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A 14.5 kg mass, fastened to the end of a...

A 14.5 kg mass, fastened to the end of a steel wire of unstretched 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross- sectional area of the wire is `0.065" cm"^(2)`. Calculate the elongation of the wire when the mass is at the lowest point of its path.

Text Solution

Verified by Experts

The cross - sectional area of each femurs
`a=10cm^(2)`
`therefore` Total cross -section area of two femurs
`A=2xx10cm^(2)`
`=2xx10^(4)m^(2)`
The mass of the upper part of a human body `m=40kg`
Acceleration of gravity `g=10ms^(-2)`
`therefore` The force acting on the upper part of a human body ,
F=mg
`=40xx10`
`thereforeF=400N` The force is acting vertically down
`therefore` Average pressure
`ltPgt=(F)/(A)=(400)/(20xx10^(-4))`
`=20xx10^(4)=2xx10^(5)Nm^(-2)`
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A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The crosssectional area of the wire is 0.065 cm. Calculate the elongation of the wire when the mass is at the lowest point of its path. [Y_("Steel") =2 xx 10 ^(11) N,m ^(-2)]

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Knowledge Check

  • The adjacent graph shows the extension (DeltaI) of a wire of length 1 m suspended from the top of a roof at one end with a load W connected to the other end. If the cross-sectional area of the wire is 10^(-6) m ^(2), calculate The Young's modulus of the material of the wire is ........

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    ` 2 xx 10 ^(11) Nm^(-2)`
    B
    `5 xx 10 ^(6) Nm ^(-2)`
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    D
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