Solutions for Homework 3
1. (a) We have h = b/tanθ and H = l b/sinθ, thus
U = mgh  MgH = 
gb sinθ

(m  M cosθ) up to a constant 

(b) dU/dθ = gb(Mmcosθ)/sinθ. If M > m this never vanishes, no equillibrium. If M = m it vanishes at θ = 0  limit of infinitely long string. If M < m equilibrium at arccos(M/m) is stable.
2. (a) This is the same as problem 1 with M = m, "doubled", in the very (infinitely) long string limit. For that system (before "doubling"),initially, θ = π/2, U = gbm. Finally θ = 0, U = 0 (evaluate 0/0 by l'Hospital's rule). Thus ^{1}/_{2}(2m)v_{¥}^{2} = gbm = mgl in terms of the l of the figure, hence
Actually it is equally simple to write U in terms of h, the distance that 2m moved, for example (for the actual, "doubled" system) U(h) = (2m)gh + 2 mg(l^{2}+h^{2})^{½}. Then U(0) = 2mgl and U(¥) = 0. And, for this system, finally a mass of 4m is moving.
(b) Using h we now have U(h) = (m)gh + 2 mg(l^{2}+h^{2})^{½}, which has a minimum at h = l/Ö3, and ΔU = U(0) U(l/Ö3) = 2mgl  [mgl/Ö3 + 2mg(2l/Ö3)] = mgl(2  Ö3).
The velocities of the outer masses at that point are half that of the inner mass, thus ^{1}/_{2}m[v^{2}+(v/2)^{2} + (v/2)^{2}] = ^{3}/_{4}v^{2} = ΔU and
v_{max} =  ^{Ö}

(4gl/3)(2  Ö3)
 

As remarked in class, this is really the velocity at the minimum of the potential, not the maximum velocity of the central mass, because the kinetic energy of the system also depends on h, T = ½mh^{·2}(3h²+l²)/(h²+l²), so ΔU should be multiplied by (h²+l²)/(3h²+l²) before it is maximized. The difference is considerable: the point of maximum velocity moves from h = 0.577l to 0.415l, and v_{max} from 0.598Ögl to 0.621Ögl  чёрт возьми!
The maximum distance h_{m} is reached where U(0) = U(h_{m}), a quadratic with the two solutions h = 0 and h_{m} = 4l/3  the latter being the one we want.
3. (a) For a small displacement dρ in ρ only, the point moves by actual distance dρ; similarly for dz; and for dφ the point moves by rdφ. Therefore the kinetic energy is
T = ^{1}/_{2} m [ (dr/dt)^{2} + (dz/dt)^{2} + r^{2} (dφ/dt)^{2}] . 

(b) We have
¶T/¶r^{·} = m 
×
r

¶T/¶z^{·} = m 
×
z
 ¶T/¶φ^{·} = m r^{2} 
×
f
 ¶T/¶r = mr 
×
f
 2



Therefore:
r equation: ma_{r} = m  ××
r

 mr 
×
f

2
 

zequation: ma_{z} = m 
××
z



... and remember that d/dt means, differentiate everything
f equation ma_{f} = mr^{2} 
××
φ
 + 2mr  ×
r
  ×
f
 

Since a unit basis is dr, dz, and rdp, a_{r} and a_{z} are already the "physical" components of the acceleration; and a_{f} df = (a_{f}/r)(rdf), so the physical pcomponent of the acceleration is
a_{f^} = a_{f}/r = r 
××
f
 + 2  ×
r
 
×
f
 . 

4. (a)
E = ^{1}/_{2} ml^{2}  ×
f
 2
 + mgl(1cos f) 

(b)
dE/dt = 0 = ml^{2}  ×
f
  ××
f
 + mgl  ×
f
 sinf or l^{2}  ××
f
 = mglsinf which is Ia = G 

(c) For small φ, sin φ » φ and we get SHM with period 2pÖl/g
5. (a) Here's what is given:
total momentum P = 
å
a

m_{a} v_{a} = k_{1} total angular momentum L_{0} = 
å
a

r_{a} ×m_{a} v_{a} = k_{2} 

The distance from a point P where r = a, is r  a, so the angular momentum is
L_{P} = 
å
a

(r_{a}  a)×m_{a} v_{a} = 
å
a

r_{a} ×m_{a} v_{a}  
å
a

a ×m_{a} v_{a} = L_{P}  a× 
å
a

m_{a} v_{a} = L_{P}  a×P = k_{1}  a ×k_{2} = const. 

(b) A particle falling in a straight line has nonconstant momentum but zero angular momentum about any two points along its trajectory.
File translated from
T_{E}X
by
T_{T}H,
version 3.67.
On 08 Feb 2006, 22:26.