A particle of mass m is attached to one end, A, of a light inelastic string of length l. The other end of the string, B,

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A particle of mass m is attached to one end, A, of a light inelastic string of length l. The other end of the string, B,

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A Particle Of Mass M Is Attached To One End A Of A Light Inelastic String Of Length L The Other End Of The String B 1
A Particle Of Mass M Is Attached To One End A Of A Light Inelastic String Of Length L The Other End Of The String B 1 (77.82 KiB) Viewed 106 times
A Particle Of Mass M Is Attached To One End A Of A Light Inelastic String Of Length L The Other End Of The String B 2
A Particle Of Mass M Is Attached To One End A Of A Light Inelastic String Of Length L The Other End Of The String B 2 (57.37 KiB) Viewed 106 times
Answer:
A Particle Of Mass M Is Attached To One End A Of A Light Inelastic String Of Length L The Other End Of The String B 3
A Particle Of Mass M Is Attached To One End A Of A Light Inelastic String Of Length L The Other End Of The String B 3 (1.51 KiB) Viewed 106 times
A particle of mass m is attached to one end, A, of a light inelastic string of length l. The other end of the string, B, is attached to a ceiling so that the particle is free to swing in a vertical plane. The angle between the string and the downward vertical is θ radians. You may assume that the air resistance on the particle is negligible. Initially, θ=3π​ and the particle is released from rest.
(i) Show that the potential energy lost by the particle since leaving its initial position is 2mgl​(2cosθ−1). Hence find an expression for v2, where v is the linear speed of the particle, in terms of l,g and θ. (ii) Show that the tension in the string at any point of the motion is mg(3cosθ−1). (iii) Find the greatest tension in the string. What is the position of the particle when the tension in the string is greatest?
v2=gl(2cosθ−1) 2mg when θ=0
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