Problem 1:

A pendulum consists of a rod of negligible mass and length L attached to a sphere of mass M of radius R = L/4 and uniform mass density.  The pendulum is released from rest from the horizontal position.  As it moves through the vertical position, what is the tension T in the rod?  (Assume a gravitational acceleration g pointing downward.)

Solution:

• Concepts:
  Energy conservation, centripetal acceleration
• Reasoning:
  The gravitation potential energy is converted into translational and rotational kinetic energy of the sphere.
  As it moves through the vertical position, the CM of the sphere has speed v.  The difference between the tension T in the rod and the weight of the sphere must equal the centripetal force.
• Details of the calculation:
  Since the rod has negligible mass, we can focus on the sphere.
  Mg5L/4 = ½Mv² + ½I_spω².
  I_sp = 2MR²/5 = ML²/40.  v = 5Lω/4.
  Mg5L/4 = (25/32)ML²ω² + ML²ω²/80.
  ω² = (g/L)200/127.  v² = 2.46 gL.
  T - Mg = Mv²/(5L/4), T = 2.97 Mg.

Problem 2:

You have a system of 4 particles of equal mass M = 1 kg.  One sits at the origin (x, y, z) = (0, 0, 0) m.  The others sit at (1, 1, 0), (0, 1, 1) m, (1, 0, 1) m.   These particles are attached to each other by massless rigid rods such that they will always maintain their positions relative to one another.
(a)  Calculate the full moment of inertia tensor for this system.
(b)  Find the principal axes for this system.

In case it's useful to you, here's a fact:
(4 − x)³ − 3(4 − x) − 2 = (2 − x)(5 − x)².

Solution:

• Concepts:
  The inertia tensor
• Reasoning:
  
  I_ij = Σ_km_k[Σ_l(x_k)_l²δ_ij - (x_k)_i(x_k)_j] is the inertia tensor.
  The I_ii are called the moments and the I_ij (i ≠ j) are called the products of inertia.
• Details of the calculation:
  (a)
  I_xx = M Σ_k(y_k² + z_k²) = M(0 + 1 + 2 + 1) = 4 kgm².
  I_yy = M Σ_k(x_k² + z_k²) = M(0 + 1 + 1 + 1) = 4 kgm².
  I_zz = M Σ_k(x_k² + y_k²) = M(0 + 2 + 2 + 1) = 4 kgm².
  I_xy = I_yx = -MΣ_kx_ky_k = -M(0 + 1 + 0 + 0) = -1 kgm².
  I_yz = I_zy = -MΣ_kz_ky_k = -M(0 + 0 + 1 + 0) = -1 kgm².
  I_xz = I_zx = -MΣ_kx_kz_k = -M(0 + 0 + 0 + 1) = -1 kgm².

  The moment of inertia tensor is

  

  (b)  L_i  = Σ_jI_ijΩ_j.  If Ω points along a principal axis then L and Ω are aligned and L_i = IΩ_i, where I is one of the principal moments of inertia.
  Then Σ_j(I_ij - I δ_ij)Ω_j = 0,   |I_ij - I δ_ij|= 0.

  

   (4 − I)³ − 3(4 − I) − 2 = (2 − I)(5 − I)² = 0.
  
  I₁ = 2, I₂ = I₃ = 5, we have a symmetric top.
  To find the corresponding principal axis we use:
  I₁ = 2:                  
  2a₁ - a₂ - a₃ = 0, -a₁ + 2a₂ - a₃ = 0, --> a₁ = a₂ = a₃, the principal axis vector is  (1, 1, 1).
  I₂ = I₃ = 5:           
  -a₁ - a₂ - a₃ = 0.
  This equation defines a plane perpendicular to (1, 1, 1).
  We can choose any orthogonal pair of axes in that plane, for example (1, -1, 0) and (1, 1, -2).

Problem 3:

Denote the radius of the earth by R_E, the mass of earth by M_E, and the angular frequency of rotation of earth by ω.
Your firm has a contract to build the first space elevator.  The space elevator consists of a cable hanging from a geo-stationary satellite of mass M_s, which orbits in a circular orbit of radius R_s, to a point just above an anchor station on the earth's surface.  A climber can then take equipment to and from a space station at R_g = (GM_E/ω²)^1/3.  (R_g is the radius for geosynchronous orbits.)  The elevator cable has length R_s - R_E to reach the surface of the earth, and a constant mass per unit length λ.

(a)  Where on earth should the anchor station be located?
(b)  Find the tension T in the cable as a function of R, assuming T(R_E) = 0. 
Hint:  The net force on each section of the cable must be zero.
For which R is T = T_max?  In addition to R_E, is there another R for which T(R) = 0?
Hint:  If one root of the cubic equation x³ + ax + b = 0 is known to be r, then the sum of the other two roots is -r, and their product is -b/r.
This leads to the quadratic equation for the other two roots, xr² + x²r - b = 0.

