Question

A stone of mass 0.25 kg tied to the end of a string is whirled in a circle of radius 1.5 m with a speed of 40 revolutions/min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

Answer 1

Given: m = 0.25 kg, r = 1.5 m, T_max = 200 N,

n = 40 rev. min^-1 = \(\cfrac{40}{60}\) rev s^-1

To find: 

i. Tension (T)

ii. Maximum speed (v_max)

Formulae: i. T = mrω²  ii. T_max = \(\cfrac{m\text v^2_{ma\text x}}r\) 

Calculation: Since, ω = 2πn = \(\cfrac{2\pi\times40}{60}\)

i. The tension in the string is 6.55 N.

ii. The maximum speed with which the stone can be whirled around is 34.64 m s^-1.