A 1.35-kg ball is tied to a 1.10-m long string is being spun in a vertical circle at a constant speed and with a period of 2.10 s. What is the maximum tension in the string?

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

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