Question

A thin rod (length = 1.09 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge. The mass of the rod may be ignored, compared to the mass of the object fixed to the top of the rod.
The rod, starting from rest, tips over and rotates downward.

(a) What is the angular speed of the rod just before it strikes the floor?

(Hint: Consider using the principle of conservation of mechanical energy.)

(b) What is the magnitude of the angular acceleration of the rod just before it strikes the floor?

Answer 1

Answer:

a) w = 4.24 rad / s , b) α  = 8.99 rad / s²

Explanation:

a) For this exercise we use the conservation of kinetic energy,

Initial. Vertical bar

        Emo = U = m g h

Final. Just before touching the floor

       Emf = K = ½ I w2

As there is no friction the mechanical energy is conserved

       Emo = emf

       mgh = ½ m w²

The moment of inertial of a point mass is

        I = m L²

       m g h = ½ (m L²) w²

       w = √ 2gh / L²

The initial height h when the bar is vertical is equal to the length of the bar

         h = L

         w = √ 2g / L

Let's calculate

       w = RA (2 9.8 / 1.09)

       w = 4.24 rad / s

b) Let's use Newton's equation for rotational motion

         τ = I α

         F L = (m L²) α

The force applied is the weight of the object, which is at a distance L from the point of gro

         mg L = m L² α  

          α  = g / L

          α  = 9.8 / 1.09

          α  = 8.99 rad / s²