A uniform rod of mass 1.0 kg and length 2.0 m is free to rotate about one end.

Question

A uniform rod of mass 1.0 kg and length 2.0 m is free to rotate about one end. If the rod is released from rest at an angle of 60° with respect to the horizontal, what is the speed of the tip of the rod as it passes the horizontal position?

Answer 1

KE_f+ PE_f= KE_i+ PE_i

Now you substitute and reduce

0.5Iω_f² + mg(0) = 0.5Iω_i² + mgh

0.5Iω_f² + mg(0) = 0.5I(0rad/s)² + mgh

0.5Iω_f² = mgh

h = 2m x sin(60º)

I = 1/3(1kg)(2m)^2

Plug in and solve for ω_f