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Rotation of a pyramid

  • #1
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Homework Statement:

Peti is the foreman of the Egyptian workers who have just finished building the great pyramid of Cheops.
L = 236 m
H = 140 m
M = 6 x 10^9 kg

It is late Friday afternoon, and everyone else has gone home for the day. The Grand Opening ceremony is scheduled for Monday.

The Vizier walks over to Peti and says, "I have good news and bad news. The good news is that it has the right size and shape. Well done."

"What's the bad news?" asks Peti.

"It's facing the wrong way. The face with the access tunnel is supposed to face NORTH, not South. The Big Guy is going to be very unhappy if you don't fix this by Monday morning."

Fortunately, Peti thought something like this might happen, so he placed a frictionless bearing under the very center of the pyramid. It will rotate freely -- if someone can push it hard enough.

Peti runs over to one corner of the pyramid and starts to push, applying a force of F = 500 N.

What is the torque applied by Peti?

How long will it take Peti to rotate the pyramid by half a rotation?

How fast will Peti be moving at the maximum rotational speed?

What is the average power supplied over that time?

Will he finish in time?

Relevant Equations:

torque=Force*radius*sin(theta)
torque=Force*radius*sin(theta)
center of mass x direction = ( 0(6 x 10^9 kg)+ (118m)(6 x 10^9 kg)+ (236m)(6 x 10^9 kg) )/(3(6 x 10^9 kg)) = 118m
center of mass y direction = ( 0(6 x 10^9 kg)+ (140m)(6 x 10^9 kg)+ (0)(6 x 10^9 kg) )/(3(6 x 10^9 kg)) = 46.7 m
radius = (118^2 + 46.7^2)^(1/2) = 126.9m
torque= 500N * 126.9m *sin(90deg)= 63.4*10^3 N*m
I am confused on how to proceed further
 

Answers and Replies

  • #2
haruspex
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center of mass x direction =
Much simpler to take the centre of the pyramid's base as origin.
center of mass y direction =
As in height? Why is that interesting?
It will rotate about a vertical axis.
 
  • #3
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Is the center of mass = (118,0) ?
torque= 500* 118 = 59*10^3 N*m?
 
  • #4
haruspex
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Is the center of mass = (118,0) ?
torque= 500* 118 = 59*10^3 N*m?
In what direction is Peti pushing? The question does not specify, so what would make sense?
 
  • #5
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To find time: power=work/time = torque * angular velocity
work=KE= (1/2)(moment of inertia)(angular velocity)^2 = torque * (Θf- Θi)
(moment of inertia)=∫ (1/12)(width of inner triangle)^2 (dh)(3*mass/ (H*L^2))
(moment of inertia)= (1/12)(width of inner triangle)^2 * (3*mass/ (H*L^2)) ∫ 1(dh)
(moment of inertia)= (1/12)(width of inner triangle)^2 * (3*mass/ (H*L^2))*(h) | (H,0)
width of inner triangle/236m = (70m) / 140m
width of inner triangle= 118m
(moment of inertia)=(1/12)(118m)^2* 3(6*10^9 kg)/ (236m^2)= 375 *10^6 kg*m^2

work=KE= (1/2)(375 *10^6 kg*m^2 )(angular velocity)^2 = (63.4*10^3 N*m) (pi)
angular velocity= 32.6 *10^-3 rad/s
KE= (1/2)(375 *10^6 kg*m^2 )(32.6 *10^-3 rad/s )^2 = 199.2 *10^3 J

power = (63.4*10^3 N*m) ( 32.6 *10^-3 rad/s ) = 2066.8 W
power = 2066.8 W= (199.2 *10^3 J ) / time
time = 96.4 seconds
 
  • #6
haruspex
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To find time: power=work/time = torque * angular velocity
work=KE= (1/2)(moment of inertia)(angular velocity)^2 = torque * (Θf- Θi)
(moment of inertia)=∫ (1/12)(width of inner triangle)^2 (dh)(3*mass/ (H*L^2))
(moment of inertia)= (1/12)(width of inner triangle)^2 * (3*mass/ (H*L^2)) ∫ 1(dh)
(moment of inertia)= (1/12)(width of inner triangle)^2 * (3*mass/ (H*L^2))*(h) | (H,0)
width of inner triangle/236m = (70m) / 140m
width of inner triangle= 118m
(moment of inertia)=(1/12)(118m)^2* 3(6*10^9 kg)/ (236m^2)= 375 *10^6 kg*m^2

work=KE= (1/2)(375 *10^6 kg*m^2 )(angular velocity)^2 = (63.4*10^3 N*m) (pi)
angular velocity= 32.6 *10^-3 rad/s
KE= (1/2)(375 *10^6 kg*m^2 )(32.6 *10^-3 rad/s )^2 = 199.2 *10^3 J

power = (63.4*10^3 N*m) ( 32.6 *10^-3 rad/s ) = 2066.8 W
power = 2066.8 W= (199.2 *10^3 J ) / time
time = 96.4 seconds
There's no point in proceeding to this part of the question until you get the earlier parts right.
Please answer my question in post #4. Think about the direction in which you would push if you were Peti.
 
