Question

A simple pendulum is formed of a rope of length L = 2.2 m and a bob of mass m. When the pendulum makes an angle θ = 10° with the vertical, the speed of the bob is 2 m/s. The angular speed, θ', at the lowest position is equal to: (g = 10 m/s^2)
0.63 rad/s
1.32 rad/s
0.98 rad/s
2.93 rad/s
1.84 rad/s

Answer 1

at 10deg
height above hanging straight down = 2.2 (1-cos 10) = 2.2* 0.0152 =0.0334
so potential energy above hanging straight down = m g h = m*10*0.0334
= 0.334 m Joules
The kinetic energy there = (1/2) m v^2 = (1/2)m *4 = 2 m
so total energy there = 2 m + 0.334 m = 2.334 m Joules

The kinetic energy at the bottom
(1/2) m v^2 = 2.334 m
so
v^2 = 4.67
v = 2.16 m/s
omega * R = v
so
omega = 2.16 / 2.2 = 0.982 rad/s