phoffric
asked on
Can Rod's Mass Density be discarded after computing Center of Mass and Moment of Inertia?
If I have a rod of varying density, then to determine the rod's motion, can that density information be discarded after computing the rod's center of mass and moment of inertia with respect to an axis of rotation at the CM? Here are more details to clarify:
I have on a frictionless, level, and flat ice skating rink a very thin, but extremely strong and rigid, rectangular rod of length=L and of mass=M and which is aligned to the y-axis; and where the rod has a density function, µ(y) kg/m. µ(y) does not have to be a continuous function. For example, it can have a constant value for a region, and in regions to the left and right of this constant value region, µ(y) could drop instantly to zero.
Given the function, µ(y), the rod's Center of Mass (y_cm) can be computed using:
http://en.wikipedia.org/wiki/Center_of_mass#Definition
Since the rod has only one dimension, I can replace the vectors R and r with the scalars, y_cm and y:
L
y_cm = (1/M) § y µ(y) dy
y=0
The rod's Moment of Inertia (I_cm) with respect to an axis of rotation at y_cm, can be computed using:
http://en.wikipedia.org/wiki/Moment_of_inertia#Definition
L
I_cm = § (y - y_cm)² µ(y) dy
y=0
There are no anchor points to hold the rod from moving, so if a level force or impulse is applied perpendicular to the stationary rod, then naturally, the rod will be set in motion. After computing y_cm and I_cm, can the motion of the rod be completely determined strictly from y_cm and I_cm, or are there some configurations where the density function, µ(y), is still needed?
If the rod's density function is not needed, then is it possible to have two different density functions, µ1(y) and µ2(y), which can yield the same y_cm and I_cm? I ask only to consider the possibility of being able to simplify the rod's behavior by coming up with a different µ(y) that yields the same two terms, y_cm and I_cm.
This is related to the questions that I asked in
https://www.experts-exchange.com/questions/26355746/Energy-imparted-to-Rod-due-to-Constant-Impulse.html
and in
https://www.experts-exchange.com/questions/26383437/Velocity-of-Rod's-Center-of-Mass-When-Struck-by-a-Hammer-Impulse.html
I have on a frictionless, level, and flat ice skating rink a very thin, but extremely strong and rigid, rectangular rod of length=L and of mass=M and which is aligned to the y-axis; and where the rod has a density function, µ(y) kg/m. µ(y) does not have to be a continuous function. For example, it can have a constant value for a region, and in regions to the left and right of this constant value region, µ(y) could drop instantly to zero.
Given the function, µ(y), the rod's Center of Mass (y_cm) can be computed using:
http://en.wikipedia.org/wiki/Center_of_mass#Definition
Since the rod has only one dimension, I can replace the vectors R and r with the scalars, y_cm and y:
L
y_cm = (1/M) § y µ(y) dy
y=0
The rod's Moment of Inertia (I_cm) with respect to an axis of rotation at y_cm, can be computed using:
http://en.wikipedia.org/wiki/Moment_of_inertia#Definition
L
I_cm = § (y - y_cm)² µ(y) dy
y=0
There are no anchor points to hold the rod from moving, so if a level force or impulse is applied perpendicular to the stationary rod, then naturally, the rod will be set in motion. After computing y_cm and I_cm, can the motion of the rod be completely determined strictly from y_cm and I_cm, or are there some configurations where the density function, µ(y), is still needed?
If the rod's density function is not needed, then is it possible to have two different density functions, µ1(y) and µ2(y), which can yield the same y_cm and I_cm? I ask only to consider the possibility of being able to simplify the rod's behavior by coming up with a different µ(y) that yields the same two terms, y_cm and I_cm.
