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 verticalcircle 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.

A block of mass 1 kg is fastended to one end of a wire of cross-sectio

1.	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 verticalcircle 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» Solution : 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`