Mathematics of Physics and Engineering

(^92) The Lagrange-Hamilton Method
Ptan = P • — + P • "J-J Q-
definition of Q agrees with (2.3.22) on page 88. Differentiating ptan with
respect to t, we get
dr d^2 r
dq~+P'df
Using (2.3.36) and p = mr = m (dr/dq) q, we find
,.r ... dr dr d?r. 2
Ptan = {N -mgn)- — +m— • J-J <T,
or, since N • (dr/dq) = 0,
(IT (IT
Ptan = Q + m-j--j^q^2 , (2.3.37)
The kinetic energy is then
^ = - 2 mvr^----e, (2.3.38)
and by (2.3.35),
£K = ^(a^2 sin^2 q + b^2 cos^2 q + (?) q^2. (2.3.39)
Hence, from (2.3.38),
8£K dr dr. dr dr
dq dq dq dq dt
From (2.3.39), using standard rules of partial differentiation,
—— = m(a^2 sin q cos q — b^2 cos q sin q) q^2.
Now, r(q) = a cos #2 + b sin qj + cqk, so that
dr— = —asmqt + bcosqj + CK, d —5- =^2 r ^ —acosqt — bsmqj.
dq dqz
Therefore,
8£K dr d^2 r. 9
dq dq dq^2
and equation (2.3.37) is equivalent to Lagrange's equation of motion
d ( 8£K\ OSK „
dt \^1 dq J dq =Q, (2.3.40)