40 Classical mechanics
there is a constraint in the form ofx^2 +y^2 =d^2. Rather than treating the constraint
via a Lagrange multiplier, we could transform to polar coordinates according to
x=dcosθ y=dsinθ. (1.11.7)The velocities are given by
x ̇=−d(sinθ)θ ̇ y ̇=d(cosθ)θ ̇ (1.11.8)so that the Lagrangian becomes
L=1
2
μ(
x ̇^2 + ̇y^2)
−U(x,y) =1
2
μd^2 θ ̇^2 −U ̃(θ), (1.11.9)where the notation,U(r) =U(x,y) =U(dcosθ,dsinθ)≡U ̃(θ), indicates that the
potential varies only according to the variation inθ. Eqn. (1.11.9) demonstrates that
the rigid body has only one degree of freedom, namely, the single angleθ. According
to the Euler–Lagrange equation (1.4.6), the equation of motion forthe angle is
μd^2 θ ̈=−∂U ̃
∂θ. (1.11.10)
In order to understand the physical content of eqn. (1.11.10), we first note that the
origin of the body-fixed frame lies at the center-of-mass positionR. We further note
the motion occurs in thexyplane and therefore consists of rotation about an axis
perpendicular to this plane, in this case, about thez-axis. The quantityω=θ ̇is called
theangular velocityabout thez-axis. The quantityI=μd^2 is a constant known as the
moment of inertiaabout thez-axis. Since the motion is purely angular, we can define
an angular momentum, analogous to the Cartesian momentum, byl=μd^2 θ ̇=Iω. In
general, angular momentum, like the Cartesian momentum, is a vector quantity given
by
l=r×p. (1.11.11)
For the present problem, in which all of the motion occurs in thexy-plane (no motion
along thez-axis),lhas only one nonzero component, namelylz, given by
lz=xpy−ypx=x(μy ̇)−y(μx ̇)=μ(xy ̇−yx ̇)=μd^2 (θ ̇cos^2 θ+θ ̇sin^2 θ)=μd^2 θ. ̇ (1.11.12)Eqn. (1.11.12) demonstrates that although the motion occursaboutthez-axis, the
direction of the angular momentum vector isalongthez-axis. Since the angular mo-
mentumlzis given as the product of the moment of inertiaIand the angular velocity