Two masses 2m and m are whirled in a horizontal
circle. The radius of path of 2m is rand of mis 2ras
shown in figure. Then the ratio of tension of two strings
are
2m
Oe
m
(1) 2:1
(3) 3:4
(2) 1:1
(4) 4:3
Answers
Given :
Two masses 2m and m are whirled in a horizontal circle. The radius of path of 2m is r and of m is 2r.
To Find :
The ratio of tension of two strings.
Solution :
❖ Centripetal force is not a new kind of force. The material forces such as tension, gravitational force, electrical force, friction, etc., may act as the centripetal force in any circular motion.
In this situation, tension acts as the centripetal force
A] Tension in string - I :
➙ T₁ = m₁ v₁² / r₁
➙ T₁ = m₁ (r₁ω)² / r₁
➙ T₁ = m₁ r₁ ω²
➙ T₁ = 2m r ω² .......... (I)
B] Tension in string - II :
➙ T₂ = m₂ v² / r₂
➙ T₂ = m₂ (r₂ω)² / r₂
➙ T₂ = m₂ r₂ ω²
➙ T₂ = m (2r) ω² .......... (II)
By taking ratio of both equations, we get
➠ T₁ / T₂ = (2mrω²) / (2mrω²)
➠ T₁ : T₂ = 1 : 1
Given :
Two masses 2m and m are whirled in a horizontal circle. The radius of path of 2m is r and of m is 2r.
To Find :
The ratio of tension of two strings.
Solution :
❖ Centripetal force is not a new kind of force. The material forces such as tension, gravitational force, electrical force, friction, etc., may act as the centripetal force in any circular motion.
In this situation, tension acts as the centripetal force
A] Tension in string - I :
➙ T₁ = m₁ v₁² / r₁
➙ T₁ = m₁ (r₁ω)² / r₁
➙ T₁ = m₁ r₁ ω²
➙ T₁ = 2m r ω² .......... (I)
B] Tension in string - II :
➙ T₂ = m₂ v² / r₂
➙ T₂ = m₂ (r₂ω)² / r₂
➙ T₂ = m₂ r₂ ω²
➙ T₂ = m (2r) ω² .......... (II)
By taking ratio of both equations, we get
➠ T₁ / T₂ = (2mrω²) / (2mrω²)
➠ T₁ : T₂ = 1 : 1