(c)  What is the range of possible values for R_s?
(d)  Can you use a cable made from carbon fiber with tensile yield strength σ = 1080 MPa and mass density ρ = 1.6 g/cm³ for the space elevator?

Given:  R_E = 6370 km,  R_g = 42300 km,  M_E = 5.98*10²⁴ kg.
Note  These are approximate values.  For more information on a space elevator, see Canvas, Modules.

Solution:

• Concepts:
  Equilibrium in a non-inertial frame.
• Reasoning:
  The space station is in a "geo-stationary" or "geosynchronous" orbit.  It is not connected to anything, it is just a satellite orbiting at the right distance to have a period of 24 hours and stay over the same point on the earth surface.  We want a cable that in the rotating earth frame goes straight up and passes by the space station, so that a climber could climb up the cable to the right height and then hop over to the space station.
  The Earth is a rotating (non-inertial) frame.  The cable is to be at rest in that frame.  The net force on each section of the cable, which includes the fictitious centrifugal force, must be zero.

• Details of the calculations:
  (a)  The anchor station should be located at the equator.
  
  (b)  Consider a section of cable of length dR at radius R.
  For this section to be in equilibrium we need T(R + dR) - T(R)  =  GM_EλdR/R² - λdRω²R.
  dT(R)/dR =  GM_Eλ/R² - λω²R  =  GM_Eλ(1/R² - R/R_g³).
  Find T_max:  dT(R)/dR = 0 if R = R_g.  The tension is T_max at R = R_g.
  Integrate to find T(R):
  T(R) = GM_Eλ∫_RE^R (1/R'² - R'/R_g³)dR'
  = GM_Eλ[1/R_E - 1/R - ½R²/R_g³  + ½R_E²/R_g³].
  T(R) = 0 --> 1/R_E - 1/R - ½R²/R_g³  + ½R_E²/R_g³ = 0,
  or R³ - R(2R_g³/R_E + R_E²) + 2R_g³ = 0.
  Since R = R_E is one solution, we find the other solutions by solving the quadratic equation
  RR_E² + R²R_E - 2R_g³ = 0.
  R = (R_E/2)[(1 + 8(R_g/R_E)³)^½ - 1] is the other positive solution.
  T = 0 at R_max ~ 150000 km.  If R_s = R_max, the cable, by itself, can be at rest in the rotating frame.
  
  (c)  R_g < R_s < R_max is the range of R_s.
  For R_s > R_g the centrifugal force F_centifugal is greater than the gravitational force F_g on the mass M_s.
  If F_centifugal - F_g = T, then the net force on each section of the cable is still zero.
  If M_s is located at R_s, we need
  F_centrifugal = M_sω²R_s = M_sGM_ER_s/R_g³ = M_sGM_E/R_s² + T,  M_s = T(R_s)R_g³R_s²/(GM_E(R_s³ - R_g³).
  Since T(R_s) > 0 , a solution for M_s can be found.
  
  (d)  T_max = GM_Eλ[1/R_E - 3/(2R_g) + ½R_E²/R_g³] = λ*4.86*10⁷ Nm/kg.
  With ρ = 1.6 g/cm³ we have
  T_max = ρA*4.86*10⁹⁷ Nm/kg = 7.77*10¹⁰ N/m² A.
  The minimum required tensile yield strength is
  σ_min = T_max/A = 7.77*10¹⁰ Pa.
  This is more than the tensile yield strength of a carbon fiber, σ = 1.080*10⁹ Pa, so a carbon fiber cable cannot be used.

Problem 4:

A simple device was used in the Pioneer III lunar probe, which was capable to reduce its spin to zero.  The device consists of a small mass m on the end of a light cord wrapped around the symmetrical spinning body.  With the satellite spinning with angular speed ω₀ about its axis of symmetry, the mass m is released.  The cord will unwind and the angular speed of the satellite will decrease.  When the cord is completely unwound, it is released and allowed to fly away.  By choosing the length of the cord properly, the spin of the satellite can be reduced to any value less than the initial value.

Note:  X and Y are body-fixed axes.  They rotate with angular velocity ωk about the symmetry axis.
The coordinate axes x and y rotates with angular velocity (ω + dφ/dt)k about the symmetry axis.