This is related to the questions that I asked in
https://www.experts-exchange.com/questions/26355746/Energy-imparted-to-Rod-due-to-Constant-Impulse.html
and in
https://www.experts-exchange.com/questions/26383437/Velocity-of-Rod's-Center-of-Mass-When-Struck-by-a-Hammer-Impulse.html
ASKER CERTIFIED SOLUTION
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>>can that density information be discarded after computing the rod's center of mass and moment of inertia with respect to an axis of rotation at the CM?
I don't see how.
The moment of inertia will be the integral of r²*f(r)*dr where r is the distance from the point of rotation and f(r) a function which gives the value of the mass at point r. The actual mass being the integral of f(r)*dr, then the moment cannot be directly expressed as some expression times m, the total mass. Unless one knows exactly what f(r) is, the "density information" is not going to cancel out.
This is a problem in astrophysics, where star formation requires the exact distribution of gas to be known a priori. See also my question regarding three masses on a thin lightweight rod.
I don't see how.
The moment of inertia will be the integral of r²*f(r)*dr where r is the distance from the point of rotation and f(r) a function which gives the value of the mass at point r. The actual mass being the integral of f(r)*dr, then the moment cannot be directly expressed as some expression times m, the total mass. Unless one knows exactly what f(r) is, the "density information" is not going to cancel out.
This is a problem in astrophysics, where star formation requires the exact distribution of gas to be known a priori. See also my question regarding three masses on a thin lightweight rod.
ASKER
>> Unless one knows exactly what f(r)
It looks like your f(r) is a density function of the rod (kg/m) that corresponds to what I was calling µ(y). In my question, µ(y) is a given, so one does know exactly what µ(y) is.
I'll give a specific example of what I was talking about, so that we're on the same page. Suppose I know that the rod's density is µ(y) = y.
Although I said in the question that I was given the rod's mass=M, I realize that this is redundant, and did not need to be stated. Instead, I should have said that I can compute M as follows:
L L
M = § µ(y) dy = y²/2 = L²/2 Kg
0 0
L
y_cm = (1/M) § y µ(y) dy = (2/L²) § y² dy = (2/L²) y³/3 = (2/L²) L³/3 = 2L/3 (m)
y=0
Not surprisingly, the rod is a little top-heavy, with y_cm = (2/3)L meters.
For working the following, let a = y_cm.
L L
I_cm = § (y - y_cm)² µ(y) dy = §(y - a)² y dy = (a² y²)/2-(2 a y³)/3+y^4/4
y=0 y=0
= y_cm²L²/2 -(2 y_cm L³)/3+L^4/4
= (2/3 L)²L²/2 -(2 (2/3 L) L³)/3+L^4/4
= (4/9 L²)L²/2 - 4L L³/9 + L^4/4
= (2/9) L^4 - 4/9 L^4 + L^4/4
= [2/9 - 4/9 + 1/4] L^4
= (1/36) L^4 (Kg.m²)
=====
Results Summarized:
M = (1/2)L² (Kg)
y_cm = (2/3)L (m)
I_cm = (1/36) L^4 (Kg.m²)
=====
So, for a specific value of L (e.g., L = 10 m), then I have these three number, M, y_cm, and I_cm. I was asking whether we can now throw away µ(y) in order to compute the motion of the rod just using these three numbers for application of a force or impulse to the rod.
If I can throw away µ(y), then furthermore, I was then wondering whether it might then be possible to find another rod having a different density distribution but which produced these same three values. And if so, then I would think that if the same force were applied to either rod, then the motion of the two rods would be identical.
I just wrote down this example for µ(y) = y, and haven't figured out whether there is another distribution (other than the mirror, µ(y) = L-y) that produces the same three numbers. Any hints on how to do this?
It looks like your f(r) is a density function of the rod (kg/m) that corresponds to what I was calling µ(y). In my question, µ(y) is a given, so one does know exactly what µ(y) is.
I'll give a specific example of what I was talking about, so that we're on the same page. Suppose I know that the rod's density is µ(y) = y.
