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Homework Statement
A box of mass M can slide horizontally on a frictionless surface. A simple pendulum of string length l and mass m, is suspended inside the block. Denote the coordinate of the centre of mass of the box by x and the angle that the pendulum makes with the vertical by θ . At t = 0 the pendulum displacement is θ = θ0 which is not equal to zero.
Find the Lagrangian and the equation of motion for the generalized coordinates x and θ . Which conservation law is obtained as a result of the cyclic coordinate?
Find the solutions for x and θ in the small angle approximation, hence show that the pendulum and the box execute SHO about their centre of mass at a frequency ω=[(M+m/M)^0.5]*(g/l)^0.5
Homework Equations
L=T-V
The EL equation
The Attempt at a Solution
The kinetic energy T=0.5M(dx/dt)2+0.5m[(dx/dt)2+l2(dθdt)2+2l(dx/dt)(dθ/dt)cosθ].
The potential energy measured from the support point of the pendulum U=-mglcosθ.
The Lagrangian L=T-V
L=0.5M(dx/dt)2+0.5m[(dx/dt)2+l2(dθdt)2+2l(dx/dt)(dθ/dt)cosθ]+mglcosθ
This gives the EOM as
(M+m)(d2x/dt2)+ml(d2θ/dt2)cosθ-ml(dθ/dt)2sinθ=0
(also x is a cyclic coordinate, which before taking the time derivative of (∂x/∂t), shows linear momentum is conserved), and
l(d2θ/dt2)+(d2x/dt2)cosθ=-gsinθ
In the small angle approximations, sinθ->θ, cosθ->1, then the EOM become
(M+m)(d2x/dt2)+ml(d2θ/dt2)-ml(dθ/dt)2θ=0
and
l(d2θ/dt2)+(d2x/dt2)=-glθ.
Then rearranging the second for (d2x/dt2)=-(gθ+l(d2θ/dt2)) and putting it into the first gives
d2θ/dt2=(gθ/l)[(m/M)+1-{ml(dθ/dt)2/Mg}]
i.e there is a nasty (dθ/dt)2 term which if not present would give me the correct answer... However I can't justify removing it, as small θ doesn't necessarily mean small (dθ/dt) does it?
Any clues? Thanks.