A simple pendulum consists of an object suspended by a string. The object is ass

Question

A simple pendulum consists of an object suspended by a string. The object is assumed to be a particle. The string, with its top end fixed, has negligible mass and does not stretch. In the absence of air friction, the system oscillates by swinging back and forth in a vertical plane. If the string is 2.55 m long and makes an initial angle of 28.5 degree with the vertical, calculate the speed of the particle at the following positions. at the lowest point in its trajectory when the angle is 15.0 degree A 2.06 kg ball is attached to the bottom end of a length of fishline with a breaking strength of 9 lb (40.1 N). The top end of the fishing line is held stationary. The ball is released from rest with the line taut and horizontal (theta = 90.0 degree). At what angle theta (measured from the vertical) will the fishline break?

Explanation / Answer

1)

L=2.55m

ø1=28.5°

ø2=15°

Let's assume that point A and point C are the highest point and the lowest point of trayectory, respectively. Using the equation for energy conservation, we get:

mg(L-Lcosø1)=(1/2)mv^2...v=sqrt(2g(L-Lcosø1))=2.461 m/s

Now, let B be the point where the string makes an angle of ø2=15°. Thus, the conservation of energy will be given by the following expression:

mg(Lcosø2-Lcosø1)=(1/2)mv^2...v=sqrt(2g(Lcosø2-Lcosø1))=2.087 m/s

2)

Radial force on ball --> T-mgcosø=Fr. Also Fr=2mgcosø. Therefore T=3mgccosø

When the rope breaks: 40.1N=3mgcosø....cosø=40.1N/(2.06kg*9.8m/s^2*3)=0.662

arccos(cosø)=ø=48.54°

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