius that the mass rotates, which is what counts.
EDIT: It's not a 2.5 kg mass at the end of a weightless rod, which is what you are calculating, but a uniform 2.5 kg rod.
I'm surprised no one has explicitely worked it out, posted the solution and demanded for pics yet.
If you spin something around its center compared to its end, it will take less effort. Try it with a pen or something.
Meh, Guild Wars WPE ended yesterday, Conan is a rerun, and I'm bored out of my mind having just dropped the only class that was actually hard this semester. Besides, if I really cared about pix, my first post would have been 70nm, 4.54*10^-11m pix?
. Anyways, I should go do my own EE homework now.Originally posted by: fawhfe
Meh, Guild Wars WPE ended yesterday, Conan is a rerun, and I'm bored out of my mind having just dropped the only class that was actually hard this semester. Besides, if I really cared about pix, my first post would have been 70nm, 4.54*10^-11m pix?
If you did, you'd be wrong. The right answer is actually 36.3 Nm
oh yeah almost forgot, pix?
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)
I'd love to see element's equations. (PM if you don't want to give the solution away)
Originally posted by: element
Originally posted by: fawhfe
Meh, Guild Wars WPE ended yesterday, Conan is a rerun, and I'm bored out of my mind having just dropped the only class that was actually hard this semester. Besides, if I really cared about pix, my first post would have been 70nm, 4.54*10^-11m pix?
If you did, you'd be wrong. The right answer is actually 36.3 Nm
oh yeah almost forgot, pix?
I'm telling you man, you're doing angular momentum wrong. Having reread your statement, sciencewhiz has quite correctly articulated that its not the same when you have a bat (uniform mass over .92 m) instead of a 2.5kg mass at the end of a string .92 meters long.
TheLonelyPhoenix
Diamond Member
- Feb 15, 2004
- 5,594
- 1
- 0
So levers, pulleys, and other simple machines that people have used to lift hundreds of times their weight were all useless because they were still lifting the same weight?
j00 fail teh physics.
The moment of inertia is greater when the weight is farther away from the fulcrum.
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?
Chaotic42
Lifer
- Jun 15, 2001
- 35,383
- 2,502
- 126
T=J*alpha
T=1/3*m*l^2*alpha
alpha=3*2*pi/.19 (convert to radians/sec)
T=1/3*2.5*.92^2*3*2*pi/.19
T=70 Nm
J=1/3*m*l^2 is the rotational inertia of a Rod of length L and mass m spun from an end.
If it was spun from the center it would be 1/12*m*l^2
T=1/3*m*l^2*alpha
alpha=3*2*pi/.19 (convert to radians/sec)
T=1/3*2.5*.92^2*3*2*pi/.19
T=70 Nm
J=1/3*m*l^2 is the rotational inertia of a Rod of length L and mass m spun from an end.
If it was spun from the center it would be 1/12*m*l^2
So much for not explicitly spelling it out. Oh well, we better get pix now.
Seriously though, to OP: I would strongly recommend doing some more of these problems. These are really basic applications of the equations and if you're having trouble with them, its only going to get worse--either the problems will get harder or you'll run out of time on exams, because the prof will expect people to be able to do them quickly.
Prob number one
I guess we pretty much establish that its force times distance. Distance is .92m
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If you apply pressure here its kinda tricky to solve, there is an easy way to solve it. key is energy used is energy regardless if what form u use. Lets say you move rotate the tip of that bar 1 inch. How does that compare say instead of rotating it you push it down the center x amount of distance? Will, if you push it down the center without rotating, then the total area your bar cover is X distance times .92 meters right? But if you push the tip and let bar rotate, then the area you moved is the cross section that the entire bar displace. SO knowing that, here's way to figure it out.
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..
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
so .92m * 2.5Kg * 15.2m/ss = 34.96 I hope i'm right
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.
I guess we pretty much establish that its force times distance. Distance is .92m
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If you apply pressure here its kinda tricky to solve, there is an easy way to solve it. key is energy used is energy regardless if what form u use. Lets say you move rotate the tip of that bar 1 inch. How does that compare say instead of rotating it you push it down the center x amount of distance? Will, if you push it down the center without rotating, then the total area your bar cover is X distance times .92 meters right? But if you push the tip and let bar rotate, then the area you moved is the cross section that the entire bar displace. SO knowing that, here's way to figure it out.
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..
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
so .92m * 2.5Kg * 15.2m/ss = 34.96 I hope i'm right
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.
In short. what i did was convert
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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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