Although I said in the question that I was given the rod's mass=M, I realize that this is redundant, and did not need to be stated. Instead, I should have said that I can compute M as follows:
L L
M = § µ(y) dy = y²/2 = L²/2 Kg
0 0
L
y_cm = (1/M) § y µ(y) dy = (2/L²) § y² dy = (2/L²) y³/3 = (2/L²) L³/3 = 2L/3 (m)
y=0
Not surprisingly, the rod is a little top-heavy, with y_cm = (2/3)L meters.
For working the following, let a = y_cm.
L L
I_cm = § (y - y_cm)² µ(y) dy = §(y - a)² y dy = (a² y²)/2-(2 a y³)/3+y^4/4
y=0 y=0
= y_cm²L²/2 -(2 y_cm L³)/3+L^4/4
= (2/3 L)²L²/2 -(2 (2/3 L) L³)/3+L^4/4
= (4/9 L²)L²/2 - 4L L³/9 + L^4/4
= (2/9) L^4 - 4/9 L^4 + L^4/4
= [2/9 - 4/9 + 1/4] L^4
= (1/36) L^4 (Kg.m²)
=====
Results Summarized:
M = (1/2)L² (Kg)
y_cm = (2/3)L (m)
I_cm = (1/36) L^4 (Kg.m²)
=====
So, for a specific value of L (e.g., L = 10 m), then I have these three number, M, y_cm, and I_cm. I was asking whether we can now throw away µ(y) in order to compute the motion of the rod just using these three numbers for application of a force or impulse to the rod.
If I can throw away µ(y), then furthermore, I was then wondering whether it might then be possible to find another rod having a different density distribution but which produced these same three values. And if so, then I would think that if the same force were applied to either rod, then the motion of the two rods would be identical.
I just wrote down this example for µ(y) = y, and haven't figured out whether there is another distribution (other than the mirror, µ(y) = L-y) that produces the same three numbers. Any hints on how to do this?
ASKER
>> See also my question regarding three masses on a thin lightweight rod
I'm hoping to work my way slowly towards it :)
If I can get the answer (even if not elegant) that would be a good baby step.
>> astrophysics, where star formation ...
Not a baby step for me.
I'm hoping to work my way slowly towards it :)
If I can get the answer (even if not elegant) that would be a good baby step.
>> astrophysics, where star formation ...
Not a baby step for me.
ASKER
I need a little hand-waving w.r.t. to using the mirror, µ(y) = L-y. The mass and I_cm should work out the same, but obviously y_cm will be (1/3)L, not (2/3)L. But these two rods are identical in that a 180 degree flip produces the other rod.
>>can that density information be discarded after computing the rod's center of mass and moment of inertia with respect to an axis of rotation at the CM?
>>I don't see how.
Let me refine what I said. Clearly once the center of mass has been determined, it is not going to change unless the density function changes, so in that sense you can "discard" the density function.
Now once the moment of inertia about the center of mass (or any other fixed point) has been calculated the same applies. But the moment of inertia is defined as the sum of the masses multiplied by the square of the distance from the axis of rotation. In your notation § µ(r)r²dr where r is the distance from the center of the object, but about a point away from the center, say distance a, this becomes § µ(r)(a-r)²dr, so I don't see how the density function can be "discarded" when taking moments about some arbitary point, not the center of the object.
>>I don't see how.
Let me refine what I said. Clearly once the center of mass has been determined, it is not going to change unless the density function changes, so in that sense you can "discard" the density function.
Now once the moment of inertia about the center of mass (or any other fixed point) has been calculated the same applies. But the moment of inertia is defined as the sum of the masses multiplied by the square of the distance from the axis of rotation. In your notation § µ(r)r²dr where r is the distance from the center of the object, but about a point away from the center, say distance a, this becomes § µ(r)(a-r)²dr, so I don't see how the density function can be "discarded" when taking moments about some arbitary point, not the center of the object.
