### 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 L-x below pulley B. Let y be the distance of the mass m below pulley A, then the monkey is l-y 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 over-dot. 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 x-equation 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 y-equation 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 vm2 = (R ×q )2 + (r × f )2 + 2Rr × q × f cos(q- f)
hence the Lagrangian is
 L = (1/4M + 1/2m)R2 ×q 2 + 1/2mr2 × f 2 + mRr ×q ×f cos(q- f) + mg(R cosq+ r cosf)
 (1/2M + m)R2 ××q + mRr( ×f cos(q-f))· + mRr ×q ×f sin(q- f) + mgRsinq

 = (1/2M + m)R2 ××q + mRr ××f cos(q-f) + mRr ×f 2 sin(q- f) + mgRsinq = 0

 mr2 ××f + mRr( ×q cos(q- f))· - mRr ×f ×q sin(q- f) + mgr cosf
 = mr2 ××f + mRr ××q cos(q- f) + mgr cosf - mRr sin(θ-φ) × 2θ = 0

3.
(a) The two masses move radially with the same speed, and m also moves in the f-direction. The potential energy comes only from M. Hence
 L = 1/2(m+M) ×r 2 + 1/2 mr2 ×f 2 - Mgr
(b) Since L does not contain f, pf is conserved, mr2f· = l. The r-equations then becomes
 (m + M) ××r - mr ×f 2 + Mg = (m + M) ××r - l2mr3 + Mg = 0.
(*)
(c) The circular orbit is at r0 = [l2/(mMg)]1/3.
(d) If r = r0 + e, Eq (*) becomes to first order
 (m+M) ××e + 3l2mr04 e = 0
which oscillates with frequency (l/r02)[3/(m2+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

md/dt[r·(1+9k2r4)] - 18mk²r³r·² - mrω² + 3mgkr² = 0

The condition for equilibrium is Feff = mrω² - 3mgkr² = 0 so ro = 0 or ω²/(3gk). It is easy to check that Feff is positive below the second r0 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 Feff were zero at all orders, we would have to consider it.
The Hamiltonian is
H = ½m(1+9k2r4)r·² - ½mr²ω² + mgkr³.

This provides another way to see that the equilibrium is stable: r0 is a minimum of the effective potential Ueff = - ½mr²ω² + mgkr³.

5. (problem 7.46)
L(ri, θi, φi + ε) = L(ri, θi, φi) implies ε ΣL/φi = 0 (since each φi changes by the same ε). Use Lagrange's equations to change this to ε d/dtΣL/φ·i = 0 so Σpφ i = const., that is, the total φ-momentum is constant.