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)`

Explanation

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`