ASKER
>> but about a point away from the center, say distance a, this becomes § µ(r)(a-r)²dr
>> so I don't see how the density function can be "discarded" when taking moments about some arbitary point
Ok, this is BRDQ#5. Let me know if I did this right.
==========================
I_a = § µ(y)(y-a)²dy = § µ(y)(y² - 2ay + a²)dy
= § ( µ(y)y² - 2ayµ(y) + a²µ(y) )dy
= § µ(y)y²dy - § 2ayµ(y)dy + § a²µ(y)dy
I_a = § µ(y)y²dy - 2a § y µ(y)dy + a² § µ(y)dy
====================
M = § µ(y) dy
y_cm = (1/M) § y µ(y) dy ==> § y µ(y) dy = M * y_cm
==> I_a = § µ(y)y²dy - 2a * M * y_cm + a² M
====================
I_cm = § (y - y_cm)² µ(y) dy
= § (y - y_cm)² µ(y) dy
= § (y² - 2 y y_cm + y_cm²) µ(y) dy
= § y² µ(y) dy - § 2 y y_cm µ(y) dy + § y_cm² µ(y) dy
= § y² µ(y) dy - 2 y_cm § y µ(y) dy + y_cm² § µ(y) dy
= § y² µ(y) dy - 2 y_cm § y µ(y) dy + y_cm² § µ(y) dy
I_cm = § y² µ(y) dy - 2 y_cm § y µ(y) dy + y_cm² § µ(y) dy
= § y² µ(y) dy - 2 y_cm (M * y_cm) + y_cm² M
= § y² µ(y) dy - 2 M * y_cm² + M * y_cm²
= § y² µ(y) dy - M * y_cm²
==========================
==> § y² µ(y) dy = I_cm + M * y_cm²
==========================
I_a = § µ(y)y²dy - 2a * M * y_cm + a² M
==========================
=> I_a = I_cm + (M * y_cm²) - 2a*M*y_cm + M*a² = I_cm + M * (y_cm² - 2a*y_cm + a²)
==> I_a = I_cm + M * (y_cm - a)² © (phoffric 15-Aug-2010)
If I did use the formulas correctly, and if I got the algebra right, then I have
I_a in terms of known constants, M, I_cm, y_cm, and 'a'
(and the density has been eliminated)
==========================
>> so I don't see how the density function can be "discarded" when taking moments about some arbitary point
Ok, this is BRDQ#5. Let me know if I did this right.
==========================
I_a = § µ(y)(y-a)²dy = § µ(y)(y² - 2ay + a²)dy
= § ( µ(y)y² - 2ayµ(y) + a²µ(y) )dy
= § µ(y)y²dy - § 2ayµ(y)dy + § a²µ(y)dy
I_a = § µ(y)y²dy - 2a § y µ(y)dy + a² § µ(y)dy
====================
M = § µ(y) dy
y_cm = (1/M) § y µ(y) dy ==> § y µ(y) dy = M * y_cm
==> I_a = § µ(y)y²dy - 2a * M * y_cm + a² M
====================
I_cm = § (y - y_cm)² µ(y) dy
= § (y - y_cm)² µ(y) dy
= § (y² - 2 y y_cm + y_cm²) µ(y) dy
= § y² µ(y) dy - § 2 y y_cm µ(y) dy + § y_cm² µ(y) dy
= § y² µ(y) dy - 2 y_cm § y µ(y) dy + y_cm² § µ(y) dy
= § y² µ(y) dy - 2 y_cm § y µ(y) dy + y_cm² § µ(y) dy
I_cm = § y² µ(y) dy - 2 y_cm § y µ(y) dy + y_cm² § µ(y) dy
= § y² µ(y) dy - 2 y_cm (M * y_cm) + y_cm² M
= § y² µ(y) dy - 2 M * y_cm² + M * y_cm²
= § y² µ(y) dy - M * y_cm²
==========================
==> § y² µ(y) dy = I_cm + M * y_cm²
==========================
I_a = § µ(y)y²dy - 2a * M * y_cm + a² M
==========================
=> I_a = I_cm + (M * y_cm²) - 2a*M*y_cm + M*a² = I_cm + M * (y_cm² - 2a*y_cm + a²)
==> I_a = I_cm + M * (y_cm - a)² © (phoffric 15-Aug-2010)
If I did use the formulas correctly, and if I got the algebra right, then I have
I_a in terms of known constants, M, I_cm, y_cm, and 'a'
(and the density has been eliminated)
==========================
ASKER
I wasn't thinking in terms of 'a' since I know that when there are no external forces on the rod, then any rotation will spin around the CM. But, I understand that the reference point when calculating the Moment of Inertia is arbitrary, so if the above can be verified, then I'll keep this formula in mind.
