dynamics
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Post Fiend
Joined: Jun 2004
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From: Joppa, Maryland
dynamics
Hello everyone,
having a little trouble with this problem and thought someone might be able to push me in the right direction...
"A rod of length L and mass m lies at rest on a horizontal sheet of ice. If the rod is given a kick at one end in a direction normal to it, find its center of rotation at that moment."
thanks for any ideas...
having a little trouble with this problem and thought someone might be able to push me in the right direction...
"A rod of length L and mass m lies at rest on a horizontal sheet of ice. If the rod is given a kick at one end in a direction normal to it, find its center of rotation at that moment."
thanks for any ideas...
MSEE
Joined: Apr 2004
Posts: 10,386
Likes: 35
From: Austin, TX
I seem to remember these kinds of problems when I took dynamics a long time ago and it seems like there is some kind of formula for moment-x and moment-y, or something like that. And I think that because the rod is on ice, they want you to assume the friction coefficients are zero, so the only force you have on the rod is the force of someone kicking it. I don't know what I did with my notes for that class, so that's about all the help I can give. Sorry!
Post Fiend
Joined: Mar 2007
Posts: 8,770
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From: The Great North~West!
Paul ~ I think it's asking for the "Moment of Inertia." If you kick the rod at the end at, perpendicular to it's length (normal) it will spin on it's axis through the center The ice does indeed let you know your coefficient of friction = zero.
And since you don't seem to have any dimensions, then it is simply a manipulation or proof of the Inertia formula,,, (I = Integral r^2 dm)
I = SUM m(sub i) r(sub i)^2
= (m)(1/2 L)^2 + (m)(1/2 L)^2
= 1/2 mL^2
Now, depending on what you've been doing in class, he may want you to integrate to find the Moment of Inertia~ If so, you'll need something more like this,,,
(I = Integral r^2 dm)
element's mass(dm)/element's length(dx)=rod's mass(M)/rod's length(L)
dm = M/L dx
and integrate from the end to end of the rod (from x=-L/2 to x = L/2) to include all the elements,,,
I = Integral of x^2(M/L) dx from x=-L/2 to x=+L/2
= M / 3L [x^3] from -L/2 to +L/2
= M / 3L [(L/2)^3 - (-L/2)^3]
= 1/12 ML^2
I would be more inclined to use this formula,,, It is for a thin rod, about the axis, through the center, perpendicular (normal) to length. That is pretty much what your original statement said.
Well,, that seems clear as mud!! Ha! Ha!! Hope it helps!!
And since you don't seem to have any dimensions, then it is simply a manipulation or proof of the Inertia formula,,, (I = Integral r^2 dm)
I = SUM m(sub i) r(sub i)^2
= (m)(1/2 L)^2 + (m)(1/2 L)^2
= 1/2 mL^2
Now, depending on what you've been doing in class, he may want you to integrate to find the Moment of Inertia~ If so, you'll need something more like this,,,
(I = Integral r^2 dm)
element's mass(dm)/element's length(dx)=rod's mass(M)/rod's length(L)
dm = M/L dx
and integrate from the end to end of the rod (from x=-L/2 to x = L/2) to include all the elements,,,
I = Integral of x^2(M/L) dx from x=-L/2 to x=+L/2
= M / 3L [x^3] from -L/2 to +L/2
= M / 3L [(L/2)^3 - (-L/2)^3]
= 1/12 ML^2
I would be more inclined to use this formula,,, It is for a thin rod, about the axis, through the center, perpendicular (normal) to length. That is pretty much what your original statement said.
Well,, that seems clear as mud!! Ha! Ha!! Hope it helps!!
Government Teat-sucker
Joined: Oct 2003
Posts: 9,748
Likes: 1
From: Virginia Beach
Lizzie. Snowbunny. My Dear.
How the heck is the young man supposed to learn anything if you give him all the answers???
And, By the way. Are you going to be able to help me with my Statistics class this fall?????
Please,Please, PLEASE????
How the heck is the young man supposed to learn anything if you give him all the answers???
And, By the way. Are you going to be able to help me with my Statistics class this fall?????
Please,Please, PLEASE????
Post Fiend
Joined: Apr 2004
Posts: 19,098
Likes: 8
From: Great State of Texas
I think the down force of the planted foot (while kicking with the other foot) will create a shear effect upon the ice which will in turn cause it to shatter. This structurally weakened ice will then separate, thereby, causing the rod and the kick engineer to spiral downward into the supposed freezing water. This will, in effect, cause both the rod and the kick engineer to be lost forever. Therefore, this experiment is flawed by it's own design and a victim of **** poor planning. Since it is scientifically proved that **** Poor Planning Provides **** Poor Results.....
....I wouldn't do it.
....I wouldn't do it.
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Post Fiend
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From: Joppa, Maryland
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Logistics Pro
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Posting Guru
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From: Location, Location.
Sorry brother, wish I'd paid attention to this before - you were being asked for the center of gravity, not the moment of inertia.
...find its center of rotation...
A free-floating body will rotate about its center of gravity.
The angular velocity it acquires is a function of the force (the kick) applied to it, and the angular equivalent of mass about the axis of rotation, the moment of intertia. But its "center of rotation" will be its center of gravity.
...find its center of rotation...
A free-floating body will rotate about its center of gravity.
The angular velocity it acquires is a function of the force (the kick) applied to it, and the angular equivalent of mass about the axis of rotation, the moment of intertia. But its "center of rotation" will be its center of gravity.
MSEE
Joined: Apr 2004
Posts: 10,386
Likes: 35
From: Austin, TX
How is that so? The center of gravity would be the middle if the mass of the bar is even throughout. Gravity acts on the entire length of the bar equally and the center of gravity is in the middle. But if it's lying flat on its side and you kick it from one end only, for the center of rotation to be in the middle you'd have to kick it equally in the opposite direction on the opposite side. If you only kick it from one end, wouldn't the center of rotation be closer to the other end? I'm not trying to start an argument but I'm curious.
This is a moments problem; you have to solve it with the integral like SnowBunny showed. I remember that same type of problem from that class. She solved it correctly - if your claim was true then the answer would have been just x = 0 (center) since she integrated from -L/2 to +L/2.
This is a moments problem; you have to solve it with the integral like SnowBunny showed. I remember that same type of problem from that class. She solved it correctly - if your claim was true then the answer would have been just x = 0 (center) since she integrated from -L/2 to +L/2.
Posting Guru
Joined: Jun 2005
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From: Location, Location.
...If you only kick it from one end, wouldn't the center of rotation be closer to the other end?...
...This is a moments problem; ..... if your claim was true then the answer would have been just x = 0 (center) ...
...find its center of rotation...
which implies finding a position. The center of gravity is a position within the object. The Moment of Inertia is a property of the object, measured in units of mass times length squared, like grams-centimeters^2, not a position. Does the phrase "It's center of rotation is 25 grams-cm^2" make sense?
If it turns out a prof has used these words to ask for the moment of inertia, he needs to take a course in engineering communication.