A pendulum consists of a 2.2 kg bob attached to a 2.8 meter long string. The bob has a velocity of 4.7 m/s as it passes the bottom.

Question

Considering the point where the string makes a 29 degree angle with the vertical, determine the following: a) the tension in the string. b) what is the angle of the string at its highest point?

Answer 1

Data:

Mass of the bob, m = 2.2 kg

Length of the string, L = 2.8 m

Velocity at the lowest point, v = 4.7 m/s

Angle at the given position, θ = 29 deg ( with vertical )

Formula/Info:

Formula for vertical height from the lowest point,

h = L ( 1 - cosθ )

Potential + kinetic energy at the given position =  Kinetic energy at lowest position.

Derivation (using kinematics):

m g h + (1/2) m v'^2 = (1/2) m v^2

g h + (1/2) v'^2 = (1/2) v^2

2 g h + v'^2 = v^2

2 g L ( 1 - cos θ ) + v'^2 = v^2

Substituting our Data:

2 * 9.8 * 2.8 * ( 1- cos 29 ) + v'^2 = 4.7^2

v'^2 = 22.09 - 6.88 = 15.21 m/s

v' = 3.9 m/s

Ans:

Velocity at the given position, v' = 3.9 m/s

ANSWER A:

Tension, T = ( mv'^2 / L ) + mg cosθ

                = ( 2.2 * 3.9^2 / 2.8 ) + 2.2 * 9.8 * cos 29

                = 11.9507 + 18.8568

                = 30.8075 N

Ans:

Tension in the string, T = 30.81 N

ANSWER B:

PE at greatest height = KE at the lowest point

m g L ( 1 - cosθ' ) = (1/2) m v^2

g L ( 1 - cosθ' ) = (1/2) v^2

9.8 * 2.8 * ( 1 - cos θ' ) = 0.5 * 4.7^2

1 - cos θ' = 0.4025

cos θ' = 0.5974

θ' = 53.31 deg

Ans:

Angle at the greatest height, θ' = 53.31 deg