Question

(Figure 1) The figure shows a simple model of a seesaw. These consist of a plank/rod of mass mr and length 2x allowed to pivot freely about its center (or central axis), as shown in the diagram. A small sphere of mass m1 is attached to the left end of the rod, and a small sphere of mass m2 is attached to the right end. The spheres are small enough that they can be considered point particles. The gravitational force acts downward. The magnitude of the acceleration due to gravity is equal to g. What is the moment of inertia I of this assembly about the axis through which it is pivoted?

Answer 1

Moment of inertia of seesaw rod about pivot = ML² /12 = m_r (2x)²/12 = m_rx² /3

Moment of inertia of mass m₁ about pivot point = m₁ x²

Moment of inertia of mass m₂ about pivot point = m₂x²

Total moment of inertia about pivot point = m_rx² /3 + m₁ x² + m₂x²  = (x²/3)*(m_r+3m₁+3m₂)