Mixing
Lagrangian Equation for two masses moving on a incline plans [closed]
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How to find equations of system dynamics using Lagrange’s approach
euler-lagrange-equation
edited Jan 28 at 11:04
Arthur
120k7120204
asked Jan 28 at 11:03
Piyumal SamarathungaPiyumal Samarathunga
62
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closed as off-topic by Shailesh, mrtaurho, Ali Caglayan, Wouter, Martin Argerami Jan 28 at 14:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, mrtaurho, Ali Caglayan, Wouter
If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
How to find equations of system dynamics using Lagrange’s approach
euler-lagrange-equation
edited Jan 28 at 11:04
Arthur
120k7120204
asked Jan 28 at 11:03
Piyumal SamarathungaPiyumal Samarathunga
62
$endgroup$
closed as off-topic by Shailesh, mrtaurho, Ali Caglayan, Wouter, Martin Argerami Jan 28 at 14:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, mrtaurho, Ali Caglayan, Wouter
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
The approach of Lagrangian mechanics is pretty straight-forward. How far did you get?
$endgroup$
– Arthur
Jan 28 at 11:05
$begingroup$
Actually, I can't understand how to write the Kinetic and Potential energies for the system.
$endgroup$
– Piyumal Samarathunga
Jan 28 at 11:24
$begingroup$
Kinetic energies can be written as, T=1/2 2m x ̇^2+1/2 m x ̇^2
$endgroup$
– Piyumal Samarathunga
Jan 28 at 11:26
$begingroup$
But the problem is about the Potential energies of the system.
$endgroup$
– Piyumal Samarathunga
Jan 28 at 11:31
$begingroup$
Potetntial energy in the spring, plus gravitational potential energy in the boxes. Trigonometry couples the height of the two boxes to $x$.
$endgroup$
– Arthur
Jan 28 at 11:33
|
show 1 more comment
$begingroup$
How to find equations of system dynamics using Lagrange’s approach
euler-lagrange-equation
edited Jan 28 at 11:04
Arthur
120k7120204
asked Jan 28 at 11:03
Piyumal SamarathungaPiyumal Samarathunga
62
$endgroup$
How to find equations of system dynamics using Lagrange’s approach
euler-lagrange-equation
euler-lagrange-equation
edited Jan 28 at 11:04
Arthur
120k7120204
asked Jan 28 at 11:03
Piyumal SamarathungaPiyumal Samarathunga
62
edited Jan 28 at 11:04
Arthur
120k7120204
asked Jan 28 at 11:03
Piyumal SamarathungaPiyumal Samarathunga
62
edited Jan 28 at 11:04
Arthur
120k7120204
edited Jan 28 at 11:04
Arthur
120k7120204
edited Jan 28 at 11:04
Arthur
120k7120204
120k7120204
asked Jan 28 at 11:03
Piyumal SamarathungaPiyumal Samarathunga
62
asked Jan 28 at 11:03
Piyumal SamarathungaPiyumal Samarathunga
62
asked Jan 28 at 11:03
Piyumal SamarathungaPiyumal Samarathunga
62
62
closed as off-topic by Shailesh, mrtaurho, Ali Caglayan, Wouter, Martin Argerami Jan 28 at 14:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, mrtaurho, Ali Caglayan, Wouter
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Shailesh, mrtaurho, Ali Caglayan, Wouter, Martin Argerami Jan 28 at 14:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, mrtaurho, Ali Caglayan, Wouter
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
The approach of Lagrangian mechanics is pretty straight-forward. How far did you get?
$endgroup$
– Arthur
Jan 28 at 11:05
$begingroup$
Actually, I can't understand how to write the Kinetic and Potential energies for the system.
$endgroup$
– Piyumal Samarathunga
Jan 28 at 11:24
$begingroup$
Kinetic energies can be written as, T=1/2 2m x ̇^2+1/2 m x ̇^2
$endgroup$
– Piyumal Samarathunga
Jan 28 at 11:26
$begingroup$
But the problem is about the Potential energies of the system.
$endgroup$
– Piyumal Samarathunga
Jan 28 at 11:31
$begingroup$
Potetntial energy in the spring, plus gravitational potential energy in the boxes. Trigonometry couples the height of the two boxes to $x$.
$endgroup$
– Arthur
Jan 28 at 11:33
|
show 1 more comment
$begingroup$
The approach of Lagrangian mechanics is pretty straight-forward. How far did you get?
$endgroup$
– Arthur
Jan 28 at 11:05
$begingroup$
Actually, I can't understand how to write the Kinetic and Potential energies for the system.
$endgroup$
– Piyumal Samarathunga
Jan 28 at 11:24
$begingroup$
Kinetic energies can be written as, T=1/2 2m x ̇^2+1/2 m x ̇^2
$endgroup$
– Piyumal Samarathunga
Jan 28 at 11:26
$begingroup$
But the problem is about the Potential energies of the system.
