[math-fun] Stuff I didn't know about pendulums
First of all, a spherical bob (on a massless string) centered L below the pivot is a compound pendulum. Its period, for a given amplitude, depends on its radius R! It acts like a point mass centered at L+(3/5)*R^2/L rather than L. Unless it's hollow, when it's just L+R^2/L. So you can tell if it's hollow by swinging it. For a unit n-cube hanging by a vertex, this "center of oscillation" is 2/3 down the main diagonal, for all n. Wikipedia claims the center of oscillation is also the center of percussion. So if you hang a bat by the knob, and match oscillations with a ball on an adjustably long string, the ball will hang down to the sweet spot. If you're the sort of person who swings bats by their knob. Mathematica 7 clams on the ODE of motion: {\[Theta]''[t] == -g*Sin[\[Theta][t]]/L, \[Theta][0] == 0} but Macsyma returns an integral for t[\Theta] which Mma knows is an EllipticF. Testing by taking the limit for small \[Theta]'[0] insists the period approaches 149/30 Sqrt[L/g] instead of 2 pi Sqrt[L/g]! The bug is in Limit. Mma then solves for Theta, \[Theta][t] -> 2*JacobiAmplitude[ (1/2)*t*Derivative[1][\[Theta]][0],(4*g)/(L*Derivative[1][\[Theta]][0]^2)] with period \[CapitalLambda] == 4*Sqrt[L/g]*EllipticK[Sin[Subscript[\[Theta], max]/2]^2] and amplitude Subscript[\[Theta], max] == 2*ArcSin[(1/2)*Sqrt[L/g]*Derivative[1][\[Theta]][0]] The differentiator and simplifiers eventually confess that these solve the ODE. But had I thought to search "pendulum" in the Mma doc, it would have led me to an "application" of JacobiAmplitude where it is claimed, without derivation, to describe an overswung pendulum. This would have been a clue. I'm surprised this all isn't more obvious in, e.g., Wikipedia or MathWorld. Or DSolve. --rwg
rwg>[A spherical bob of radius R] acts
like a point mass centered at L+(3/5)*R^2/L rather than L. Unless it's hollow, when it's just L+R^2/L.
How can this be right? It says a hollow sphere suspended by its north pole is like a point mass at its south pole. More generally, for a hollow shell of radii r1<r2 centered at L, I get In[17]:= Assuming[L >= 0, Integrate[ EuclideanDistance[{0, 0, L}, {r*Cos[\[Theta]]*Sin[\[Phi]], r*Sin[\[Theta]]*Sin[\[Phi]], r*Cos[\[Phi]]}]^2*r^2* Sin[\[Phi]], {r, r1, r2}, {\[Theta], 0, 2*\[Pi]}, {\[Phi], 0, \[Pi]}]/ Integrate[(L - r*Cos[\[Phi]])*r^2*Sin[\[Phi]], {r, r1, r2}, {\[Theta], 0, 2*\[Pi]}, {\[Phi], 0, \[Pi]}]]; In[18]:= Factor /@ Apart[%] Out[18]= 2 3 3 r2 3 r1 (r1 + r2) L + ----- + ----------------------- 5 L 2 2 5 L (r1 + r1 r2 + r2 ) assuming uniform density and the formula Int |P-z|^2 dV / Int (P-z).{0,0,1} dV, where P is the pivot {0,0,L} and z scans the volume. This came from an analysis (which I trust) that collinear point masses m1 and m2 hanging at L1 and L2 on a massless stick have an effective length of m1*L1^2+m2*L2^2 ---------------, m1*L1+m2*L2 the weighted "contraharmonic mean". The .{0,0,1} was about potential energy. What is my sin, Lanvin? --rwg On Sun, Nov 28, 2010 at 5:54 PM, Bill Gosper <billgosper@gmail.com> wrote:
First of all, a spherical bob (on a massless string) centered L below the pivot is a compound pendulum. Its period, for a given amplitude, depends on its radius R! It acts like a point mass centered at L+(3/5)*R^2/L rather than L. Unless it's hollow, when it's just L+R^2/L. So you can tell if it's hollow by swinging it. For a unit n-cube hanging by a vertex, this "center of oscillation" is 2/3 down the main diagonal, for all n.
Wikipedia claims the center of oscillation is also the center of percussion. So if you hang a bat by the knob, and match oscillations with a ball on an adjustably long string, the ball will hang down to the sweet spot. If you're the sort of person who swings bats by their knob.