(a)  Let R, ω, φ be defined in the figure above.  Assume that the mass m is released when φ = 0.
Let i and j be unit vectors pointing along the x- and y-axis, respectively.
Find the components of r in the i and j direction and show that the velocity of the mass may be written as
v = -Rω i + (ω + dφ/dt)Rφ j.
(b)  In terms of R, ω, φ, dφ/dt, and I, the moment of inertia of the satellite, find expressions for the kinetic energy and the angular momentum of the system.
(c)  Use conservation of energy and angular momentum to show that
C(ω₀² - ω²) = φ²(ω + dφ/dt)² and C(ω₀ - ω) = φ²(ω + dφ/dt), and find the constant C.
(d)  Solve for dφ/dt and ω in terms of ω₀ and C.
(e)  Find the length of the cord required for a mass m to spin a satellite with moment of inertia I and radius R down to zero.

Solution:

• Concepts:
  Conservation of mechanical energy and angular momentum
• Reasoning:
  There are no external torques, so angular momentum is conserved. No internal energy is released and no mechanical energy is dissipated through friction, so mechanical energy is conserved. 
• Details of the calculation:
  (a)  Let i and j be unit vectors pointing along the x- and y-axis, respectively.
  Then the position of the mass is given by  r = Rφ i + R j.  
  The velocity of the mass is
  v = dr/dt + (ω + dφ/dt)k × r = Rdφ/dt i + (ω + dφ/dt)Rφ j - (ω + dφ/dt)R i
  = -Rω i + (ω + dφ/dt)Rφ j.

  (b)  The angular momentum of the mass is
  L_m = r × mv = mR²[ω + φ²(ω + dφ/dt)] k, and the angular momentum of the system is
  L_s = [Iω + mR²[ω + φ²(ω + dφ/dt)]] k.
  The kinetic energy of the system is
  T_s = ½mv² + ½Iω² = ½Iω² + ½mR²[ω² + φ²(ω + dφ/dt)²].
  
  (c)  Energy and angular momentum are conserved.
  T_s = ½(I + mR²)ω₀² = ½Iω² + ½mR²[ω² + φ²(ω + dφ/dt)²].
  L_s = (I + mR²)ω₀ = Iω + mR²[ω + φ²(ω + dφ/dt)].
  Dividing by mR² we obtain
  C(ω₀² - ω²) = φ²(ω + dφ/dt)² and
  C(ω₀ - ω) = φ²(ω + dφ/dt), with C = 1 + I/(mR²).

  (d)  Dividing C(ω₀² - ω²) = φ²(ω + dφ/dt) by C(ω₀ - ω) = φ²(ω + dφ/dt) we obtain
  ω₀ + ω = ω + dφ/dt,  ω₀ = dφ/dt,  ω₀t = φ.
  C(ω₀ - ω) = φ²(ω + ω₀),  ω = ω₀(C - φ²)/(C + φ²) = ω₀(C - ω₀²t ²)/(C + ω₀²t ²).

  (e)  We want ω = 0 when φ = l/R, where l is the length of the cord.
  ω₀(C - (l/R)²)/(C + (l/R)²) = 0   C = (l/R)²,  l = R√C = (R² + I/m)^½.

Problem 5:

A uniform disk with mass M and radius R starts from rest and moves down an inclined at an angle from the horizontal φ.  The center of the disk has dropped a vertical distance h when it reaches the bottom of the incline.  The incline and the horizontal floor are perfectly rough and the disk rolls without slipping on the incline and on the floor.  What is the velocity of the CM of the disk on the horizontal floor?

Solution:

• Concepts:
  ΔP = FΔt,  ΔL = τΔt,  relationship between linear and angular momentum for rolling motion
• Reasoning:
  The disk makes an inelastic collision with the floor.  Its momentum and angular momentum change.  The changes in momentum and angular momentum are not independent of each other but are connected by the fact that the same force provides the linear and angular impulse.
• Details of the calculation:
  Energy conservation yields the speed of the disk when it contacts the floor.
  Mgh = ½Iω² + ½Mv² = (3/4)Mv²/R. v = (4gh/3)^½.
  Choose your coordinate system such that the x-axis points to the right and the y-axis points up.
  
  The disk has momentum p = mv cosφ - mv sinφ j and angular momentum L = -½MR²v/R k = -½MRv k.
  As the disk contacts the ground a vertical impulse reduces the y-component of its momentum to zero and changes the x-component of its momentum to
  p' = Mv' = Mv cosφ + F_xΔt. 
  The force F_x also results in a torque which changes the magnitude of the disks angular momentum to 
  L' = Iω^' = ½MRv - F_xRΔt = ½MRv - RMv'+ RMv cosφ.
  For rolling we need ω' = v'/R.
  ½MRv' = ½MRv - RMv'+ RMv cosφ.
  ½v' = ½v - v'+ v cosφ.
  (3/2)v' = (½ + cosφ)v.
  v' = (2/3)(½ + cosφ)v.
  v' =  (16gh/27)^½(½ + cosφ).