Assignment 5 solutions
1.
Let L be the length of the "upper" rope, and l(t) the length of the monkey's rope; let x be the distance of the 2m mass below pulley B. Then pulley A is Lx below pulley B. Let y be the distance of the mass m below pulley A, then the monkey is ly below A. So we have
distance of m below B = z = L  x + y distance of monkey below B = s = L  x + l  y
The kinetic energy is T = ½(2mx'² + mz'² + ms'²) and the potential energy is U = – mg(2x + z + s), where we use ' instead of an overdot. Substitute for z and s to find
T = ½ m[4x'² + 2y'² + l'² – 2l'(x' + y')] U = – mg(2L + l)
Note that U is independent of x and y: if l is constant, the system is at equilibrium for any configuration.
We are lucky with these coordinates: the xequation is simply 4mx" – ml" = 0. So if the monkey hauls in the rope, l" is negative, and x decreases at 1/4 the rate of hauling, u/4. The yequation and the constraints that give z and u then show that all objects move up an equal distance, 1/4 the amount that the monkey has hauled in.
2. (Problem 7.29)
Let θ be the angle of the disk, and r (rather than l) be the lngth of the pendulum. Then the velocity of m is the vector sum of the velocity of P and the velocity of m with respect to P. The angle between these is θ  φ (if we measure θ from the vertical), so we have
w = 
×
q

v_{m}^{2} = (R 
×
q

)^{2} + (r  ×
f

)^{2} + 2Rr  ×
q

 ×
f
 cos(q f) 

hence the Lagrangian is
L = (^{1}/_{4}M + ^{1}/_{2}m)R^{2} 
×
q
 2

+ ^{1}/_{2}mr^{2}  ×
f
 2
 + mRr 
×
q
 
×
f
 cos(q f) + mg(R cosq+ r cosf) 

This leads to the EsOM:
(^{1}/_{2}M + m)R^{2}  ××
q

+ mRr(  ×
f

cos(qf))^{·} + mRr  ×
q
  ×
f

sin(q f) + mgRsinq 

= (^{1}/_{2}M + m)R^{2}  ××
q

+ mRr  ××
f

cos(qf) + mRr  ×
f
 2

sin(q f) + mgRsinq = 0 

mr^{2}  ××
f

+ mRr(  ×
q

cos(q f))^{·}  mRr  ×
f
  ×
q

sin(q f) + mgr cosf 
= mr^{2}  ××
f

+ mRr  ××
q

cos(q f) + mgr cosf  mRr sin(θφ)  × 2 θ = 0 

3.
(a) The two masses move radially with the same speed, and m also moves in the fdirection. The potential energy comes only from M. Hence
L = ^{1}/_{2}(m+M) 
×
r

2
 + ^{1}/_{2} mr^{2}  ×
f

2
  Mgr 

(b) Since L does not contain f, p_{f} is conserved, mr^{2}f^{·} = l. The requations then becomes
(m + M)  ××
r
  mr  ×
f
 2

+ Mg = (m + M)  ××
r
  
l^{2} mr^{3}

+ Mg = 0. 
 (*) 
(c) The circular orbit is at r_{0} = [l^{2}/(mMg)]^{1/3}.
(d) If r = r_{0} + e, Eq (*) becomes to first order
(m+M)  ××
e

+  3l^{2} mr_{0}^{4}
 e = 0 

which oscillates with frequency (l/r_{0}^{2})[3/(m^{2}+mM)]^{½} = [^{3Mg}/_{r0(M+m)}]^{½}.
4. (problem 7.41)
Use r instead of ρ. T = ½m(r^{·}² + z^{·}² + r²ω²) = ½m(r^{·}² + (3kr²)²r^{·}² + r²ω²); U = mgz = mgkr³. Put these together in L = T  U and get the EOM
m^{d}/_{dt}[r^{·}(1+9k^{2}r^{4})]  18mk²r³r^{·}²  mrω² + 3mgkr² = 0
The condition for equilibrium is F_{eff} = mrω²  3mgkr² = 0 so r_{o} = 0 or ω²/(3gk). It is easy to check that F_{eff} is positive below the second r_{0} and negative above, a restoring force, so the equilibrium is stable. Here we have neglected the term in r^{·}² since it is second order. If F_{eff} were zero at all orders, we would have to consider it.
The Hamiltonian isH = ½m(1+9k^{2}r^{4})r^{·}²  ½mr²ω² + mgkr³.
This provides another way to see that the equilibrium is stable: r_{0} is a minimum of the effective potential U_{eff} =  ½mr²ω² + mgkr³.
5. (problem 7.46)
L(r_{i}, θ_{i}, φ_{i} + ε) = L(r_{i}, θ_{i}, φ_{i}) implies ε Σ¶L/¶φ_{i} = 0 (since each φ_{i} changes by the same ε). Use Lagrange's equations to change this to
ε ^{d}/_{dt}Σ¶L/¶φ^{·}_{i} = 0
so Σp_{φ i} = const., that is, the total φmomentum is constant.