Mathematica 7 clams on the ODE of motion: {\[Theta]''[t] == -g*Sin[\[Theta][t]]/L, \[Theta][0] == 0} but Macsyma returns an integral for t[\Theta] which Mma knows is an EllipticF. Testing by taking the limit for small \[Theta]'[0] insists the period approaches 149/30 Sqrt[L/g] instead of 2 pi Sqrt[L/g]! The bug is in Limit.
Mma then solves for Theta, \[Theta][t] -> 2*JacobiAmplitude[
(1/2)*t*Derivative[1][\[Theta]][0],(4*g)/(L*Derivative[1][\[Theta]][0]^2)] with period \[CapitalLambda] == 4*Sqrt[L/g]*EllipticK[Sin[Subscript[\[Theta], max]/2]^2] and amplitude Subscript[\[Theta], max] == 2*ArcSin[(1/2)*Sqrt[L/g]*Derivative[1][\[Theta]][0]] The differentiator and simplifiers eventually confess that these solve the ODE.
But had I thought to search "pendulum" in the Mma doc, it would have led me to an "application" of JacobiAmplitude where it is claimed, without derivation, to describe an overswung pendulum. This would have been a clue.
I'm surprised this all isn't more obvious in, e.g., Wikipedia or MathWorld. Or DSolve. --rwg
Bill, you can work out many of these moments of inertia mentally if you use the "parallel axis theorem": http://en.wikipedia.org/wiki/Parallel_axis_theorem Veit On Nov 29, 2010, at 5:21 PM, Bill Gosper wrote:
rwg>[A spherical bob of radius R] acts
like a point mass centered at L+(3/5)*R^2/L rather than L. Unless it's hollow, when it's just L+R^2/L.
How can this be right? It says a hollow sphere suspended by its north pole is like a point mass at its south pole. More generally, for a hollow shell of radii r1<r2 centered at L, I get
In[17]:= Assuming[L >= 0, Integrate[ EuclideanDistance[{0, 0, L}, {r*Cos[\[Theta]]*Sin[\[Phi]], r*Sin[\[Theta]]*Sin[\[Phi]], r*Cos[\[Phi]]}]^2*r^2* Sin[\[Phi]], {r, r1, r2}, {\[Theta], 0, 2*\[Pi]}, {\[Phi], 0, \[Pi]}]/ Integrate[(L - r*Cos[\[Phi]])*r^2*Sin[\[Phi]], {r, r1, r2}, {\[Theta], 0, 2*\[Pi]}, {\[Phi], 0, \[Pi]}]];
In[18]:= Factor /@ Apart[%]
Out[18]= 2 3 3 r2 3 r1 (r1 + r2) L + ----- + ----------------------- 5 L 2 2 5 L (r1 + r1 r2 + r2 )
assuming uniform density and the formula
Int |P-z|^2 dV / Int (P-z).{0,0,1} dV,
where P is the pivot {0,0,L} and z scans the volume. This came from an analysis (which I trust) that collinear point masses m1 and m2 hanging at L1 and L2 on a massless stick have an effective length of
m1*L1^2+m2*L2^2 ---------------, m1*L1+m2*L2
the weighted "contraharmonic mean". The .{0,0,1} was about potential energy.
What is my sin, Lanvin? --rwg
On Sun, Nov 28, 2010 at 5:54 PM, Bill Gosper <billgosper@gmail.com> wrote:
First of all, a spherical bob (on a massless string) centered L below the pivot is a compound pendulum. Its period, for a given amplitude, depends on its radius R! It acts like a point mass centered at L+(3/5)*R^2/L rather than L. Unless it's hollow, when it's just L+R^2/L. So you can tell if it's hollow by swinging it. For a unit n-cube hanging by a vertex, this "center of oscillation" is 2/3 down the main diagonal, for all n.
Wikipedia claims the center of oscillation is also the center of percussion. So if you hang a bat by the knob, and match oscillations with a ball on an adjustably long string, the ball will hang down to the sweet spot. If you're the sort of person who swings bats by their knob.
Mathematica 7 clams on the ODE of motion: {\[Theta]''[t] == -g*Sin[\[Theta][t]]/L, \[Theta][0] == 0} but Macsyma returns an integral for t[\Theta] which Mma knows is an EllipticF. Testing by taking the limit for small \[Theta]'[0] insists the period approaches 149/30 Sqrt[L/g] instead of 2 pi Sqrt[L/g]! The bug is in Limit.
