Question

A simple pendulum with mass m = 2.1 kg and length L = 2.3 m hangs from the ceiling. It is pulled back to an small angle of θ = 11.9° from the vertical and released at t = 0.

4)What is the angular displacement at t = 3.56 s? (give the answer as a negative angle if the angle is to the left of the vertical)

6)What is the magnitude of the radial acceleration as the pendulum passes through the equilibrium position?

Answer 1

w = sqrt [g/L] = sqrt [9.8 / 2.3] = 2.06 s^-1

theta = theta_max = 11.9 deg * pi rad / 180 = 0.208 rad

v_max = 2.3 * 2.06 * 0.208 = 0.986 m/s

theta = theta_max * cos wt

= 11.9 * cos (2.06 * 3.56) = 5.92 deg

6) a_r = v_max² / L = 0.986² / 2.3

= 0.423 m/s²