Laws of motion NCERT Solutions Class 11 Physics - Solved Exercise Question 5.21

Question 5.21:
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:
Mass of the stone, m = 0.25 kg
Radius of the circle, r = 1.5 m
Number of revolution per second, n = 40 / 60 = 2 / 3 rps
Angular velocity, ω = v / r = 2πn
The centripetal force for the stone is provided by the tension T, in the string, i.e.,
T = FCentripetal
= mv2 / r = mrω = mr(2πn)2
= 0.25 X 1.5 X (2 X 3.14 X (2/3) )2
= 6.57 N
Maximum tension in the string, Tmax = 200 N
Tmax = mv2max / r
∴ vmax = (Tmax X r  / m)1/2
= (200 X 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.

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