A concentrated mass A of 2 kg is attached to the end of a 3 kg uniform slender rod. The rod is attached to the vertical wall with a spring 15 cm from the bottom pivot point. When the rod is vertical, it has an angular velocity, ω, of 2 rad/s. The undeformed length of the spring is 10 cm. Determine the spring modulus k such that the rod's angular velocity will be zero when it is fall to the horizontal.

Force Diagram at
Arbitrary Position

The system is conservative so the energy has be tot he same when vertical (1) and when horizontal (2),
     T₁ + V₁ = T₂ + V₂

Moment of inertia of the rod and mass about pivot point, B is
     I_B = m_rod L²/3 + m_concL²
        = 3(0.2)²/3 + 2(0.2)²
        = 0.120 kg m²

The initial kinetic energy,
     T₁ = 0.5(0.12)(2)² = 0.24 J

The final kinetic energy,
     T₂ = 0

Using a datum line through point B, the initial potential energy is
     V₁ = 2g (0.2) + 3g (0.1) + kδ² / 2
         = (0.4 + 0.3) 9.81 + k(0.05 - 0.1)² / 2
         = 6.867 + k 0.00125 J 

The final length of the spring is
     [(0.15)² + (0.15 + 0.05)²]^0.5 = 0.25 m

The final potential energy is,
     V₂ = 0 + 0 + k(0.25 - 0.1)²/2
          = k 0.01125 J

The conservation of energy equation becomes,
     0.24 + ( 6.867 + 0.00125 k) = 0 + 0.01125 k

k = 710.7 N/m