A stone of mass 0.25 kg tied to the end of a string is whirled round i

1.	A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in thestring? What is the maximum speed with which the stone can be whirled around ifthe string can withstand a maximum tension of 200 N ?Answer me fast✌✌✌
Answer» 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.

A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in thestring? What is the maximum speed with which the stone can be whirled around ifthe string can withstand a maximum tension of 200 N ?Answer me fast✌✌✌

Answer»

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.