Tough integral from a falling clock. $int_0^{2pi} sqrt{g^2t^2 + 2rgtsin(t) + r^2} {rm d}t$












2












$begingroup$


A clock is under free fall for $60$ seconds and its second hand makes exactly one revolution during that period of time. It begins at rest with its second hand facing upwards. Given that the second hand has a length of $r>0$ and that the downward gravitational acceleration is some constant $g>0$, what is the total distance traveled by the tip of the second hand?



I came up with this little problem myself and got stuck at the end with a possibly impossible integral to evaluate. Here's a quick graphical interpretation of the problem: https://www.desmos.com/calculator/ck8gxjughf



So, I started with the parametric equations representing the path of the tip of the second hand:



$$
begin{align}
x = rsin left( frac{pi t}{30} right), quad
y = rcos left( frac{pi t}{30} right) - frac12 g t^2
&& && text{for } 0 leq t leq 60
end{align}
$$



Then the total distance traveled can immediately be written as
$$
begin{align}
int_0^{60} sqrt{ dot x ^2 + dot y ^2} dt
& = int_0^{60} sqrt{ left( frac{pi r}{30} cos left(frac{pi t}{30} right) right)^2 + left( - frac{pi r}{30} sin left(frac{pi t}{30} right) - gt right)^2} {rm d}t \
& = int_0^{60} sqrt{ g^2t^2 + frac{pi r g}{15} t sin left( frac{pi t}{30} right) + left( frac{pi r}{30} right)^2} {rm d}t
end{align}
$$



Which you can quickly approximate, giving a semi-satisfying answer to the problem, but I doubt has a closed-form solution in terms of $r$ and $g$.



Since having the revolution take $60$ seconds seems to make things messy, and that we're more interested in evaluating these integrals, consider instead
$$
begin{align}
x=r sin(t), quad y = r cos(t) - frac12 gt^2 && && text{ for } 0 leq t leq 2pi
end{align}
$$




Are there any closed-form solutions of this integral for specific values of $r$ and $g$?
$$
int_0^{2pi} sqrt{g^2t^2 + 2rgtsin(t) + r^2} {rm d}t
$$







integration definite-integrals parametric classical-mechanics arc-length







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    As this is a physics problem, I'm tempted to examine the limits $r ll 1$, for which the $r^{2}$ in the square root is suppressed and a Taylor expansion of the square root to first order leads to $2pi^{2}g$. While $r gg 1$ suppresses the $g^{2}t^{2}$ term and leads to $2pi(r-g)$. Mathematica is unable to evaluate the general case, nor the related integrals $int_{0}^{2pi}sqrt{1+tsin t},mathrm{d}t$ and $int_{0}^{2pi}sqrt{t^{2}+tsin t},mathrm{d}t$.
    $endgroup$
    – Ininterrompue
    Dec 28 '18 at 19:25


















2












$begingroup$


A clock is under free fall for $60$ seconds and its second hand makes exactly one revolution during that period of time. It begins at rest with its second hand facing upwards. Given that the second hand has a length of $r>0$ and that the downward gravitational acceleration is some constant $g>0$, what is the total distance traveled by the tip of the second hand?



I came up with this little problem myself and got stuck at the end with a possibly impossible integral to evaluate. Here's a quick graphical interpretation of the problem: https://www.desmos.com/calculator/ck8gxjughf



So, I started with the parametric equations representing the path of the tip of the second hand:



$$
begin{align}
x = rsin left( frac{pi t}{30} right), quad
y = rcos left( frac{pi t}{30} right) - frac12 g t^2
&& && text{for } 0 leq t leq 60
end{align}
$$



Then the total distance traveled can immediately be written as
$$
begin{align}
int_0^{60} sqrt{ dot x ^2 + dot y ^2} dt
& = int_0^{60} sqrt{ left( frac{pi r}{30} cos left(frac{pi t}{30} right) right)^2 + left( - frac{pi r}{30} sin left(frac{pi t}{30} right) - gt right)^2} {rm d}t \
& = int_0^{60} sqrt{ g^2t^2 + frac{pi r g}{15} t sin left( frac{pi t}{30} right) + left( frac{pi r}{30} right)^2} {rm d}t
end{align}
$$



Which you can quickly approximate, giving a semi-satisfying answer to the problem, but I doubt has a closed-form solution in terms of $r$ and $g$.



