Question
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./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?
Solution
We have,
Mass of the stone, m = 0.25 kg
Radius of the circle, r = 1.5 mMass of the stone, m = 0.25 kg
Number of revolution per second, n =
40 / 60 = 2 / 3 rpsAngular velocity, ω =
= 2πn The tension in the string provides the centripetal force.
i.e.,
T = Fcentripetal
=
= mrω
= mr(2πn)2
= 0.25 × 1.5 × (2 × 3.14 × (
)2= 6.57 N
Maximum tension in the string,
Tmax = 200 N
max = mv2max / r
∴ vmax = (Tmax × r / m)1/2
= (200 × 1.5 / 0.25)1/2
= (1200)1/2
= 34.64 m/s
Therefore, the maximum speed of the stone is 34.64 m/s.
