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) |
|
This leads to the EsOM:
|
(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) | × 2
θ = 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
| - |
l2
mr3
|
+ Mg = 0. |
| (*) |
(c) The circular orbit is at r0 = [l2/(mMg)]1/3.
(d) If r = r0 + e, Eq (*) becomes to first order
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 isH = 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 Σ¶L/¶φi = 0 (since each φi changes by the same ε). Use Lagrange's equations to change this to
ε d/dtΣ¶L/¶φ·i = 0
so Σpφ i = const., that is, the total φ-momentum is constant.