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A 14.5kg mass, fastened to the end of a steel wire of unstretched length 1.0m, is whirled in a vertical circle with an angular velocity of 2rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.


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Solution

Step 1: Given Data

Mass of the body m=14.5kg

Length of the steel wire l=1.0m

Angular velocity ω=2rev/s

The cross-sectional area of the wire A=0.065cm2

We know that Young's modulus Ysteel=2×1011Nm-2

Step 2: Calculate the Force

The wire is fastened to a given mass and is rotating with the given angular velocity.

The forces acting on the wire are the weight of the mass and the centrifugal force due to the circular motion with a radius l.

Therefore, the total force assuming g=9.8m/s2 is given as,

F=mg+mlω2

=14.59.8+1×2×2π2

=14.5×169.8

=2462N

Step 3: Calculate the Elongation

We know that Young's modulus is given by

Y=StressStrain

where Stress=FA, where F is the force,

Strain=ll

Y=FAll

Y=FlAl

l=FlAY

Upon substituting the values we get,

l=2462×10.065×10-4×2×1011

=246265×2×104

=18.9×10-4m

=1.89mm

Hence, the elongation of the wire when the mass is at the lowest point of its path is 1.89mm.


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