Question

A uniform rod with a mass of m = 1.94 kg and a length of l = 2.10 m is attached to a horizontal surface with a hi=nge. The rod can rotate around the hi=nge without friction. (See figure.)

Initially the rod is held at rest at an angle of θ = 70.4° with respect to the horizontal surface. Then the rod is released.
1. What is the angular speed of the rod, when it lands on the horizontal surface?
2. What is the angular acceleration of the rod, just before it touches the horizontal surface?

Answer 1

H = L sin θ / 2 height of center of mass

E = m g H = m g L sin θ / 2

I = m L^2 / 3 moment of inertia about end of rod

1/2 I ω^2 = m g L sin θ / 2 KE of rod when it hits surface

1/2 m L^2 / 3 ω^2 = m g L sin θ / 2

ω^2 = 3 g sin θ / L

ω = (3 g sin θ / L)^1/2 angular speed of rod striking surface

P = I α if P = torque

α = m g L/2 / (m L^2 / 3) = 3 g / (2 L) angular acceleration of rod