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 ?

Answer 1

Given: The mass of stone is 0.25 kg, radius of circle is 1.5 m, speed of stone is 40 rev/min and the maximum tension in the string is 200 N.

The speed of the stone in rps is given as,

n=N× 1 60

Where, n is the speed in rps and N is the speed in rpm.

By substituting the values in the above expression, we get

n=40× 1 60 = 2 3  rps

The angular speed of stone is given as,

ω=2πn

Where, ω is the angular velocity of stone.

By substituting the values in the above expression, we get

ω=2π× 2 3

The tension in the string is given as,

T=mr ω 2

Where, T is the tension in the string and r is the radius of circle.

By substituting the values in the above expression, we get

T=0.25×1.5× ( 2×π× 2 3 ) 2 =6.6 N

The maximum speed with which the stone can be whirled around is given as,

T max = m v max 2 r v max = r T max m

Where, v max is the maximum speed with which the stone can be whirled around, T max is the maximum tension in the string and m is the mass of stone.

By substituting the values in the above expression, we get

v max = 1.5×200 0.25 =34.6 m/s ≈35 m/s

Thus, the tension in the string is 6.6 N and the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N is 35 m/s.