Mma then solves for Theta, \[Theta][t] -> 2*JacobiAmplitude[
(1/2)*t*Derivative[1][\[Theta]][0],(4*g)/(L*Derivative[1][\[Theta]][0]^2)] with period \[CapitalLambda] == 4*Sqrt[L/g]*EllipticK[Sin[Subscript[\[Theta], max]/2]^2] and amplitude Subscript[\[Theta], max] == 2*ArcSin[(1/2)*Sqrt[L/g]*Derivative[1][\[Theta]][0]] The differentiator and simplifiers eventually confess that these solve the ODE.
But had I thought to search "pendulum" in the Mma doc, it would have led me to an "application" of JacobiAmplitude where it is claimed, without derivation, to describe an overswung pendulum. This would have been a clue.
I'm surprised this all isn't more obvious in, e.g., Wikipedia or MathWorld. Or DSolve. --rwg
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Veit>Bill, you can work out many of these moments of inertia mentally if you use the "parallel axis theorem": http://en.wikipedia.org/wiki/Parallel_axis_theorem Veit Veit, the center of oscillation seems to be the ratio of moment of inertia to some other moment. The math isn't the problem--it's my physics. --Bill On Mon, Nov 29, 2010 at 2:21 PM, Bill Gosper <billgosper@gmail.com> wrote:
rwg>[A spherical bob of radius R] acts
like a point mass centered at L+(3/5)*R^2/L rather than L. Unless it's hollow, when it's just L+R^2/L.
How can this be right? It says a hollow sphere suspended by its north pole is like a point mass at its south pole. More generally, for a hollow shell of radii r1<r2 centered at L, I get
In[17]:= Assuming[L >= 0, Integrate[ EuclideanDistance[{0, 0, L}, {r*Cos[\[Theta]]*Sin[\[Phi]], r*Sin[\[Theta]]*Sin[\[Phi]], r*Cos[\[Phi]]}]^2*r^2* Sin[\[Phi]], {r, r1, r2}, {\[Theta], 0, 2*\[Pi]}, {\[Phi], 0, \[Pi]}]/ Integrate[(L - r*Cos[\[Phi]])*r^2*Sin[\[Phi]], {r, r1, r2}, {\[Theta], 0, 2*\[Pi]}, {\[Phi], 0, \[Pi]}]];
In[18]:= Factor /@ Apart[%]
Out[18]= 2 3 3 r2 3 r1 (r1 + r2) L + ----- + ----------------------- 5 L 2 2 5 L (r1 + r1 r2 + r2 )
assuming uniform density and the formula
Int |P-z|^2 dV / Int (P-z).{0,0,1} dV,
where P is the pivot {0,0,L} and z scans the volume. This came from an analysis (which I trust) that collinear point masses m1 and m2 hanging at L1 and L2 on a massless stick have an effective length of
m1*L1^2+m2*L2^2 ---------------, m1*L1+m2*L2
the weighted "contraharmonic mean". The .{0,0,1} was about potential energy.
What is my sin, Lanvin? --rwg
On Sun, Nov 28, 2010 at 5:54 PM, Bill Gosper <billgosper@gmail.com> wrote:
First of all, a spherical bob (on a massless string) centered L below the pivot is a compound pendulum. Its period, for a given amplitude, depends on its radius R! It acts like a point mass centered at L+(3/5)*R^2/L rather than L. Unless it's hollow, when it's just L+R^2/L. So you can tell if it's hollow by swinging it. For a unit n-cube hanging by a vertex, this "center of oscillation" is 2/3 down the main diagonal, for all n.
Wikipedia claims the center of oscillation is also the center of percussion. So if you hang a bat by the knob, and match oscillations with a ball on an adjustably long string, the ball will hang down to the sweet spot. If you're the sort of person who swings bats by their knob.
Mathematica 7 clams on the ODE of motion: {\[Theta]''[t] == -g*Sin[\[Theta][t]]/L, \[Theta][0] == 0} but Macsyma returns an integral for t[\Theta] which Mma knows is an EllipticF. Testing by taking the limit for small \[Theta]'[0] insists the period approaches 149/30 Sqrt[L/g] instead of 2 pi Sqrt[L/g]! The bug is in Limit.