What is interesting to me is that the final form looks like the result when the pivot point of the rod is at point 'a'. But I thought what I was doing was computing I_a even if (1) there is no anchor (as in the OP) or (2) the pivot is at the CM, but I_a is taken relative to point 'a' on the rod. So, I have my doubts now.
What is interesting to me is that the final form looks like the result when the pivot point of the rod is at point 'a'. But I thought what I was doing was computing I_a even if (1) there is no anchor (as in the OP) or (2) the pivot is at the CM, but I_a is taken relative to point 'a' on the rod. So, I have my doubts now.
>>I_a = § µ(y)y²dy - 2a * M * y_cm + a² M
OK
>>(and the density has been eliminated)
The density function has not been eliminated but hidden in the center of mass. There is now a coupling between the mass and the position of where the moment is to be taken - namely a, whereas previously when the density is a constant the parallel theorem applies in calculating the moment of inertia about some arbitary point. The center of mass and the axis of rotation (if that's what one calls it) are no longer identical.
OK
>>(and the density has been eliminated)
The density function has not been eliminated but hidden in the center of mass. There is now a coupling between the mass and the position of where the moment is to be taken - namely a, whereas previously when the density is a constant the parallel theorem applies in calculating the moment of inertia about some arbitary point. The center of mass and the axis of rotation (if that's what one calls it) are no longer identical.
ASKER
>> density function has not been eliminated but hidden in the center of mass
Yes, hidden in y_cm, I_cm, and M. And, if I can come up with a different density function that somehow yields these same three values, then I am led to believe that even though the two rods that have different weight distributions, they will both behave identically. You threw in a value 'a'; and it may be that the two different hidden density functions can be not used any longer once the other three values are known.
So, equivalently, I am led to believe that if I were given y_cm, I_cm, and M that were identical for two rods; and was also told that the rod's had different density distributions, then I can totally ignore that latter fact.
>> There is now a coupling between the mass and the position of where the moment is to be taken
Is there some cases where there is not a coupling between the mass and the position of where the moment is to be taken?
I just looked up the parallel axis (although I had reviewed it a few weeks ago). I don't see anything about any restrictions that the "density is a constant ". (But I am glancing at this too fast, so maybe I am missing something; or did not understand the point you were making.)
http://en.wikipedia.org/wiki/Moment_of_inertia#Parallel_axis_theorem
>> The center of mass and the axis of rotation (if that's what one calls it) are no longer identical.
Sorry, I am not sure I understand. Is this what you mean: If I stick a pin in a rod to create a pivot at some point that is different than the CM, then the rod's CM is rotating about the pivot.
Yes, hidden in y_cm, I_cm, and M. And, if I can come up with a different density function that somehow yields these same three values, then I am led to believe that even though the two rods that have different weight distributions, they will both behave identically. You threw in a value 'a'; and it may be that the two different hidden density functions can be not used any longer once the other three values are known.
So, equivalently, I am led to believe that if I were given y_cm, I_cm, and M that were identical for two rods; and was also told that the rod's had different density distributions, then I can totally ignore that latter fact.
>> There is now a coupling between the mass and the position of where the moment is to be taken
Is there some cases where there is not a coupling between the mass and the position of where the moment is to be taken?
