879
Theorem C.0.2 Let yα (x, t) denote the position of an object of mass mα where x is afunction of t. Let the kinetic energy be defined by
T ≡ 12 ∑
α
mα ẏα ·ẏα .
Let the mass mα be acted on by a force F α . Then Newton’s second law implies
ddt
(∂T∂ ẋk
)− ∂T
∂xk = ∑α
F α ·∂yα
∂xk (3.3)
In case F α = ∇Φα +gα where gα is a force of constraint so the total force comes fromforces of constraint and the gradient of a potential function, then
ddt
(∂ (T −Φ)
∂ ẋk
)− ∂ (T −Φ)
∂xk = 0
Also, the above 3.3 implies Newton’s second law.
Proof: The above derivation shows that Newton’s law implies the above two formulas.On the other hand, if 3.3 holds, then in the case of one mass, the first part of the derivationwhich depended only on the chain rule and product rule shows
ddt
(∂T∂ ẋk
)− ∂T
∂xk = m ÿ· ∂y
∂xk
Thus if 3.3 and there is no force of constraint, then F = mÿ which is Newton’s second law.■
Example C.0.3 In the case of the simple pendulum, x= θ as shown in the picture and(y1
y2
)=
(l sinθ
l− l cosθ
)
the force acting on weight being mg(−j) = ∇(−mgy2
). Find the equation of motion of
this pendulum.
T =12
m
(l cos(θ)θ
′
l sin(θ)θ′
)·
(l cos(θ)θ
′
l sin(θ)θ′
)=
12
ml2 (θ′)2
Then Φ = −mg(l− l cosθ) . T −Φ = 12 ml2
(θ′)2
+mg(l− l cosθ). Thus the equation ofmotion of this pendulum is
ddt
(ml2
θ′)−mgl (−sin(θ)) = 0
soθ′′+
gl
sinθ = 0
This is an equation which doesn’t have a simple analytic solution in terms of standardcalculus type functions.