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Another (mostly) calculus question
#1
Another (mostly) calculus question
In one of the solutions to the problems in my physics textbooks, it's written (without further explanation, as if it were somehow obvious) that the gravitational potential energy of a vertical rod is given by the formula U=1/2*m*g*l. How does that make any sense? The sum of the gravitational energy of the infinitesimally small parts of the rod is obviously U=integral(m*g*l,l,0,l)=1/2*m*g*l^2.
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#2
RE: Another (mostly) calculus question
Don't need calculus unless you're a masochist. If the rod is standing on the ground, its center of mass is 1/2 way up the bar. PE=mgl for the center of mass, thus mgl/2 for the bar.
I never thought I'd live long enough to become a grumpy old bastard. Here I am, killing it!
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#3
RE: Another (mostly) calculus question
(April 26, 2019 at 2:13 pm)Fireball Wrote: Don't need calculus unless you're a masochist. If the rod is standing on the ground, its center of mass is 1/2 way up the bar. PE=mgl for the center of mass, thus mgl/2 for the bar.

Practically yes.


But - if picking nits --

Earth' s gravitational attraction is stronger at the " bottom" ( in relation to the earth) than it is at the top - so the center of mass will be below the centerpoint of the rod...

Tongue
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#4
RE: Another (mostly) calculus question
What if the rod is made of Viagra?
God(s) and religions are man made and the bane of humanity. 

Of all the things I've lost, I miss my mind the most. Ozzy or Twain/take your pick
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#5
RE: Another (mostly) calculus question
(April 26, 2019 at 2:25 pm)onlinebiker Wrote:
(April 26, 2019 at 2:13 pm)Fireball Wrote: Don't need calculus unless you're a masochist. If the rod is standing on the ground, its center of mass is 1/2 way up the bar. PE=mgl for the center of mass, thus mgl/2 for the bar.

Practically yes.


But - if picking nits --

Earth' s gravitational attraction is stronger at the " bottom" ( in relation to the earth) than it is at the top - so the center of mass will be below the centerpoint of the rod...

Tongue

That difference is orders of magnitude smaller than the error in the measurement of the length of the rod. Hmph
I never thought I'd live long enough to become a grumpy old bastard. Here I am, killing it!
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#6
RE: Another (mostly) calculus question
(April 26, 2019 at 5:37 pm)Fireball Wrote:
(April 26, 2019 at 2:25 pm)onlinebiker Wrote: Practically yes.


But - if picking nits --

Earth' s gravitational attraction is stronger at the " bottom" ( in relation to the earth) than it is at the top - so the center of mass will be below the centerpoint of the rod...

Tongue

That difference is orders of magnitude smaller than the error in the measurement of the length of the rod. Hmph

I've been told the length of the rod doesn't matter.
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#7
RE: Another (mostly) calculus question
(April 26, 2019 at 5:41 pm)IWNKYAAIMI Wrote:
(April 26, 2019 at 5:37 pm)Fireball Wrote: That difference is orders of magnitude smaller than the error in the measurement of the length of the rod. Hmph

I've been told the length of the rod doesn't matter.

Girth! Amirite!? Naughty
I never thought I'd live long enough to become a grumpy old bastard. Here I am, killing it!
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#8
RE: Another (mostly) calculus question
(April 26, 2019 at 5:37 pm)Fireball Wrote:
(April 26, 2019 at 2:25 pm)onlinebiker Wrote: Practically yes.


But - if picking nits --

Earth' s gravitational attraction is stronger at the " bottom" ( in relation to the earth) than it is at the top - so the center of mass will be below the centerpoint of the rod...

Tongue

That difference is orders of magnitude smaller than the error in the measurement of the length of the rod. Hmph
Agreed...

But - somebody had to pick the nits...

Tongue
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#9
RE: Another (mostly) calculus question
(April 26, 2019 at 10:30 pm)onlinebiker Wrote:
(April 26, 2019 at 5:37 pm)Fireball Wrote: That difference is orders of magnitude smaller than the error in the measurement of the length of the rod. Hmph
Agreed...

But - somebody had to pick the nits...

Tongue

You been looking at the back of my head, again? Huh
I never thought I'd live long enough to become a grumpy old bastard. Here I am, killing it!
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#10
RE: Another (mostly) calculus question
(April 26, 2019 at 12:57 pm)FlatAssembler Wrote: In one of the solutions to the problems in my physics textbooks, it's written (without further explanation, as if it were somehow obvious) that the gravitational potential energy of a vertical rod is given by the formula U=1/2*m*g*l. How does that make any sense? The sum of the gravitational energy of the infinitesimally small parts of the rod is obviously U=integral(m*g*l,l,0,l)=1/2*m*g*l^2.
You've made a mistake in assuming the potential energy of an infinitesimally small element of the rod in the integral. It actually is the following:
dU=dm*g*x
where dm is an infinitely small element of mass and x is its height above an arbitrary zero energy level. Since dm=rho(x)*dx where rho(x) is lineal density of the rod, it can be rewritten as
dU=rho(x)*g*x*dx .
If the rod is uniform (rho(x)=m/l=const) we arrive at
dU=(m/l)*g*x*dx
and the integral is the following:
U = int((m/l)*g*x dx,  x=0 to x=l) = (m*g*l^2)/(2*l) = 1/2*m*g*l.

Also it's a good practice to give distinct names to your parameters and variables. In
Quote:U=integral(m*g*l , l ,0,l)
you use l for both the length of the rod (a constant parameter in this context) and the height above the ground (an integration variable). Even though in this very case they are measured along the same line they are different things. While it may not look as a big deal it will inevitably lead to confusion sooner or later.

P. S. I have to note that for an inclined rod the calculation will not be quite the same bacause dm=rho(x)*dx=m/l*dx is only valid if the rod is parallel to the axis x. Otherwise you have to introduce an inclination angle.
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