velocity and acceleration in a parabola
$begingroup$
A particle moves with constant speed along a parabola of equation $y^{2}=2px$ with $p=constant$. I want to find its velocity and acceleration vector.
Since the velocity magnitude is constant I know the acceleration will be along the normal of the parabola and of the form
$$vec{a}=frac{v^{2}}{R(t)}vec{n}, v=constant$$
where $R(t)$ is the time dependent radius of curvature of the parabola and $vec{n}$ is the normal.
To find the velocity I parametrize the parabola with $(y^{2}(t)/(2p),y(t))$. The velocity is the tangent vector given by deriving this parametric position
Then $vec{v}=left(frac{y}{p}frac{dy}{dt},frac{dy}{dt}right)$ and the constraint of constant speed leads to
$$|vec{v}|=constant=frac{dy}{dt}sqrt{left(frac{y}{p}right)^{2}+1}$$
This seems to be a first order differential equation. Is it solvable?
How do I find $vec{n}$ and $R(t)$?
polar-coordinates tangent-line kinematics
asked Dec 5 '18 at 14:08
JennyToyJennyToy
25018
$endgroup$
add a comment |
$begingroup$
A particle moves with constant speed along a parabola of equation $y^{2}=2px$ with $p=constant$. I want to find its velocity and acceleration vector.
Since the velocity magnitude is constant I know the acceleration will be along the normal of the parabola and of the form
$$vec{a}=frac{v^{2}}{R(t)}vec{n}, v=constant$$
where $R(t)$ is the time dependent radius of curvature of the parabola and $vec{n}$ is the normal.
To find the velocity I parametrize the parabola with $(y^{2}(t)/(2p),y(t))$. The velocity is the tangent vector given by deriving this parametric position
Then $vec{v}=left(frac{y}{p}frac{dy}{dt},frac{dy}{dt}right)$ and the constraint of constant speed leads to
$$|vec{v}|=constant=frac{dy}{dt}sqrt{left(frac{y}{p}right)^{2}+1}$$
This seems to be a first order differential equation. Is it solvable?
How do I find $vec{n}$ and $R(t)$?
polar-coordinates tangent-line kinematics
asked Dec 5 '18 at 14:08
JennyToyJennyToy
25018
$endgroup$
add a comment |
$begingroup$
A particle moves with constant speed along a parabola of equation $y^{2}=2px$ with $p=constant$. I want to find its velocity and acceleration vector.
Since the velocity magnitude is constant I know the acceleration will be along the normal of the parabola and of the form
$$vec{a}=frac{v^{2}}{R(t)}vec{n}, v=constant$$
where $R(t)$ is the time dependent radius of curvature of the parabola and $vec{n}$ is the normal.
To find the velocity I parametrize the parabola with $(y^{2}(t)/(2p),y(t))$. The velocity is the tangent vector given by deriving this parametric position
Then $vec{v}=left(frac{y}{p}frac{dy}{dt},frac{dy}{dt}right)$ and the constraint of constant speed leads to
$$|vec{v}|=constant=frac{dy}{dt}sqrt{left(frac{y}{p}right)^{2}+1}$$
This seems to be a first order differential equation. Is it solvable?
How do I find $vec{n}$ and $R(t)$?
polar-coordinates tangent-line kinematics
asked Dec 5 '18 at 14:08
JennyToyJennyToy
25018
$endgroup$
A particle moves with constant speed along a parabola of equation $y^{2}=2px$ with $p=constant$. I want to find its velocity and acceleration vector.
Since the velocity magnitude is constant I know the acceleration will be along the normal of the parabola and of the form
$$vec{a}=frac{v^{2}}{R(t)}vec{n}, v=constant$$
where $R(t)$ is the time dependent radius of curvature of the parabola and $vec{n}$ is the normal.
