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

5311417




5311417








  • 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











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1 Answer
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oldest

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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

19.6k31339




19.6k31339












  • $begingroup$
    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


















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