Answer:
T = 13.3 N
Explanation:
In this exercise we use Newton's second law, in the lower birth of the circle the tension and force are opposite, so the tension has its maximum value
T - W = m a
The acceleration is centripetal
a = v² / r = w² r
We replace
T = mg + m w² r
The angular velocity is related to the period
w = 2π f = 2π / T₀
T₀ = 2.10 s
T = m (g + 4π² r / T₀²)
Let's calculate
T = 1.35 (9.8 + 4π² 1.10 /2.10²)
T = 13.3 N
Answer:
26.51 N
Explanation:
Using
F = mω²r + mg ....................... Equation 1
Where F = maximum tension in the string, ω = angular velocity, r = radius of the circle formed/ Length of the ball, g = acceleration due to gravity
Also,
ω = 2π/T........................ Equation 2
Where T = Period, π = Pie
Substitute equation 2 into equation 1
F = [mr(2π/T)²]+mg
F = [mr4π²/T²]+mg.................... Equation 3
Given: m = 1.35 kg, r = 1.10 m, T = 2.10 s, g = 9.8 m/s², π = 3.14
Substitute into equation 3
F = 1.35(1.1)(4)(3.14²)/2.1² + 1.35(9.8)
F = 13.28+13.23
F = 26.51 N.
Hence the maximum tension in the string = 26.51 N