Mma then solves for Theta, \[Theta][t] -> 2*JacobiAmplitude[
(1/2)*t*Derivative[1][\[Theta]][0],(4*g)/(L*Derivative[1][\[Theta]][0]^2)] with period \[CapitalLambda] == 4*Sqrt[L/g]*EllipticK[Sin[Subscript[\[Theta], max]/2]^2] and amplitude Subscript[\[Theta], max] == 2*ArcSin[(1/2)*Sqrt[L/g]*Derivative[1][\[Theta]][0]] The differentiator and simplifiers eventually confess that these solve the ODE.
But had I thought to search "pendulum" in the Mma doc, it would have led me to an "application" of JacobiAmplitude where it is claimed, without derivation, to describe an overswung pendulum. This would have been a clue.
I'm surprised this all isn't more obvious in, e.g., Wikipedia or MathWorld. Or DSolve. --rwg
On Nov 29, 2010, at 8:24 PM, Bill Gosper wrote:
Veit, the center of oscillation seems to be the ratio of moment of inertia to some other moment. The math isn't the problem--it's my physics. --Bill
it's all just natural philosophy ... Here's the equation of motion, I phi'' = T I = moment of inertia about point of support phi = angle about point of support T = torque due to gravity The torque is the same as if all the mass were located at the center of mass. (That's because the gravitational acceleration is a constant vector field, approximately). Let R be the distance between the point of support and the center of mass. Also, let I_0 be the moment of inertia about the center of mass. By the parallel axis theorem I = I_0 + MR^2, where M is the mass of the pendulum. The equation of motion is therefore (I_0 + MR^2) phi'' = - MgR sin(phi). The ratio I_0/M = R_0^2 defines the "radius of gyration". Divide through by MR and you get (R_0^2/R + R) phi'' = - g sin{phi). This is the same as the equation for a point-mass pendulum on a massless string of length L = R_0^2/R + R. As an example, a spherical shell of radius r supported at its north pole has R_0 = r, R = r, L = 2r (as if all the mass were placed at the south pole). Veit
On Mon, Nov 29, 2010 at 5:24 PM, Bill Gosper <billgosper@gmail.com> wrote:
Veit>Bill, you can work out many of these moments of inertia mentally if you
use the "parallel axis theorem": http://en.wikipedia.org/wiki/Parallel_axis_theorem
Veit
Veit, the center of oscillation seems to be the ratio of moment of inertia to some other moment. The math isn't the problem--it's my physics. --Bill
Well, the math CAN be a problem-- Proof that pi = 149/60 :) Out[312]= 2 1 w 4 4 EllipticF[- ArcCos[1 - --], --] 2 2 2 w --------------------------------- w
In[313]:= Limit[%, w -> 0]
Out[313]= 149/60
In[314]:= $MaxExtraPrecision = \[Infinity]; N[%% /. w -> 10^-69, 69]
Out[314]= 3.14159265358979323846264338327950288419716939937510582097494459230782 + 0.*10^-69 I
And luckily for me, the Doc search for "pendulum" omitted EllipticK, whose first two applications give the period formula with Sin[alpha/2] instead of Sin[alpha/2]^2.
And 7.0 doesn't seem to know that
3/4 3/2 2 (-1) Pi EllipticK[2] == -------------------- . 1 3 Gamma[-(-)] Gamma[-] 4 4 --rwg
Thought for today (simplfied by Julian) sum(sech(((%pi * n)/(2 * sqrt(3)))),n, - inf,inf) = ((3 * 2^(1/6) * G^3((1/3)))/(2 * 3^(1/4) * (sqrt(3) + 1)^(5/2) * sqrt(5 * sqrt(3) - 4 * sqrt(2) - 3) * (%pi)^2)) inf ==== \ %pi n
sech(---------) =
/ 2 sqrt(3) ==== n = - inf 1/6 3 1 3 2 gamma (-) 3 ------------------------------------------------------------ 1/4 5/2 2 2 3 (sqrt(3) + 1) sqrt(5 sqrt(3) - 4 sqrt(2) - 3) %pi Empirical conjecture: (1/2)*Integrate[(r^2*(Pi*r*Cos[\[Theta]]* Hypergeometric2F1[1/4, 3/4, 2, (4*r^2*Cos[\[Theta]]^2)/(1 + r^2)^2] + 4*(1 + r^2)*HypergeometricPFQ[{-(1/4), 1/4, 1}, {1/2, 3/2}, (4*r^2*Cos[\[Theta]]^2)/(1 + r^2)^2]))/ Sqrt[1 + r^2], {r, 0, 1}, {\[Theta], 0, 2*Pi}] == (8*Pi)/5
participants (2)
-
Bill Gosper -
Veit Elser