A mass of 15 kg is tied at the end of a steel wire of the length lm. It is whirled in a vertical plane with angular velocity 1 rad/s. Cross sectional area of the wire is 0.06 cm^(2). Calculate the elongation of the wire when the mass is at its lowest position. Y_("steel”) = 2 xx 10 ^(11) Nm ^(-2)

A mass of 15 kg is tied at the end of a steel wire of the length lm. I

1.	A mass of 15 kg is tied at the end of a steel wire of the length lm. It is whirled in a vertical plane with angular velocity 1 rad/s. Cross sectional area of the wire is 0.06 cm^(2). Calculate the elongation of the wire when the mass is at its lowest position. Y_("steel”) = 2 xx 10 ^(11) Nm ^(-2)
Answer» Solution :Here `l = 1m ` `m = 15 kg ` `omega = 1 rad s ^(-1)` `A = 0.06 CM ^(2)` `= 6 xx 10 ^(-6) m ^(2)` `Y_("steel") = 2 xx 10 ^(11) Nm ^(-2)` `g =9.8 ms ^(-2)` centrifugal force in circular motion F = centrepetal force `=(mv ^(2))/(l) = m l omega ^(2) [ because V = r omega ]` Atthe lowest position total force acting on the body is the sum of gravitational force and centrifugal force `((mv ^(2))/(r ))` `= mg + m l omega ^(2)` `= m ( g + l omega ^(2))` `= 15 (9.8 + 1 xx (1) ^(2))=15 xx 10 .8 = 162 N` Now stress`sigma = (F)/(A) = (162)/( 6 xx 10 ^(-6)) = 27 xx 10 ^(6) Nm ^(-2)` Now Young.s MODULUS `Y = (sigma )/( epsi _(l)) = (sigma _(l))/( (Delta l )/( l )) = (sigma _(1) l )/(Delta l ) ` `therefore Delta l = (sigma _(1) l)/( Y) = ( 27 xx 10 ^(6) xx 1 )/(2 xx 10 ^(11))=13.5 xx 10 %(-5) m` `therefore Delta l =0.135 mm`