I just looked up the parallel axis (although I had reviewed it a few weeks ago). I don't see anything about any restrictions that the "density is a constant ". (But I am glancing at this too fast, so maybe I am missing something; or did not understand the point you were making.)
http://en.wikipedia.org/wiki/Moment_of_inertia#Parallel_axis_theorem
>> The center of mass and the axis of rotation (if that's what one calls it) are no longer identical.
Sorry, I am not sure I understand. Is this what you mean: If I stick a pin in a rod to create a pivot at some point that is different than the CM, then the rod's CM is rotating about the pivot.
ASKER
Now, I got: I_a = I_cm + M * (y_cm - a)²
And this is for the case (I was thinking) where there was no anchor, but the I was to be taken w.r.t. an arbitrary point a. Yet it looks like the parallel axis theorem where r is the "distance from the center of mass axis of rotation".
When you were talking about 'a', were you talking about the mass spinning on this point 'a' ? This is the case as I understood it where the parallel axis theorem becomes useful.
If I let r = (y_cm - a), then I_a = I_cm + M * r², which has the identical form as the parallel axis theorem. Yet in my case there is no anchor about 'a'. So, I am not sure what big picture I am missing.
And this is for the case (I was thinking) where there was no anchor, but the I was to be taken w.r.t. an arbitrary point a. Yet it looks like the parallel axis theorem where r is the "distance from the center of mass axis of rotation".
When you were talking about 'a', were you talking about the mass spinning on this point 'a' ? This is the case as I understood it where the parallel axis theorem becomes useful.
If I let r = (y_cm - a), then I_a = I_cm + M * r², which has the identical form as the parallel axis theorem. Yet in my case there is no anchor about 'a'. So, I am not sure what big picture I am missing.
ASKER
Here is what I figured out. Given any µ(y) for our rod (or even any distributions of point masses on the rod, which can be closely approximated by a µ(y) ), then from this input, we can figure out M, y_cm, and I_cm.
Then I_a = I_cm + M * (y_cm - a)²
If I create another rod with only two point masses, m1 and m2, where
m1 = m2 = M/2
place at positions (y_cm ± d)
where d = sqrt( I_cm / M )
then this second rod will behave identically to the first rod when a force or impulse is applied as described in the OP.
Not sure if this has any value in solving problems, but it seems interesting to me.
Then I_a = I_cm + M * (y_cm - a)²
If I create another rod with only two point masses, m1 and m2, where
m1 = m2 = M/2
place at positions (y_cm ± d)
where d = sqrt( I_cm / M )
then this second rod will behave identically to the first rod when a force or impulse is applied as described in the OP.
Not sure if this has any value in solving problems, but it seems interesting to me.
ASKER
In http:#33429435, I wrote:
>> there was no anchor, but the I was to be taken w.r.t. an arbitrary point 'a'
After some review, I believe this makes no sense. If there is no anchor or constraints on the rod, then the rod will spin on it y_cm point, so I believe that only I_cm can be used to describe the rod's motion and not I_a for another arbitrary point.
On the other hand, if the rod has a pin at some point 'a' other than y_cm, then to describe the rod's motion, we need to use the value I_a, which is derived from the parallel axis theorem:
I_a = I_cm + M * (y_cm - a)²
In http:#33428047 I now realize that I wasn't deriving anything new. I_a is defined to be the rod's Moment of Inertia when the rod spins around point 'a', and if 'a' is not y_cm, then that must mean that the rod has a pin in it at point 'a' to force the rod to spin around point 'a'. So, all I was doing was deriving the parallel axis theorem (without realizing it at the time).
Do you agree or disagree with my new conclusions?
>> there was no anchor, but the I was to be taken w.r.t. an arbitrary point 'a'
After some review, I believe this makes no sense. If there is no anchor or constraints on the rod, then the rod will spin on it y_cm point, so I believe that only I_cm can be used to describe the rod's motion and not I_a for another arbitrary point.
