Two masses, one swinging
The Lagrangian is
L = 1 mr˙² + 1 m(r˙² + r²θ˙²) − mgr + mgr cos θ. (6.114)
2 2
The last two terms are the (negatives of the) potentials of each mass, relative to where
they would be if the right mass were located at the right pulley. The equations of motion
obtained from varying r and θ are
2r¨ = rθ˙² − g(1 − cos θ),
(6.115)
d (r²θ˙) = −gr sin θ.
dt
The first equation deals with the forces and accelerations along the direction of the
string. The second equation equates the torque from gravity with the change in angular
momentum of the right mass. If we do a (coarse) small-angle approximation and keep
only terms up to first order in θ, we find that at t = 0 (using the initial condition,
ṙ = 0), Eqs. (6.115) become
r¨ = 0,
θ¨ + g θ = 0. (6.116)
r
258 The Lagrangian method
These equations say that the left mass stays still, and the right mass behaves just like
a pendulum.
If we want to find the leading term in the initial acceleration of the left mass (that
is, the leading term in r¨), we need to be a little less coarse in our approximation. So
let's keep terms in Eq. (6.115) up to second order in θ. We then have at t = 0 (using
the initial condition, ṙ = 0)
2r¨ = rθ˙² − 1 gθ²,
2
(6.117)
θ¨ + g θ = 0.
r
The second equation still says that the right mass undergoes harmonic motion. We are
told that the amplitude is A, so we have
θ(t) = A cos(ωt + φ), (6.118)
where ω = √g/r. Plugging this into the first equation gives
2r¨ = A²g [sin²(ωt + φ) − 1 cos²(ωt + φ)]. (6.119)
2
If we average this over a few periods, both sin² α and cos² α average to 1/2, so we find
r¨ avg = A²g. (6.120)
8
This is a small second-order effect. It is positive, so the left mass slowly begins to
climb.