Answer:
(a) The tension in the pendulum cable at the lowest part is 12.125 N
(b) The angle the cable make with the vertical is 75⁰
(c) The tension in the pendulum cable at the highest point is 1.29 N
Explanation:
Part (a) the tension in the pendulum cable
T = m(a+g)
Where;
T is the tension on the pendulum cable (N)
m is the mass of the pendulum bob ( 0.5 kg)
a is the centripetal acceleration (m/s²)
g is acceleration due to gravity ( 9.8 m/s²)
a = v²/r
Given v = 3.4 m/s and r = length of the pendulum = 80cm = 0.8m
a = (3.4)²/0.8 = 14.45 m/s²
T = m(a+g) = 0.5(14.45 +9.8) = 12.125 N
Part (b) angle the cable make with the vertical
initial velocity (V₁) = 3.4 m/s
final velocity (V₂) = 0
initial height (h₁) = 0
final height (h₂) = ?
Applying the principle of conservation of mechanical energy
P.E₁ + K.E₁ = P.E₂ + K.E₂
mgh₁ + 0.5mv₁² = mgh₂ + 0.5mv₂²
mg(0) + 0.5*0.5(3.4²) = 0.5*9.8(h₂) + 0.5m(0)²
0.5*0.5(3.4²) = 0.5*9.8(h₂)
2.89 = 4.9h₂
h₂ = 2.89/4.9 = 0.59 m
Note: if the pendulum bob swings to highest point, the vertical displacement will be length of the pendulum minus maximum height attained by the pendulum.
Vertical displacement = 0.8 - 0.59 = 0.21 m
let the angle between this Vertical displacement and length of pendulum = θ
cos θ = vertical displacement/ length of pendulum
cos θ = 0.21/0.8
cos θ = 0.2625
θ = cos⁻¹(0.2625) = 74.8⁰ = 75⁰
Part (c) Tension in the pendulum when it reaches highest point
At maximum height, centripetal acceleration (a) = 0
T = mgcosθ
T = 0.5*9.8*cos(75)
T = 0.5*9.8*0.2625 = 1.29 N
T = 1.29 N
