Question

A simple pendulum of length 0.50 m has a 0.30-kg bob. At t = 0, the bob passes through the lowest position in its motion, and at this instant it has a horizontal speed of 0.39 m/s . A. What is the maximum angular displacement ϑmax away from vertical that the pendulum reaches? B. What is the magnitude of the bob's linear speed v when it has angular displacement ϑmax/2?

Answer 1

Length of pendulum = L = 0.5 m

Mass of bob = 0.3 kg

Speed of the bob at the lowest point = V₀ = 0.39 m/s

Gravitational acceleration = g = 9.81 m/s²

Maximum angular displacement = _max

Height gained by the bob when it has maximum angular displacement = h₁

Height gained by the bob when the angular displacement is (_max/2) = h₂

Speed of the bob when the angular displacement is (_max/2) = V

From the figure,

O₁C = LCos(_max)

CA = h₁ = O₁A - O₁C

h₁ = L(1 - Cos(_max))

O₂F = LCos(_max/2)

FD = h₂ = O₂D - O₂F

h₂ = L(1 - Cos(_max/2))

When the bob has maximum angular displacement away from the vertical the speed of the bob will be zero, therefore the kinetic energy of the bob at the bottom most point is converted into potential energy.

mV₀²/2 = mgh₁

V₀²/2 = gL(1 - Cos(_max))

(0.39²)/2 = (9.81)(0.5)(1 - Cos(_max))

1 - Cos(_max) = 0.0155

Cos(_max) = 0.9845

_max = 10.1^o

_max/2 = 5.05^o

When the bob has an angular displacement of (_max/2) away from the vertical the speed of the bob will be V, therefore by energy balance,

mV₀²/2 = mV²/2 + mgh₂

V₀²/2 = V²/2 + gL(1 - Cos(_max/2))

(0.39²)/2 = V²/2 + (9.81)(0.5)(1 - Cos(5.05))

V² = 0.114

V = 0.337 m/s

A) Maximum angular displacement from the vertical that the pendulum reaches = _max = 10.1^o

B) Magnitude of speed of the bob when it has angular displacement (_max/2) = 0.337 m/s