  • #7
collinsmark
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torque= 500* 118 = 59*10^3 N*m?
I think this might go along with what @haruspex is asking.

But are you sure Peti is pushing at the corner of the pyramid? The problem statement says that Peti ran over to a corner of the pyramid. You seem to have placed him on the center of an edge of the pyramid.
 
  • #8
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torque= 83000 N*m
 
  • #9
collinsmark
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torque= 83000 N*m
That looks better (although you might want to increase your precision a little. But it looks like you've got the right idea.)
 
  • #10
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power=work/time = torque * angular velocity
work=KE= (1/2)(moment of inertia)(angular velocity)^2 = torque * (Θf- Θi)
(moment of inertia)=∫ (1/12)(width of inner triangle)^2 (dh)(3*mass/ (H*L^2))
(moment of inertia)= (1/12)(width of inner triangle)^2 * (3*mass/ (H*L^2)) ∫ 1(dh)
(moment of inertia)= (1/12)(width of inner triangle)^2 * (3*mass/ (H*L^2))*(h) | (H,0)
width of inner triangle/236m = (70m) / 140m
width of inner triangle= 118m
(moment of inertia)=(1/12)(118m)^2* 3(6*10^9 kg)/ (236m^2)= 375 *10^6 kg*m^2

work=KE= (1/2)(375 *10^6 kg*m^2 )(angular velocity)^2 = (83000 N*m) (pi)
angular velocity= 37.3 *10^-3 rad/s
KE= (1/2)(375 *10^6 kg*m^2 )(37.3 *10^-3 rad/s )^2 = 260.8 *10^3 J

power = (83000 N*m) ( 37.3 *10^-3 rad/s ) = 3095.9 W
power = 3095.9 W= (260.8 *10^3 J ) / time
time = 84.2 seconds
 
  • #11
collinsmark
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(moment of inertia)=∫ (1/12)(width of inner triangle)^2 (dh)(3*mass/ (H*L^2))
(moment of inertia)= (1/12)(width of inner triangle)^2 * (3*mass/ (H*L^2)) ∫ 1(dh)
(moment of inertia)= (1/12)(width of inner triangle)^2 * (3*mass/ (H*L^2))*(h) | (H,0)
width of inner triangle/236m = (70m) / 140m
width of inner triangle= 118m
(moment of inertia)=(1/12)(118m)^2* 3(6*10^9 kg)/ (236m^2)= 375 *10^6 kg*m^2
We should probably figure out the moment of inertia before continuing. The moment of inertia that I calculated is about 5 orders of magnitude larger than the result that you got.

Could you maybe walk us through your logic when calculating the moment of inertia? What's the "inner triangle"?
 
  • #12
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I drew a horizontal line through the middle of the triangle. It 'cut' the pyramid into an upper, smaller triangle and a lower trapezoid. The base of the upper triangle is (width of inner triangle).
 
  • #13
collinsmark
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I drew a horizontal line through the middle of the triangle. It 'cut' the pyramid into an upper, smaller triangle and a lower trapezoid. The base of the upper triangle is (width of inner triangle).
I'm sorry, but I still don't quite understand. You drew a horizontal line through which triangle?
 
  • #14
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d(mass) = d(v) (density)
density= mass/volume
d(v)= (w^2 *dh)
d(mass) = (w^2 *dh) (M/(1/3(H*L^2))
d(moment of inertia)= (1/12) *d(mass)
moment of inertia=∫ (1/12) (w^2 *dh) (M/ (1/3(H*L^2))
 
  • #15
collinsmark
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d(mass) = d(v) (density)
density= mass/volume
d(v)= (w^2 *dh)
d(mass) = (w^2 *dh) (M/(1/3(H*L^2))
d(moment of inertia)= (1/12) *d(mass)
moment of inertia=∫ (1/12) (w^2 *dh) (M/ (1/3(H*L^2))
I agree that the density of the pyramid is

[itex] \rho = \frac{3 M}{L^2 H} [/itex],

where [itex] H [/itex] is the height of the pyramid and [itex] L [/itex] is the width of the pyramid.