$endgroup$
– Piyumal Samarathunga
Jan 28 at 11:31
$begingroup$
Potetntial energy in the spring, plus gravitational potential energy in the boxes. Trigonometry couples the height of the two boxes to $x$.
$endgroup$
– Arthur
Jan 28 at 11:33
$begingroup$
The approach of Lagrangian mechanics is pretty straight-forward. How far did you get?
$endgroup$
– Arthur
Jan 28 at 11:05
$begingroup$
The approach of Lagrangian mechanics is pretty straight-forward. How far did you get?
$endgroup$
– Arthur
Jan 28 at 11:05
$begingroup$
Actually, I can't understand how to write the Kinetic and Potential energies for the system.
$endgroup$
– Piyumal Samarathunga
Jan 28 at 11:24
$begingroup$
Actually, I can't understand how to write the Kinetic and Potential energies for the system.
$endgroup$
– Piyumal Samarathunga
Jan 28 at 11:24
$begingroup$
Kinetic energies can be written as, T=1/2 2m x ̇^2+1/2 m x ̇^2
$endgroup$
– Piyumal Samarathunga
Jan 28 at 11:26
$begingroup$
Kinetic energies can be written as, T=1/2 2m x ̇^2+1/2 m x ̇^2
$endgroup$
– Piyumal Samarathunga
Jan 28 at 11:26
$begingroup$
But the problem is about the Potential energies of the system.
$endgroup$
– Piyumal Samarathunga
Jan 28 at 11:31
$begingroup$
But the problem is about the Potential energies of the system.
$endgroup$
– Piyumal Samarathunga
Jan 28 at 11:31
$begingroup$
Potetntial energy in the spring, plus gravitational potential energy in the boxes. Trigonometry couples the height of the two boxes to $x$.
$endgroup$
– Arthur
Jan 28 at 11:33
$begingroup$
Potetntial energy in the spring, plus gravitational potential energy in the boxes. Trigonometry couples the height of the two boxes to $x$.
$endgroup$
– Arthur
Jan 28 at 11:33
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
You have
$$
T = frac32mdot x^2\
V = frac12kx^2 + 2mgxsin(phi_2) - mgxsin(phi_1)
$$
and this all looks good. This gives
$$
L(t, x, dot x) = frac32mdot x^2 - frac12kx^2 - mgx(2sin(phi_2) - sin(phi_1))
$$
Note that $x = 0$ represents the neutral position for the spring, and the positive $x$ direction depends on $phi_1$ and $phi_2$: If you remove the spring, whichever way the blocks will accelerate under gravity, that's the negative $x$-direction.
Using the Euler lagrange equation, this gives us
$$
frac{mathrm dfrac{partial L}{partial dot x}}{mathrm dt}(t, x(t), dot x(t)) = frac{partial L}{partial x}(t, x(t), dot x(t))\
3mddot x = -kx - mg(2sin(phi_2) - sin(phi_1))\
ddot x = -frac{k}{3m}x - frac g3(2sin(phi_2) - sin(phi_1))
$$
which basically indicates harmonic motion about a point off the neutral position for the spring (i.e. the effect of gravity is in practice to move the center of the spring a bit).
answered Jan 28 at 14:22
ArthurArthur
120k7120204
$endgroup$
$begingroup$
Thank you very much.
$endgroup$
– Piyumal Samarathunga
Jan 28 at 18:34
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have
$$
T = frac32mdot x^2\
V = frac12kx^2 + 2mgxsin(phi_2) - mgxsin(phi_1)
$$
and this all looks good. This gives
$$
L(t, x, dot x) = frac32mdot x^2 - frac12kx^2 - mgx(2sin(phi_2) - sin(phi_1))
$$
Note that $x = 0$ represents the neutral position for the spring, and the positive $x$ direction depends on $phi_1$ and $phi_2$: If you remove the spring, whichever way the blocks will accelerate under gravity, that's the negative $x$-direction.
Using the Euler lagrange equation, this gives us
$$
frac{mathrm dfrac{partial L}{partial dot x}}{mathrm dt}(t, x(t), dot x(t)) = frac{partial L}{partial x}(t, x(t), dot x(t))\
3mddot x = -kx - mg(2sin(phi_2) - sin(phi_1))\
ddot x = -frac{k}{3m}x - frac g3(2sin(phi_2) - sin(phi_1))
$$
which basically indicates harmonic motion about a point off the neutral position for the spring (i.e. the effect of gravity is in practice to move the center of the spring a bit).
answered Jan 28 at 14:22
ArthurArthur
120k7120204
$endgroup$
$begingroup$
Thank you very much.
$endgroup$
– Piyumal Samarathunga
Jan 28 at 18:34
add a comment |
$begingroup$
You have
$$
T = frac32mdot x^2\
V = frac12kx^2 + 2mgxsin(phi_2) - mgxsin(phi_1)
$$
and this all looks good. This gives
$$
L(t, x, dot x) = frac32mdot x^2 - frac12kx^2 - mgx(2sin(phi_2) - sin(phi_1))
$$
Note that $x = 0$ represents the neutral position for the spring, and the positive $x$ direction depends on $phi_1$ and $phi_2$: If you remove the spring, whichever way the blocks will accelerate under gravity, that's the negative $x$-direction.
