A wire of length L and cross section area A is kept on a horizontal surface and one of its end is fixed at point 0. A ball of mass m is tied to its other end and the system is rotated with angular velocity omega.Show that increase in its length. Delta l = (m omega ^(2) L ^(2))/(AY). Y is young’s modulus.

A wire of length L and cross section area A is kept on a horizontal su

1.	A wire of length L and cross section area A is kept on a horizontal surface and one of its end is fixed at point 0. A ball of mass m is tied to its other end and the system is rotated with angular velocity omega.Show that increase in its length. Delta l = (m omega ^(2) L ^(2))/(AY). Y is young’s modulus.
Answer» Solution : By rotating the ball of MASS m with angular speed W, the pseudo centifugal force acting on it due to centripetal force `F = ( mv ^(2))/() = ( ML ^(2) omega ^(2))/( L )` `F = mL iomega^(2)` `therefore` ongitudinal stress `sigma = (F)/(A) = (mL omega ^(2))/(A) ` Let increase in length of rod `= Dela L` `therefore` strain `= (Delta L )/(L)` Young modulus `Y = ("stress")/("strain") = ( m Lomega ^(2))/(A) //(Delta L )/(L)` `therefore Delta L = (m L ^(2) omega ^(2))/(A.Y)`