Then we can build up the pyamid by summing many, many thin cuboids. These cuboids start large at the base, and keep getting smaller as we increase height.

But the moment of inertia of a square, thin cuboid rotated around its center (in its plane) is

[itex] I = \frac{1}{6}m w^2 [/itex] (not [itex] \frac{1}{12} m w^2) [/itex].

[Edit: Since we're going to be treating these cuboids as being infinitesimally thin, you could write the differential moment of inertia as [itex] dI = \frac{1}{6}(dm)w^2 [/itex]. The point I was making above is the moment of intertia formula for a square cuboid has a [itex] \frac{1}{6} [/itex] in it, not [itex] \frac{1}{12} [/itex].]

These cuboids that we're building up have both width and length of [itex] w [/itex] (it's a square) and a height of [itex] dh [/itex].

So the [differential] mass of a given cuboid is density times its volume,

[itex] dm = \rho w^2 dh [/itex].

But as keep building up the pyramid, [itex] w [/itex] changes. When [itex] h = 0 , \ w = L [/itex]. But when [itex] h = H, \ w =0. [/itex] You haven't applied this scaling in your equation.

Note that you need to apply this scaling in two places. Once for the mass calculation, and again for the moment of inertia formulation.

[Edit: in other words, you need to provide an expression for [itex] w [/itex], based on [itex] L, H, [/itex] and [itex] h [/itex].]

[Edit: Still in other words, you'll be integrating over height [itex] h [/itex]. The width of the cuboid [itex] w [/itex] is a function of [itex] h [/itex]. So you can't leave the variable "[itex] w [/itex]" in your equation when you integrate because it's not constant. You need to reformulate it as a function of [itex] h [/itex] and use that instead.]
 
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  • #16
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w= (L- (L/H)h )
 
  • #17
jbriggs444
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w= (L- (L/H)h )
You can simplify that expression by measuring "h" from the top down.
 
  • #18
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W =( H-h)L /H
 
  • #19
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If this w is substituted into I, I =0
 
  • #20
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  • #21
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W =( H-h)L /H
That will work too, and is mathematically equivalent.
 
  • #22
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dI=(1/6)(dm)w2
dI=(1/6)(ρ*w2*dh) w2
I = (1/6)((3M/(L^2 *H)) w2*h) w2
I=(1/6)(3M/(L^2 *H)) ( (H-h)L /H )^4 *h |(H,0)
I=0 -0
 
  • #23
collinsmark
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dI=(1/6)(dm)w2
dI=(1/6)(ρ*w2*dh) w2
I = (1/6)((3M/(L^2 *H)) w2*h) w2
I=(1/6)(3M/(L^2 *H)) ( (H-h)L /H )^4 *h |(H,0)
I=0 -0
It's really tough to read that without [itex] \LaTeX [/itex]. Anyway, it's important to substitute the [itex] w = \frac{H - h}{H}l [/itex] into both places in the equation before you integrate.

It will show up in one place in the differential mass [itex] dm [/itex] (which it looks like you've done), but don't forget about the second [itex] w [/itex] in the [itex] dI = \frac{1}{6}dm w^2 [/itex] part.

[Edit: Okay I think I see what you did. You integrated first and then made the [itex] w [/itex] substitution later. But the [itex] w [/itex] substitution must be done first, before integration.]

This all has to be done before integration.

[Edit: In other words, by the time you actually start the integration, there shouldn't be any [itex] w [/itex]'s left.]
 
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  • #24
jbriggs444
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W =( H-h)L /H
Simpler. W = h L/H

At height 0 (top of pyramid), W = 0 L/H = 0. This checks out. The width of the pyramid at the top is zero.
At height H (bottom of the pyramid), W = L. This checks out. The width of the pyramid at the bottom is the full L.

You integrate from 0 to H either way. But this way the integrand is simpler.

Edit: Think about it this way. The moment of inertia of a pyramid rotating on its base about a vertical axis is the same as for an inverted pyramid rotating on its tip about the same vertical axis.
 
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  • #25
collinsmark
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Simpler. W = h L/H

At height 0 (top of pyramid), W = 0 L/H = 0. This checks out. The width of the pyramid at the top is zero.
At height H (bottom of the pyramid), W = L. This checks out. The width of the pyramid at the bottom is the full L.

You integrate from 0 to H either way. But this way the integrand is simpler.
That will work too. It just requires a re-definition of h.

The important thing is to pick one model and stick with it consistently. Whichever floats your boat. (Although you are correct in that redefining h in this way will make the integral a little simpler.)
 
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