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A 6 kg weight is fastened to the end of...

A `6 kg ` weight is fastened to the end of a steel wire of unstretched length `60 cm `. It is whirled in a vertical circle and has an angular velocity of ` 2 rev//s` at the bottom of the circle. The area of cross - section of the wire is `0.05 cm ^(2)` . Calculate the elongation of the wire when the weight is at the lowest point of the path . Young's modulus of steel `= 2xx10 ^(11) N //m^(2)` .

Text Solution

Verified by Experts

The correct Answer is:
2.87
`T- mg = m omega^(2)r`
`therefore T= mg + m omega^(2)r`
`T= 30 + 3 xx 16pi^(2)`
`=30 + 48 pi^(2)= 30 + 48 xx 3.14`
= 180.72N
`delta L= ("Stress" xx L)/(Y)= ((180.72)/(pi xx (10^(-3))^(2)) xx 1)/(20 xx 10^(10))`
`(180.72)/(3.14 xx 20 xx 10^(4))= 2.87 xx 10^(-4)m`
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