I propose an answer to my own question. Would there not an error in Sympy
? I can read here
http://docs.sympy.org/dev/modules/physics/mechanics/api/part_bod.html#sympy.physics.mechanics.rigidbody.RigidBody.angular_momentum
that:
*Angular momentum of the rigid body.*
The angular momentum H, about some point O, of a rigid body B, in a frame N
is given by
H = I* . omega + r* x (M * v)
where I* is the central inertia dyadic of B, omega is the angular velocity
of body B in the frame, N, r* is the position vector from point O to the
mass center of B, and v is the velocity of point O in the frame, N.
I think v is not the velocity of point O, but velocity of the center of
mass ?
Am I wrong ?
Thanks for your help
Philippe
Le samedi 24 octobre 2015 18:00:59 UTC+2, Philippe Fichou a écrit :
>
> Hey, all
> Consider a pendulum of length OA = 2a, of mass m as a rigid body of center
> of mass G (OG = a) which turn around (O,z). The angle between the reference
> frame R and the rod is q.
> The inertia of the body is I = (G,0,ma^2/3,ma^2/3)
> When I ask Sympy for the angular momentum about point O, it says
> m*a**2/3*q'*R.z, the same as point G.
> I should have 4*m*a**2/3*q'*R.z.
> Anybody can help me ?
>
> Here is the code:
>
> from sympy import symbols
> from sympy.physics.mechanics import *
>
> m,a = symbols('m a')
> q = dynamicsymbols('q')
>
> R = ReferenceFrame('R')
> R1 = R.orientnew('R1', 'Axis', [q, R.z])
> R1.set_ang_vel(R,q.diff() * R.z)
>
> I = inertia(R1,0,m*a**2/3,m*a**2/3)
>
> O = Point('O')
>
> A = O.locatenew('A', 2*a * R1.x)
> G = O.locatenew('G', a * R1.x)
>
> S = RigidBody('S',G,R1,m,(I,G))
>
> O.set_vel(R, 0)
> A.v2pt_theory(O, R, R1)
> G.v2pt_theory(O, R, R1)
>
> print(S.angular_momentum(O,R))
> print(S.angular_momentum(G,R))
>
> Thanks !
>
> Philippe
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