Problem 1:

Consider a rectangular lamina with dimensions A and B, and mass M.  The lamina is attached to a wall with a nail placed at a point P = P(x,y), where the coordinates x and y are referred to the center of the lamina.
(a)  Determine the position of stable equilibrium.
(b)  Describe the motion of the lamina when it is slightly displaced from this equilibrium position.

Solution:

• Concepts:
  Stable equilibrium
• Reasoning:
  The potential energy U has a minimum at the position of stable equilibrium.
• Details of the calculation:
  (a)  Potential energy:  U = -Mgdcosθ = -Mg(x² + y²)^½cosθ.
  Extreme U:  dU/dθ = 0, sinθ = 0, θ = 0 or π.
  θ = π, CM lies directly above the pivot point -->  unstable equilibrium.
  θ = 0, CM lies directly below the pivot point -->  stable equilibrium.
  
  (b)  For small angles θ near θ = 0 we have U(θ) = U(0) + dU/dθ|₀θ + ½ d²U/dθ²|₀θ² + ... .
  d²U/dθ² = Mgdcosθ.  U(θ) = U(0) + ½Mgdθ², keeping only terms up to second order in θ.
  U(θ) = U(0) + ½kθ² with k = Mgd.
  τ = I d²θ/dt²  = -dU/dθ =  - Mgdθ.  We have a restoring torque.
  
  To find I we use the parallel axis theorem.
  I_CM = (1/12)M(A² + B²).  I(x,y) = I_CM + M(x² + y²)=  I_CM + Md² = M(A²/12 + B²/12 + x² + y²). 
  d²θ/dt²  = -g(x² + y²)^½(A²/12 + B²/12 + x² + y²)θ.
  Frequency of oscillations:  ω² = g(x² + y²)^½(A²/12 + B²/12 + x² + y²).

Problem 2:

A cylinder with mass m, radius R, and uniform density starts rolling without slipping from rest in a train that accelerates with constant acceleration a₀.  The axis of the cylinder is perpendicular to the motion of the train as shown below.  Find the acceleration of the cylinder in the lab frame (rest frame) and in the train frame.

Solution:

• Concepts:
  Motion in an accelerating frame.
• Reasoning:
  In an accelerating frame fictitious forces appear.  The net force in such a frame is
  F = F_inertial - ma, where a is the acceleration of the frame.
• Details of the calculation:
  Let the acceleration of the train point in the x-direction.
  In the train frame, the net force on the ball is F = -ma₀ + f = ma' where f is the force of static friction preventing the contact point from slipping and a' is the acceleration of the CM of the cylinder.  Friction exerts a torque on the cylinder, |τ| = fR = I|α| = I|a'|/R.
  If the z-axis points up this torque points in the -y direction.
  f = -Ia'/R² since f points in the positive and a' in the negative x-direction.
  -ma₀ = ma' + Ia'/R².  a' = -a₀/(I/mR² + 1).
  For the cylinder I = ½mR²,  a' = -(2/3)a₀ is the acceleration of the cylinder in the train frame.
  
  The acceleration of the cylinder in the frame of the stationary floor is a'' = a₀ + a' = a₀/3.

Problem 3:

A square plate with uniform density, mass m, and of side length 2a, lies on a frictionless table as shown in the figure.  A force F = F i is applied for a very short time Δt and delivers an impulse Δp to the upper corner.  Find the position of this corner as a function of time.

Solution:

• Concepts:
  Δp = FΔt,  ΔL = τΔt
• Reasoning:
  The same force provides the linear and angular impulse.  The force is pointing in the x-direction.
• Details of the calculation:
  FΔt = mΔv = mv i,  √2 a FΔt = ΔL = Iω. 
  L and ω point in the -z direction, the rotation is clockwise.
  v i = i Δp/m = is the velocity of th CM and ω is angular velocity about the CM after the impulse.
  The moment of inertia of the plate about an axis normal to the plate through its its CM is
  I = 2ma²/3.
  [I = 4∫₀^adx∫₀^ady ρ(x² + y²), with ρ = m/4.]
  We therefore have
  ω = √2 a mv/I = (3/√2) v/a.
  The position of the corner as a function of time is
  x(t) = vt + √2 a sin(ωt),  y(t) = √2 a cos(ωt),  or
  x(t) = Δpt/m + √2 a sin(3Δpt/(am√2)), 
  y(t) = √2 a cos(3Δpt/(am√2)t).

Problem 4:

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).