A block of mass 1 kg is fastended to one end of a wire of cross-sectional area 2 `mm^(2)` and is rotated in a vertical circle of radius 20 cm. The speed of the block at the bottom of the circle is 3.5 `ms^(-1)`. Find the elongation of the wire when the block is at the top of the circle.

Explanation

i) Tension at the bottom of the circle,
`T = (mv^2)/(r ) + mg = (1 xx (3.5)^2)/(0.2) + 1 xx 9.8`
` = 61.25 + 9.8 = 71.05 N`
This tension in the string is equal to the force, F i.e., F=71.05N, L =r=0.2m.
The increase in length,
` e = (FL)/(AY) = (71.05 xx 0.2)/(2 xx 10^(-6) xx 2 xx 10^11)`
` = 3.553 xx 10^(-5) m`
ii) Tension at the top of the circle, T = Tension at the bottom - 6 mg
`= 71.05 - 6xx 1 xx 9.8 = 71.05 - 58.8 = 12.25 N.`
F= 12.25 N, L = 0.2 m.
The increase in length ` = (FL)/(AY)`
` = (12.25 xx 0.2)/(2 xx 10^(-6) xx 2 xx 10^11)`
` = 0.6125 xx 10^(-5) m`