Since having the revolution take $60$ seconds seems to make things messy, and that we're more interested in evaluating these integrals, consider instead
$$
begin{align}
x=r sin(t), quad y = r cos(t) - frac12 gt^2 && && text{ for } 0 leq t leq 2pi
end{align}
$$




Are there any closed-form solutions of this integral for specific values of $r$ and $g$?
$$
int_0^{2pi} sqrt{g^2t^2 + 2rgtsin(t) + r^2} {rm d}t
$$







integration definite-integrals parametric classical-mechanics arc-length







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    As this is a physics problem, I'm tempted to examine the limits $r ll 1$, for which the $r^{2}$ in the square root is suppressed and a Taylor expansion of the square root to first order leads to $2pi^{2}g$. While $r gg 1$ suppresses the $g^{2}t^{2}$ term and leads to $2pi(r-g)$. Mathematica is unable to evaluate the general case, nor the related integrals $int_{0}^{2pi}sqrt{1+tsin t},mathrm{d}t$ and $int_{0}^{2pi}sqrt{t^{2}+tsin t},mathrm{d}t$.
    $endgroup$
    – Ininterrompue
    Dec 28 '18 at 19:25
















2












2








2


0



$begingroup$


A clock is under free fall for $60$ seconds and its second hand makes exactly one revolution during that period of time. It begins at rest with its second hand facing upwards. Given that the second hand has a length of $r>0$ and that the downward gravitational acceleration is some constant $g>0$, what is the total distance traveled by the tip of the second hand?



I came up with this little problem myself and got stuck at the end with a possibly impossible integral to evaluate. Here's a quick graphical interpretation of the problem: https://www.desmos.com/calculator/ck8gxjughf



So, I started with the parametric equations representing the path of the tip of the second hand:



$$
begin{align}
x = rsin left( frac{pi t}{30} right), quad
y = rcos left( frac{pi t}{30} right) - frac12 g t^2
&& && text{for } 0 leq t leq 60
end{align}
$$



Then the total distance traveled can immediately be written as
$$
begin{align}
int_0^{60} sqrt{ dot x ^2 + dot y ^2} dt
& = int_0^{60} sqrt{ left( frac{pi r}{30} cos left(frac{pi t}{30} right) right)^2 + left( - frac{pi r}{30} sin left(frac{pi t}{30} right) - gt right)^2} {rm d}t \
& = int_0^{60} sqrt{ g^2t^2 + frac{pi r g}{15} t sin left( frac{pi t}{30} right) + left( frac{pi r}{30} right)^2} {rm d}t
end{align}
$$



Which you can quickly approximate, giving a semi-satisfying answer to the problem, but I doubt has a closed-form solution in terms of $r$ and $g$.



Since having the revolution take $60$ seconds seems to make things messy, and that we're more interested in evaluating these integrals, consider instead
$$
begin{align}
x=r sin(t), quad y = r cos(t) - frac12 gt^2 && && text{ for } 0 leq t leq 2pi
end{align}
$$




Are there any closed-form solutions of this integral for specific values of $r$ and $g$?
$$
int_0^{2pi} sqrt{g^2t^2 + 2rgtsin(t) + r^2} {rm d}t
$$







integration definite-integrals parametric classical-mechanics arc-length







share|cite|improve this question











$endgroup$




A clock is under free fall for $60$ seconds and its second hand makes exactly one revolution during that period of time. It begins at rest with its second hand facing upwards. Given that the second hand has a length of $r>0$ and that the downward gravitational acceleration is some constant $g>0$, what is the total distance traveled by the tip of the second hand?



I came up with this little problem myself and got stuck at the end with a possibly impossible integral to evaluate. Here's a quick graphical interpretation of the problem: https://www.desmos.com/calculator/ck8gxjughf



So, I started with the parametric equations representing the path of the tip of the second hand:



$$
begin{align}
x = rsin left( frac{pi t}{30} right), quad
y = rcos left( frac{pi t}{30} right) - frac12 g t^2
&& && text{for } 0 leq t leq 60
end{align}
$$



Then the total distance traveled can immediately be written as
$$
begin{align}
int_0^{60} sqrt{ dot x ^2 + dot y ^2} dt
& = int_0^{60} sqrt{ left( frac{pi r}{30} cos left(frac{pi t}{30} right) right)^2 + left( - frac{pi r}{30} sin left(frac{pi t}{30} right) - gt right)^2} {rm d}t \
& = int_0^{60} sqrt{ g^2t^2 + frac{pi r g}{15} t sin left( frac{pi t}{30} right) + left( frac{pi r}{30} right)^2} {rm d}t
end{align}
$$



Which you can quickly approximate, giving a semi-satisfying answer to the problem, but I doubt has a closed-form solution in terms of $r$ and $g$.