To find the velocity I parametrize the parabola with $(y^{2}(t)/(2p),y(t))$. The velocity is the tangent vector given by deriving this parametric position
Then $vec{v}=left(frac{y}{p}frac{dy}{dt},frac{dy}{dt}right)$ and the constraint of constant speed leads to
$$|vec{v}|=constant=frac{dy}{dt}sqrt{left(frac{y}{p}right)^{2}+1}$$
This seems to be a first order differential equation. Is it solvable?
How do I find $vec{n}$ and $R(t)$?
polar-coordinates tangent-line kinematics
polar-coordinates tangent-line kinematics
asked Dec 5 '18 at 14:08
JennyToyJennyToy
25018
asked Dec 5 '18 at 14:08
JennyToyJennyToy
25018
asked Dec 5 '18 at 14:08
JennyToyJennyToy
25018
asked Dec 5 '18 at 14:08
JennyToyJennyToy
25018
asked Dec 5 '18 at 14:08
JennyToyJennyToy
25018
25018
add a comment |
add a comment |
1 Answer
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The diferential equation is solvable, the integral $intsqrt{1+x^2}dx$ can be found in the tables (or found via integration by parts).
What you actually found is the mapping between $y$ and $t$. For the radius of curvature, it's much easier to just get it geometrically, with respect to position (parameterized with $y$). The standard differential geometry of curves gives you:
$$frac{1}{R(x(y))}=frac{x''(y)}{sqrt{1+x'(y)^2}^3}$$
where $x=frac{y^2}{2p}$.
So, the acceleration, expressed with $y$, is straight forward and does not require integration. However, converting $a(y)$ into $a(t)$ requires the function $y(t)$ which will be problematic, because the solution of your differential equation gives you $t(y)$ that is transcendental and thus not invertible in closed form.
answered Dec 5 '18 at 14:22
orionorion
13.3k11836
$endgroup$
$begingroup$
The normal $vec{n}$ can be obtained by picking any vector of unit length whose dot product with $vec{v}$ is zero?
$endgroup$
– JennyToy
Dec 5 '18 at 15:02
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There are only two options - one points in, one points out. One is wrong.
$endgroup$
– orion
Dec 5 '18 at 15:04
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The diferential equation is solvable, the integral $intsqrt{1+x^2}dx$ can be found in the tables (or found via integration by parts).
What you actually found is the mapping between $y$ and $t$. For the radius of curvature, it's much easier to just get it geometrically, with respect to position (parameterized with $y$). The standard differential geometry of curves gives you:
$$frac{1}{R(x(y))}=frac{x''(y)}{sqrt{1+x'(y)^2}^3}$$
where $x=frac{y^2}{2p}$.
So, the acceleration, expressed with $y$, is straight forward and does not require integration. However, converting $a(y)$ into $a(t)$ requires the function $y(t)$ which will be problematic, because the solution of your differential equation gives you $t(y)$ that is transcendental and thus not invertible in closed form.
answered Dec 5 '18 at 14:22
orionorion
13.3k11836
$endgroup$
$begingroup$
The normal $vec{n}$ can be obtained by picking any vector of unit length whose dot product with $vec{v}$ is zero?
$endgroup$
– JennyToy
Dec 5 '18 at 15:02
$begingroup$
There are only two options - one points in, one points out. One is wrong.
$endgroup$
– orion
Dec 5 '18 at 15:04
add a comment |
$begingroup$
The diferential equation is solvable, the integral $intsqrt{1+x^2}dx$ can be found in the tables (or found via integration by parts).
What you actually found is the mapping between $y$ and $t$. For the radius of curvature, it's much easier to just get it geometrically, with respect to position (parameterized with $y$). The standard differential geometry of curves gives you:
$$frac{1}{R(x(y))}=frac{x''(y)}{sqrt{1+x'(y)^2}^3}$$
where $x=frac{y^2}{2p}$.
