| Topic: |
Science > Physics |
| User: |
"dmathd" |
| Date: |
02 Dec 2005 06:19:47 PM |
| Object: |
rotational kinetic energy question? |
I am a high school math teacher trying to assist a student with a
physics problem, so you will have to bear with me. I understand the
formulas for rotational kinetic energy with no linear motion as being
E=(1/2) x(I) x (w)^2. with I= (1/2) x m x r^2.
When we look at the solution that the book gives they only use (1/2)(m
x r^2)(v/r)^2.
My problem is that when you substitute in for I, what happens to the
other 1/2? The book doesn't use it to come up with their solution. Any
help would be appreciated.
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| User: "Old Man" |
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| Title: Re: rotational kinetic energy question? |
02 Dec 2005 08:40:06 PM |
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"dmathd" <dmathd@excite.com> wrote in message
news:1133569187.024087.152840@g47g2000cwa.googlegroups.com...
I am a high school math teacher trying to assist a student with a
physics problem, so you will have to bear with me. I understand the
formulas for rotational kinetic energy with no linear motion as being
E=(1/2) x(I) x (w)^2
Generally correct.
with I= (1/2) x m x r^2.
Not generally correct. "I" depends upon the mass distribution.
Generally: I = Integral[ ( r^2 ) dm ]
for a ring: I = M * R^2
For a disk I = ( M / 2 ) * R^2
[Old Man]
When we look at the solution that the book gives they only use (1/2)(m
x r^2)(v/r)^2.
My problem is that when you substitute in for I, what happens to the
other 1/2? The book doesn't use it to come up with their solution. Any
help would be appreciated.
.
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| User: "Ken S. Tucker" |
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| Title: Re: rotational kinetic energy question? |
05 Dec 2005 02:34:30 AM |
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Old Man wrote:
"dmathd" <dmathd@excite.com> wrote in message
news:1133569187.024087.152840@g47g2000cwa.googlegroups.com...
I am a high school math teacher trying to assist a student with a
physics problem, so you will have to bear with me. I understand the
formulas for rotational kinetic energy with no linear motion as being
E=(1/2) x(I) x (w)^2
Generally correct.
with I= (1/2) x m x r^2.
Not generally correct. "I" depends upon the mass distribution.
Generally: I = Integral[ ( r^2 ) dm ]
for a ring: I = M * R^2
For a disk I = ( M / 2 ) * R^2
[Old Man]
When we look at the solution that the book gives they only use (1/2)(m
x r^2)(v/r)^2.
My problem is that when you substitute in for I, what happens to the
other 1/2? The book doesn't use it to come up with their solution. Any
help would be appreciated.
OLD MAN has it *almost* right, in my text it's called
"INERTIAL PROPERTIES OF HOMOGENEOUS BODIES"
and I do hope your body is homogeneous (snicker). If the density
varies you'll need to do triple integral.
For a "homo" sphere,
"I" = (2/5) mr^2
which is the easiest because it's "moment of inertia" is the
same for X, Y and Z axes passing threw the center.
However for a disk the inertial moments depend on
orientation,
I(xx) = I(yy) = (m/4)*r^2
I(zz) = (m/2)*r^2
where z is perpendicular to the disk center.
If the OP would give the shape of the object then the
Angular Kinetic Energy can be determined.
Another thing is the reference for the AKE. Although that's
a detail, IIRC, the AKE of the earth is zero if I'm standing
on the north pole.
Regards
Ken
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| User: "Bruce Scott TOK" |
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| Title: Re: rotational kinetic energy question? |
04 Dec 2005 11:09:37 AM |
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I am a high school math teacher trying to assist a student with a
physics problem, so you will have to bear with me. I understand the
formulas for rotational kinetic energy with no linear motion as being
E=(1/2) x(I) x (w)^2. with I= (1/2) x m x r^2.
When we look at the solution that the book gives they only use (1/2)(m
x r^2)(v/r)^2.
My problem is that when you substitute in for I, what happens to the
other 1/2? The book doesn't use it to come up with their solution. Any
help would be appreciated.
Unfortunately the other responses weren't very friendly.
Basically, your question is about the same as one I asked my first
semester at university... the answer is that to get I you have to
integrate the mass distribution... the formula for a single point is
dI = dm r^2, so that for a solid object you integrate over the mass
elements. I always scales as mass times size squared, but the numerical
constant depends on the shape _and_ on which axis it is being rotated
around.