Since having the revolution take $60$ seconds seems to make things messy, and that we're more interested in evaluating these integrals, consider instead
$$
begin{align}
x=r sin(t), quad y = r cos(t) - frac12 gt^2 && && text{ for } 0 leq t leq 2pi
end{align}
$$




Are there any closed-form solutions of this integral for specific values of $r$ and $g$?
$$
int_0^{2pi} sqrt{g^2t^2 + 2rgtsin(t) + r^2} {rm d}t
$$







integration definite-integrals parametric classical-mechanics arc-length




integration definite-integrals parametric classical-mechanics arc-length






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 17:09







Mint

















asked Dec 28 '18 at 17:53









MintMint

5411417




5411417








  • 2




    $begingroup$
    As this is a physics problem, I'm tempted to examine the limits $r ll 1$, for which the $r^{2}$ in the square root is suppressed and a Taylor expansion of the square root to first order leads to $2pi^{2}g$. While $r gg 1$ suppresses the $g^{2}t^{2}$ term and leads to $2pi(r-g)$. Mathematica is unable to evaluate the general case, nor the related integrals $int_{0}^{2pi}sqrt{1+tsin t},mathrm{d}t$ and $int_{0}^{2pi}sqrt{t^{2}+tsin t},mathrm{d}t$.
    $endgroup$
    – Ininterrompue
    Dec 28 '18 at 19:25
















  • 2




    $begingroup$
    As this is a physics problem, I'm tempted to examine the limits $r ll 1$, for which the $r^{2}$ in the square root is suppressed and a Taylor expansion of the square root to first order leads to $2pi^{2}g$. While $r gg 1$ suppresses the $g^{2}t^{2}$ term and leads to $2pi(r-g)$. Mathematica is unable to evaluate the general case, nor the related integrals $int_{0}^{2pi}sqrt{1+tsin t},mathrm{d}t$ and $int_{0}^{2pi}sqrt{t^{2}+tsin t},mathrm{d}t$.
    $endgroup$
    – Ininterrompue
    Dec 28 '18 at 19:25










2




2




$begingroup$
As this is a physics problem, I'm tempted to examine the limits $r ll 1$, for which the $r^{2}$ in the square root is suppressed and a Taylor expansion of the square root to first order leads to $2pi^{2}g$. While $r gg 1$ suppresses the $g^{2}t^{2}$ term and leads to $2pi(r-g)$. Mathematica is unable to evaluate the general case, nor the related integrals $int_{0}^{2pi}sqrt{1+tsin t},mathrm{d}t$ and $int_{0}^{2pi}sqrt{t^{2}+tsin t},mathrm{d}t$.
$endgroup$
– Ininterrompue
Dec 28 '18 at 19:25






$begingroup$
As this is a physics problem, I'm tempted to examine the limits $r ll 1$, for which the $r^{2}$ in the square root is suppressed and a Taylor expansion of the square root to first order leads to $2pi^{2}g$. While $r gg 1$ suppresses the $g^{2}t^{2}$ term and leads to $2pi(r-g)$. Mathematica is unable to evaluate the general case, nor the related integrals $int_{0}^{2pi}sqrt{1+tsin t},mathrm{d}t$ and $int_{0}^{2pi}sqrt{t^{2}+tsin t},mathrm{d}t$.
$endgroup$
– Ininterrompue
Dec 28 '18 at 19:25












1 Answer
1






active

oldest

votes


















2












$begingroup$

So, the tip of the seconds hand will generate an "accelerated" cycloid.



Being accelerated, it is no longer periodic, and the length will depend on the starting position.



The presence of the term $t cdot sin t$ under the square root seems to make very unprobable
that the integral might be evaluated in closed form.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is any way to relate it to an unaccelerated one? (Intuitively there ought to be a way to take account of the "stretching", but I'm guessing that will be pretty messy too.)
    $endgroup$
    – timtfj
    Dec 28 '18 at 19:17










  • $begingroup$
    well, that was my first tought also, but the stretching combines with rotation, making it no longer linear..
    $endgroup$
    – G Cab
    Dec 28 '18 at 20:20






  • 1




    $begingroup$
    Yes, that's why I think the result would be horrible to deal with—it would have to take account of the slope of the un-stretched curve at every point.
    $endgroup$
    – timtfj
    Dec 28 '18 at 20:28












Your Answer








StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055135%2ftough-integral-from-a-falling-clock-int-02-pi-sqrtg2t2-2rgt-sint%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

So, the tip of the seconds hand will generate an "accelerated" cycloid.



Being accelerated, it is no longer periodic, and the length will depend on the starting position.



The presence of the term $t cdot sin t$ under the square root seems to make very unprobable
that the integral might be evaluated in closed form.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is any way to relate it to an unaccelerated one? (Intuitively there ought to be a way to take account of the "stretching", but I'm guessing that will be pretty messy too.)
    $endgroup$
    – timtfj
    Dec 28 '18 at 19:17










  • $begingroup$
    well, that was my first tought also, but the stretching combines with rotation, making it no longer linear..
    $endgroup$
    – G Cab
    Dec 28 '18 at 20:20






  • 1




    $begingroup$
    Yes, that's why I think the result would be horrible to deal with—it would have to take account of the slope of the un-stretched curve at every point.
    $endgroup$
    – timtfj
    Dec 28 '18 at 20:28
















2












$begingroup$

So, the tip of the seconds hand will generate an "accelerated" cycloid.



