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A bob of mass 1kg is moving in a uniform circular path in a vertical plane of radius 1m. If it is whirled at...

A bob of mass 1kg is moving in a uniform circular path in a vertical plane of radius 1m. If it is whirled at a
frequency of 2 cycles per second, calculate:
a) The tension in the string when the bob is at the top part of the circle
b) Tension when it is at the bottom
c) At what position is the string likely to break? Why?
d) What is the minimum speed required to maintain the string under tension.

Answers

Moraa
a) The tension in the string when the bob is at the top part of the circle
T=mv2/r –mg= mr?2-mg=mr(2p/f)2-mg
= (1*1*4*p2/22)-(1*10)
= 147.95N
b) Tension when it is at the bottom
T= mr(2p/f)2+mg
= (1*1*4*p2/22) + (1*10)
= 167.95N
c) At what position is the string likely to break? Why?
At the bottom of the circle. The string experiences maximum tension at this point.
d) What is the minimum speed required to maintain the string under tension.
V=(rg)1/2=(1*10)1/2= 3.162m/s
Moraa orina answered the question on April 11, 2018 at 17:08

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