The formula I = (1/2) m r^2 is only valid for a uniform disk or cylinder
rotating about its center, with rotation axis perpendicular to the disk
plane. For rods and other shapes _and_ also other _orientations_ you
get a different constant multiplier. The text below has the details.
The formula for energy in terms of I is correct however, E = I w^2/2, as
for rigid rotation v = w r and hence you do the same integral to sum up
the elements of m v^2 you do to get m r^2.
See Goldstein, _Classical Mechanics_, pp 188-198 (2nd ed).
--
ciao,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
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| User: "Sbharris[atsign]ix.netcom.com" |
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| Title: Re: rotational kinetic energy question? |
04 Dec 2005 05:50:26 PM |
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Bruce Scott TOK wrote:
I am a high school math teacher trying to assist a student with a
physics problem, so you will have to bear with me. I understand the
formulas for rotational kinetic energy with no linear motion as being
E=(1/2) x(I) x (w)^2. with I= (1/2) x m x r^2.
When we look at the solution that the book gives they only use (1/2)(m
x r^2)(v/r)^2.
My problem is that when you substitute in for I, what happens to the
other 1/2? The book doesn't use it to come up with their solution. Any
help would be appreciated.
Unfortunately the other responses weren't very friendly.
Hey! I resent that! Particularly since I said pretty much what you
did, and before you did, too.
Unfriendily,
SBH
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| User: "Bruce Scott TOK" |
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| Title: Re: rotational kinetic energy question? |
05 Dec 2005 09:04:09 AM |
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SBH wrote:
Hey! I resent that! Particularly since I said pretty much what you
did, and before you did, too.
Unfriendily,
You could have talked a little less "down" to him. One reason this
group is the way it is, is the sharpness with which people respond to
newbies, perhaps with too much frustration over the trolls. The way to
deal with trolls is not to respond to them, period, so when you get
someone (and I agree it is a minority) who is sincere you aren't
preconditioned.
It is important not to shout at people like this OP. S/He is a sincere
math teacher not used to thinking about physics.
--
ciao,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
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| User: "" |
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| Title: Re: rotational kinetic energy question? |
02 Dec 2005 06:59:48 PM |
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Try to repost your problem more clearly, first showing the equation
that you are using, and next a definition for what each of the
variables is.
Realize that the kinetic energy of a rotating 3-dimensional mass is not
exactly rocket science, but when you post a blurry equation with no
clear cut definition of what the variables and constants are, it
becomes a mystery.
Harry C.
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| User: "Sam Wormley" |
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| Title: Re: rotational kinetic energy question? |
02 Dec 2005 06:40:51 PM |
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dmathd wrote:
I am a high school math teacher trying to assist a student with a
physics problem, so you will have to bear with me. I understand the
formulas for rotational kinetic energy with no linear motion as being
E=(1/2) x(I) x (w)^2. with I= (1/2) x m x r^2.
When we look at the solution that the book gives they only use (1/2)(m
x r^2)(v/r)^2.
My problem is that when you substitute in for I, what happens to the
other 1/2? The book doesn't use it to come up with their solution. Any
help would be appreciated.
Kinetic Energy
http://scienceworld.wolfram.com/physics/KineticEnergy.html
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| User: "" |
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| Title: Re: rotational kinetic energy question? |
02 Dec 2005 07:12:11 PM |
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Sam, you devil you. The OP is obviously in over his head with simple
algebraic equations, and you throw calculus at him.
What the original poster need to learn is that in introductory physics,
the first step is to draw a sketch of the physical sitation.
The second step is to identifiy the symbols used for varaibles, and the
symbols (if any) used for constants.
Then, write the equation for the kinetic energy of the above
configuration.
If, at that point you make an error, anyone with an introductory
knowledge of mechanics can step in immediately, comprehend, the
situation, and hopefully provide a solution or correct the error.
The shortfalls of the current situation are:
The teacher did not document the configuartion of the rotating mass.
The teacher failed to define what his terms mean.
The teacher produced a "hat job" of an equation that has little
meaning.
The teacher has no clue as to why his answer is incorrect, but at least
he is, to his credit, looking for help.
Sam, I am not so heartless as to throw this guy a citation to a website
filled with calculus, and I am a little irritated with you for doing
so.
This guy needs help, so let's try to help him if he can clearly express
the problem that he is attempting to solve. Fair enough?
Harry C.
