Laws of Motion
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Question : 21
Total: 40
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:
Frequency of revolution of stone v = 40 rev . ∕ min =
rev . ∕ s
mass of stone,m = 0.25 k g
radius of circle,r = 1.5 m
angular speed of the stoneω = 2 π v = 2 π ×
=
rad s − 1
T = tension in the string = ?
T max =
v max =
The centripetal force is provided by the tension (T) in the string
T =
= m r ω 2 ( ∵ v = r ω )
= 0.25 × 1.5 × (
) 2 N
= 6.58 N = 6.6 N
As the string can withstand a maximum tension of 200 N
∴ T max =
v max = √
= √
= √ 1200
= 34.64 m s − 1 = 35.0 m s − 1
∴ T = 6.6 N , v max = 35.0 m s − 1
mass of stone,
radius of circle,
angular speed of the stone
T = tension in the string = ?
The centripetal force is provided by the tension (T) in the string
As the string can withstand a maximum tension of 200 N
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