On the other hand, if the rod has a pin at some point 'a' other than y_cm, then to describe the rod's motion, we need to use the value I_a, which is derived from the parallel axis theorem:
I_a = I_cm + M * (y_cm - a)²
In http:#33428047 I now realize that I wasn't deriving anything new. I_a is defined to be the rod's Moment of Inertia when the rod spins around point 'a', and if 'a' is not y_cm, then that must mean that the rod has a pin in it at point 'a' to force the rod to spin around point 'a'. So, all I was doing was deriving the parallel axis theorem (without realizing it at the time).
Do you agree or disagree with my new conclusions?
ASKER
>> moment of inertia is defined as the sum of the masses multiplied by the square of the distance from the axis of rotation. In your notation § µ(r)r²dr where r is the distance from the center of the object...
I didn't have a form of § µ(r)r²dr. When you say "center", are you referring "center of mass"?
In my accidental derivation of the Parallel Axis Theorem, I did have an intermediate term:
§ y² µ(y) dy = I_cm + M * y_cm²
But I don't see how this is useful by itself.
>> whereas previously when the density is a constant the parallel theorem applies in calculating the moment of inertia about some arbitary point.
Now that I did derive this theorem, I can say that there were no restrictions on µ(y). In particular, µ(y) did not have to be constant. In fact, IMO µ(y) could even have point masses defined within it by making µ(y) extremely large over an extremely tiny interval around the point mass location. So, in this derivation of I_a, µ(y) has been eliminated:
I_a = I_cm + M * (y_cm - a)²
There is only <I_cm, M, y_cm> and the rod's spin point 'a'.
In http:#33442223, it looks to me like I can have a rod of many point masses arbitrarily placed, as one instance of µ(y), compute the triplet, <I_cm, M, y_cm>, and then replace that set of point masses with just two identical point masses at at positions (y_cm ± d) where d = sqrt( I_cm / M ).
I didn't have a form of § µ(r)r²dr. When you say "center", are you referring "center of mass"?
In my accidental derivation of the Parallel Axis Theorem, I did have an intermediate term:
§ y² µ(y) dy = I_cm + M * y_cm²
But I don't see how this is useful by itself.
>> whereas previously when the density is a constant the parallel theorem applies in calculating the moment of inertia about some arbitary point.
Now that I did derive this theorem, I can say that there were no restrictions on µ(y). In particular, µ(y) did not have to be constant. In fact, IMO µ(y) could even have point masses defined within it by making µ(y) extremely large over an extremely tiny interval around the point mass location. So, in this derivation of I_a, µ(y) has been eliminated:
I_a = I_cm + M * (y_cm - a)²
There is only <I_cm, M, y_cm> and the rod's spin point 'a'.
In http:#33442223, it looks to me like I can have a rod of many point masses arbitrarily placed, as one instance of µ(y), compute the triplet, <I_cm, M, y_cm>, and then replace that set of point masses with just two identical point masses at at positions (y_cm ± d) where d = sqrt( I_cm / M ).
ASKER
Hi BigRat,Just wondering whether you've had the time to wade through my results, and see whether you have any issues with it. Thanks, Paul
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ASKER
I split points 250 each between http:#33483110 and http:#33405550. I'll contact mod to fix this. Thanks again!
ASKER
I split points 250 each between http:#33483110 and http:#33405550. I'll contact mod to fix this. Thanks again!
ASKER
So, if I have two different rods, both having the same mass, M, and L, but different density distributions, it may turn out that the two rods can have the same y_cm and I_cm; and therefore, the equations of motion for both rods will be identical for the same application of force or impulse.
Given this assurance, I will see if I can come up with some method of coming up with a µ2(y) given a µ1(y) that yields the same y_cm and I_cm as the µ1(y), other than using a simple 180 degree rotation of the rod, of course.
Thanks.