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| User: "Phil Holman piholmanc@yourservice" |
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| Title: Re: rotational kinetic energy question? |
02 Dec 2005 09:14:28 PM |
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<hhc314@yahoo.com> wrote in message
news:1133572331.117521.217860@g49g2000cwa.googlegroups.com...
Sam, you devil you. The OP is obviously in over his head with simple
algebraic equations, and you throw calculus at him.
What the original poster need to learn is that in introductory
physics,
the first step is to draw a sketch of the physical sitation.
The second step is to identifiy the symbols used for varaibles, and
the
symbols (if any) used for constants.
Then, write the equation for the kinetic energy of the above
configuration.
If, at that point you make an error, anyone with an introductory
knowledge of mechanics can step in immediately, comprehend, the
situation, and hopefully provide a solution or correct the error.
The shortfalls of the current situation are:
The teacher did not document the configuartion of the rotating mass.
The teacher failed to define what his terms mean.
The teacher produced a "hat job" of an equation that has little
meaning.
The teacher has no clue as to why his answer is incorrect, but at
least
he is, to his credit, looking for help.
Sam, I am not so heartless as to throw this guy a citation to a
website
filled with calculus, and I am a little irritated with you for doing
so.
This guy needs help, so let's try to help him if he can clearly
express
the problem that he is attempting to solve. Fair enough?
Over his/her head with simple algebraic equations is not true nor is it
the problem. Knowing that I is geometry dependant is the only thing
lacking. For a math high school teacher, knowledge of calculus is a
prerequisite.......multivariable even.
Phil H
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| User: "" |
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| Title: Re: rotational kinetic energy question? |
02 Dec 2005 07:12:51 PM |
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http://scienceworld.wolfram.com/physics/KineticEnergy.html
***************
Boy, that Wolfram has info on just about everthing under the Sun,
doesn't he. It's great to be alive.
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| User: "PD" |
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| Title: Re: rotational kinetic energy question? |
05 Dec 2005 10:33:38 AM |
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dmathd wrote:
I am a high school math teacher trying to assist a student with a
physics problem, so you will have to bear with me. I understand the
formulas for rotational kinetic energy with no linear motion as being
E=(1/2) x(I) x (w)^2. with I= (1/2) x m x r^2.
Most likely, the problem is in your second equation.
The correct formula for moment of inertia is I = C * m * r^2.
What C is, is determined by the geometry of the rotating thing.
If it's a solid disk, then C = 1/2.
If it's a rim with spokes, so that essentially all the mass is at the
outside, then C = 1.
If it's a yardstick, with r = length of stick, then C = 1/12.
If it's a sphere, then C = 2/5.
PD
When we look at the solution that the book gives they only use (1/2)(m
x r^2)(v/r)^2.
My problem is that when you substitute in for I, what happens to the
other 1/2? The book doesn't use it to come up with their solution. Any
help would be appreciated.
.
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| User: "Sbharris[atsign]ix.netcom.com" |
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| Title: Re: rotational kinetic energy question? |
02 Dec 2005 07:17:58 PM |
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dmathd wrote:
I am a high school math teacher trying to assist a student with a
physics problem, so you will have to bear with me. I understand the
formulas for rotational kinetic energy with no linear motion as being
E=(1/2) x(I) x (w)^2. with I= (1/2) x m x r^2.
No, that I is a SPECIFIC moment, and only holds for specific objects,
like disks. For rings, I really is mr^2, and no 1/2.
When we look at the solution that the book gives they only use (1/2)(m
x r^2)(v/r)^2.
My problem is that when you substitute in for I, what happens to the
other 1/2? The book doesn't use it to come up with their solution. Any
help would be appreciated.
The problem is that I (moment of inertia) has many solutions in terms
of k MR^2, depending on how the mass is placed in the spinning object.
The k will be a dimentionless coefficient which comes out of
integrating (from r= 0 to R) r^2dm to get I, the moment of interia,
where dm is the differential mass distribution as a function of r.
For a ring, I = MR2, and there's no "1/2". So you get exactly the
answer you expect from the linear problem--- all the mass is at radius
R and is moving at v, so energy is 1/2mv^2
However, for a rotating uniform disk, I = 1/2 MR^2, and the total
energy comes out to 1/4mv^2 because v is the velocity only at the rim,
and not all the mass is moving that fast. You see? The integration
already takes care of the radial distribution problem beforehand.
SBH
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