Question

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.​

Answer 1

$\huge {\pink{\underline {\underline {\bf {Question : }}}}}$

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 .

$\huge {\green{\underline {\underline {\bf {Answer : }}}}}$

(i) Tension at the bottom of the circle,

${\bf {T = \frac {mv^2}{r} + mg = \frac {1×(3.5)^2}{0.2} + 1 × 9.8 = 61 . 25 + 9.8 = 71.05 N }}$

The tension in the string is equal to the force F

So , F = 71 . 05 N , L = r = 0.2 m

The increase in length ,

$e = \frac {FL}{AY}$

${\bf{\implies e = \frac {71.05 × 0.2}{2× 10^-6 × 2 × 10^11} = 3.553 × 10^-5}}$

(ii) Tension on top of the circle ,

T = Tension at the bottom - 6mg

T = 71.05 - 6 × 1 × 9.8 = 71 .05 - 58.8 = 12.25 N

F = 12.25 ; L = 0.2 m

${\mathtt{\: increase\: in \:length\: (e) = \frac {FL}{AY}}}$

$= {\bf{ \frac {12.25 × 0.2 }{2× 10^-6 × 2 × 10 ^ 11} = 0.6125 × 10 ^ -5 m }}$