So, the acceleration, expressed with $y$, is straight forward and does not require integration. However, converting $a(y)$ into $a(t)$ requires the function $y(t)$ which will be problematic, because the solution of your differential equation gives you $t(y)$ that is transcendental and thus not invertible in closed form.
answered Dec 5 '18 at 14:22
orionorion
13.3k11836
$endgroup$
$begingroup$
The normal $vec{n}$ can be obtained by picking any vector of unit length whose dot product with $vec{v}$ is zero?
$endgroup$
– JennyToy
Dec 5 '18 at 15:02
$begingroup$
There are only two options - one points in, one points out. One is wrong.
$endgroup$
– orion
Dec 5 '18 at 15:04
add a comment |
$begingroup$
The diferential equation is solvable, the integral $intsqrt{1+x^2}dx$ can be found in the tables (or found via integration by parts).
What you actually found is the mapping between $y$ and $t$. For the radius of curvature, it's much easier to just get it geometrically, with respect to position (parameterized with $y$). The standard differential geometry of curves gives you:
$$frac{1}{R(x(y))}=frac{x''(y)}{sqrt{1+x'(y)^2}^3}$$
where $x=frac{y^2}{2p}$.
So, the acceleration, expressed with $y$, is straight forward and does not require integration. However, converting $a(y)$ into $a(t)$ requires the function $y(t)$ which will be problematic, because the solution of your differential equation gives you $t(y)$ that is transcendental and thus not invertible in closed form.
answered Dec 5 '18 at 14:22
orionorion
13.3k11836
$endgroup$
The diferential equation is solvable, the integral $intsqrt{1+x^2}dx$ can be found in the tables (or found via integration by parts).
What you actually found is the mapping between $y$ and $t$. For the radius of curvature, it's much easier to just get it geometrically, with respect to position (parameterized with $y$). The standard differential geometry of curves gives you:
$$frac{1}{R(x(y))}=frac{x''(y)}{sqrt{1+x'(y)^2}^3}$$
where $x=frac{y^2}{2p}$.
So, the acceleration, expressed with $y$, is straight forward and does not require integration. However, converting $a(y)$ into $a(t)$ requires the function $y(t)$ which will be problematic, because the solution of your differential equation gives you $t(y)$ that is transcendental and thus not invertible in closed form.
answered Dec 5 '18 at 14:22
orionorion
13.3k11836
answered Dec 5 '18 at 14:22
orionorion
13.3k11836
answered Dec 5 '18 at 14:22
orionorion
13.3k11836
answered Dec 5 '18 at 14:22
orionorion
13.3k11836
13.3k11836
$begingroup$
The normal $vec{n}$ can be obtained by picking any vector of unit length whose dot product with $vec{v}$ is zero?
$endgroup$
– JennyToy
Dec 5 '18 at 15:02
$begingroup$
There are only two options - one points in, one points out. One is wrong.
$endgroup$
– orion
Dec 5 '18 at 15:04
add a comment |
$begingroup$
The normal $vec{n}$ can be obtained by picking any vector of unit length whose dot product with $vec{v}$ is zero?
$endgroup$
– JennyToy
Dec 5 '18 at 15:02
$begingroup$
There are only two options - one points in, one points out. One is wrong.
$endgroup$
– orion
Dec 5 '18 at 15:04
$begingroup$
The normal $vec{n}$ can be obtained by picking any vector of unit length whose dot product with $vec{v}$ is zero?
$endgroup$
– JennyToy
Dec 5 '18 at 15:02
$begingroup$
The normal $vec{n}$ can be obtained by picking any vector of unit length whose dot product with $vec{v}$ is zero?
$endgroup$
– JennyToy
Dec 5 '18 at 15:02
$begingroup$
There are only two options - one points in, one points out. One is wrong.
$endgroup$
– orion
Dec 5 '18 at 15:04
$begingroup$
There are only two options - one points in, one points out. One is wrong.
$endgroup$
– orion
Dec 5 '18 at 15:04
add a comment |
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