Using the Euler lagrange equation, this gives us
$$
frac{mathrm dfrac{partial L}{partial dot x}}{mathrm dt}(t, x(t), dot x(t)) = frac{partial L}{partial x}(t, x(t), dot x(t))\
3mddot x = -kx - mg(2sin(phi_2) - sin(phi_1))\
ddot x = -frac{k}{3m}x - frac g3(2sin(phi_2) - sin(phi_1))
$$
which basically indicates harmonic motion about a point off the neutral position for the spring (i.e. the effect of gravity is in practice to move the center of the spring a bit).
answered Jan 28 at 14:22
ArthurArthur
120k7120204
$endgroup$
$begingroup$
Thank you very much.
$endgroup$
– Piyumal Samarathunga
Jan 28 at 18:34
add a comment |
$begingroup$
You have
$$
T = frac32mdot x^2\
V = frac12kx^2 + 2mgxsin(phi_2) - mgxsin(phi_1)
$$
and this all looks good. This gives
$$
L(t, x, dot x) = frac32mdot x^2 - frac12kx^2 - mgx(2sin(phi_2) - sin(phi_1))
$$
Note that $x = 0$ represents the neutral position for the spring, and the positive $x$ direction depends on $phi_1$ and $phi_2$: If you remove the spring, whichever way the blocks will accelerate under gravity, that's the negative $x$-direction.
Using the Euler lagrange equation, this gives us
$$
frac{mathrm dfrac{partial L}{partial dot x}}{mathrm dt}(t, x(t), dot x(t)) = frac{partial L}{partial x}(t, x(t), dot x(t))\
3mddot x = -kx - mg(2sin(phi_2) - sin(phi_1))\
ddot x = -frac{k}{3m}x - frac g3(2sin(phi_2) - sin(phi_1))
$$
which basically indicates harmonic motion about a point off the neutral position for the spring (i.e. the effect of gravity is in practice to move the center of the spring a bit).
answered Jan 28 at 14:22
ArthurArthur
120k7120204
$endgroup$
You have
$$
T = frac32mdot x^2\
V = frac12kx^2 + 2mgxsin(phi_2) - mgxsin(phi_1)
$$
and this all looks good. This gives
$$
L(t, x, dot x) = frac32mdot x^2 - frac12kx^2 - mgx(2sin(phi_2) - sin(phi_1))
$$
Note that $x = 0$ represents the neutral position for the spring, and the positive $x$ direction depends on $phi_1$ and $phi_2$: If you remove the spring, whichever way the blocks will accelerate under gravity, that's the negative $x$-direction.
Using the Euler lagrange equation, this gives us
$$
frac{mathrm dfrac{partial L}{partial dot x}}{mathrm dt}(t, x(t), dot x(t)) = frac{partial L}{partial x}(t, x(t), dot x(t))\
3mddot x = -kx - mg(2sin(phi_2) - sin(phi_1))\
ddot x = -frac{k}{3m}x - frac g3(2sin(phi_2) - sin(phi_1))
$$
which basically indicates harmonic motion about a point off the neutral position for the spring (i.e. the effect of gravity is in practice to move the center of the spring a bit).
answered Jan 28 at 14:22
ArthurArthur
120k7120204
edited Jan 28 at 15:15
answered Jan 28 at 14:22
ArthurArthur
120k7120204
answered Jan 28 at 14:22
ArthurArthur
120k7120204
answered Jan 28 at 14:22
ArthurArthur
120k7120204
120k7120204
$begingroup$
Thank you very much.
$endgroup$
– Piyumal Samarathunga
Jan 28 at 18:34
add a comment |
$begingroup$
Thank you very much.
$endgroup$
– Piyumal Samarathunga
Jan 28 at 18:34
$begingroup$
Thank you very much.
$endgroup$
– Piyumal Samarathunga
Jan 28 at 18:34
$begingroup$
Thank you very much.
$endgroup$
– Piyumal Samarathunga
Jan 28 at 18:34
add a comment |
$begingroup$
The approach of Lagrangian mechanics is pretty straight-forward. How far did you get?
$endgroup$
– Arthur
Jan 28 at 11:05
$begingroup$
Actually, I can't understand how to write the Kinetic and Potential energies for the system.
$endgroup$
– Piyumal Samarathunga
Jan 28 at 11:24
$begingroup$
Kinetic energies can be written as, T=1/2 2m x ̇^2+1/2 m x ̇^2
$endgroup$
– Piyumal Samarathunga
Jan 28 at 11:26
$begingroup$
But the problem is about the Potential energies of the system.
$endgroup$
– Piyumal Samarathunga
Jan 28 at 11:31
$begingroup$
Potetntial energy in the spring, plus gravitational potential energy in the boxes. Trigonometry couples the height of the two boxes to $x$.
$endgroup$
– Arthur
Jan 28 at 11:33