A simple pendulum consists of a light rod of length L = 0.9 m anchored at a fixed point (like the ceiling), and a small mass m = 0.5 kg attached to the free end. The configuration is moved to an angle θ = 39 degrees from the vertical and released from rest. Due to the presence of dissipative nonconservative forces (i.e. air resistance and friction at the anchor), 0.17 J of mechanical energy is removed from the system in its swing from θ to the vertical. i.e. Wnc = -0.17 J. What is the tension (in Newtons) in the light rod at the instant the pendulum is vertical?

Answer 1

ΔE=0.17 J,
h=L(1-cosα)
PE=KE+ΔE
mgh=mv²/2 + ΔE
mv² =2{mgL(1-cosα)- ΔE}
T= mv²/L + mg = 2{mgL(1-cosα)- ΔE}/L +mg

Answer 2

Thank you! Very helpful.