Being accelerated, it is no longer periodic, and the length will depend on the starting position.



The presence of the term $t cdot sin t$ under the square root seems to make very unprobable
that the integral might be evaluated in closed form.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is any way to relate it to an unaccelerated one? (Intuitively there ought to be a way to take account of the "stretching", but I'm guessing that will be pretty messy too.)
    $endgroup$
    – timtfj
    Dec 28 '18 at 19:17










  • $begingroup$
    well, that was my first tought also, but the stretching combines with rotation, making it no longer linear..
    $endgroup$
    – G Cab
    Dec 28 '18 at 20:20






  • 1




    $begingroup$
    Yes, that's why I think the result would be horrible to deal with—it would have to take account of the slope of the un-stretched curve at every point.
    $endgroup$
    – timtfj
    Dec 28 '18 at 20:28














2












2








2





$begingroup$

So, the tip of the seconds hand will generate an "accelerated" cycloid.



Being accelerated, it is no longer periodic, and the length will depend on the starting position.



The presence of the term $t cdot sin t$ under the square root seems to make very unprobable
that the integral might be evaluated in closed form.






share|cite|improve this answer











$endgroup$



So, the tip of the seconds hand will generate an "accelerated" cycloid.



Being accelerated, it is no longer periodic, and the length will depend on the starting position.



The presence of the term $t cdot sin t$ under the square root seems to make very unprobable
that the integral might be evaluated in closed form.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 28 '18 at 18:51

























answered Dec 28 '18 at 18:32









G CabG Cab

20.5k31342




20.5k31342












  • $begingroup$
    Is any way to relate it to an unaccelerated one? (Intuitively there ought to be a way to take account of the "stretching", but I'm guessing that will be pretty messy too.)
    $endgroup$
    – timtfj
    Dec 28 '18 at 19:17










  • $begingroup$
    well, that was my first tought also, but the stretching combines with rotation, making it no longer linear..
    $endgroup$
    – G Cab
    Dec 28 '18 at 20:20






  • 1




    $begingroup$
    Yes, that's why I think the result would be horrible to deal with—it would have to take account of the slope of the un-stretched curve at every point.
    $endgroup$
    – timtfj
    Dec 28 '18 at 20:28


















  • $begingroup$
    Is any way to relate it to an unaccelerated one? (Intuitively there ought to be a way to take account of the "stretching", but I'm guessing that will be pretty messy too.)
    $endgroup$
    – timtfj
    Dec 28 '18 at 19:17










  • $begingroup$
    well, that was my first tought also, but the stretching combines with rotation, making it no longer linear..
    $endgroup$
    – G Cab
    Dec 28 '18 at 20:20






  • 1




    $begingroup$
    Yes, that's why I think the result would be horrible to deal with—it would have to take account of the slope of the un-stretched curve at every point.
    $endgroup$
    – timtfj
    Dec 28 '18 at 20:28
















$begingroup$
Is any way to relate it to an unaccelerated one? (Intuitively there ought to be a way to take account of the "stretching", but I'm guessing that will be pretty messy too.)
$endgroup$
– timtfj
Dec 28 '18 at 19:17




$begingroup$
Is any way to relate it to an unaccelerated one? (Intuitively there ought to be a way to take account of the "stretching", but I'm guessing that will be pretty messy too.)
$endgroup$
– timtfj
Dec 28 '18 at 19:17












$begingroup$
well, that was my first tought also, but the stretching combines with rotation, making it no longer linear..
$endgroup$
– G Cab
Dec 28 '18 at 20:20




$begingroup$
well, that was my first tought also, but the stretching combines with rotation, making it no longer linear..
$endgroup$
– G Cab
Dec 28 '18 at 20:20




1




1




$begingroup$
Yes, that's why I think the result would be horrible to deal with—it would have to take account of the slope of the un-stretched curve at every point.
$endgroup$
– timtfj
Dec 28 '18 at 20:28




$begingroup$
Yes, that's why I think the result would be horrible to deal with—it would have to take account of the slope of the un-stretched curve at every point.
$endgroup$
– timtfj
Dec 28 '18 at 20:28


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055135%2ftough-integral-from-a-falling-clock-int-02-pi-sqrtg2t2-2rgt-sint%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Bundesstraße 106

Image
Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
Read more

Verónica Boquete

Image
Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more