a stone of mass 1.3kg tied tied to the end of string in a horizontal plane is whirled around in a circle of radius 1 mts with a frequency 40rpm.what is the tension in the string.what is the speed with which the stone can be whirled around,if the string can withstand the maximum tension of 200N
Answers
m = 1.3 kg. r = 1 m. v = r w.
w = 2 pi f.
f = frequency = 40 rpm = 2/3 Hz.
T = 1.3 * 1 * (4 pi^2 * 4/9) Newton.
= 20.8 pi^2 /9 Newton.
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T max = 200 N = m r w^2
max w = sqrt(200/(1.3*1)) red/sec.
Max frequency = 600/pi * sqrt(5/13) rpm.
Explanation:
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
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 × 1.5 × (2 × 3.14 × (2/3) )2
= 6.57 N
Maximum tension in the string, Tmax = 200 N
Tmax = 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.