A small body of mass m hangs at one end of a string of length a, the other end of which is fixed. It is given a horizontal velocity U at its lowest position so that the string would just become slack, when it make an angle of 60° with the upward drawn vertical line. find the tension in the string at point of projection
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See the enclosed picture.
Length of string = a (= L).
Mass = m.
1. At the point of projection:
Speed = u. Tension = T1.
T1 = mg + m u^2/a
2. At the point the string becomes slack..
Angle with the vertical = θ.
Tension = T2. Speed = v.
Balance forces and energy.
T2 + mg cosθ = mv^2/a
mu^2 /2 = mv^2 /2 + mga (1+ cosθ)
Solving these T2 = mu^2/a - mg* (2+3 cosθ).
If the string becomes slack then T2 = 0.
u^2 = g*a *(2+3 cosθ)
If θ = 60° then, u^2 = 7 g a /2.
Finally, tension T1 = 9 m g / 2.
Length of string = a (= L).
Mass = m.
1. At the point of projection:
Speed = u. Tension = T1.
T1 = mg + m u^2/a
2. At the point the string becomes slack..
Angle with the vertical = θ.
Tension = T2. Speed = v.
Balance forces and energy.
T2 + mg cosθ = mv^2/a
mu^2 /2 = mv^2 /2 + mga (1+ cosθ)
Solving these T2 = mu^2/a - mg* (2+3 cosθ).
If the string becomes slack then T2 = 0.
u^2 = g*a *(2+3 cosθ)
If θ = 60° then, u^2 = 7 g a /2.
Finally, tension T1 = 9 m g / 2.
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