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?
Chapter Laws Of Motion Page 111
Answers
Radius of circular path = 1.5m
Angular frequency (f)= 40 rev/min
= 40 /60 rev /sec
= 2/3 rev/s
We know, in uniform circular motion , centripital force is provided by tension in string.
e.g Tension in string = mω²r
Angular speed (ω) = 2πf
So,
Tension = m×4πr²f²
= 0.25 × 1.5 × 4 × (22/7)² × (2/3)²
= 6.6 N
Given,
Maximum tension which can be withstand by the string .
Tmax = 200 N
Tmax = mVmax²/r
Vmax = √{Tmax × r /m}
Vmax = √1200 m/s = 34.6m/s
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.