# Pendulum motion lagrange's equation

i have been trying to solve this past exam problem, a simple pendulum of length l and bob with mass m is attracted to a massless support moving horizontally with constant acceleration a. Determine the lagrange's equations of motion and the period of small oscillations.

here's what i solved for lagrange's equation:
Coordinates of mass
x = l cos θ
y = l sin θ + f(t)
Velocity of mass
x˙ = l(−sin θ)θ˙
y˙ = (cosθ)θ˙ + f˙
Kinetic energy
T =1/2m( ˙ x2 + ˙y2)
=1/2m[l2θ˙2 + (2lf˙ cos θ)θ˙ + f˙2]
Potential energy
U = −mgx
= −mgl cos θ
∂T/∂θ=1/2m · 2lf˙θ˙(−sin θ) = −mlf˙θ˙ sin θ
∂T/∂θ˙=1/2m[2l2θ˙ + 2lf˙ cos θ] = ml2θ˙ + mlf˙ cos θ
∂U/∂θ= −mgl(−sin θ) = mgl sin θ
Lagrangean
L = T − U =1/2m[l2 ˙ θ2 + (2l ˙ f cos θ)˙θ + f˙2] + mgl cos θ
Lagrange’s eqs.
∂L/∂θ −d/dt(∂L/∂θ)= 0
∂T/∂θ −∂U/∂θ −d/dt∂T/∂θ˙= 0
−mlf˙θ˙ sin θ − mgl sin θ −d/dt[ml2˙θ + mlf˙ cos θ] = 0
mlf˙θ˙ sin θ + mgl sin θ + ml2θ¨+ mlf¨cos θ + mlf˙(−sin θ)θ˙ = 0
Finally,
ml2 ¨θ + mgl sin θ + ml ¨ f cos θ = 0

I am not sure if i have it done correctly and aslo still trying to figure out the next part... Thanks for your concern...

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Is it attracted to a massless support or attached to a massless support? It makes a difference?

You never stated how you set up your coordinates, but they look a bit strange. It is pretty common in pendulum problems to measure the angle from the downward vertical, in which case we get something like
x = L sin(theta)
y = L cos(theta)
where x is pos to the right, y is pos downward.
I'm having a lot of difficulty visualizing how your coordinates work, so perhaps you could explain them.

I rather think this is a moving support problem, in which case, you really need something like
Xm = Xs + L*sin(theta)
where
Xm is the position of the mass
Xs is the position of the support
and the last term is the motion of the pendulum bob relative to the support.

Once the kinematics is cleared up, then we can talk about the Lagrange equation. Without the correct kinematics, Lagrange is just a case of GIGO.