Centrifugal force developed due to rotation

P = w2 r M where M = mass of length ( ½ – x)

Now M = p x A x (1/2 – x)

and r = x + (1/2 – x) ½

p = pA w2 (1/2 – x) [ x + ( 1/2 – x) 1/2 ]

Applying formula δ =PL / AE , we have

dδ = p dx / AE

= P w^{2} / E (1/2 – X) {x + 1/4 – x /2} dx

= 2 pw^{2} / E (1/4 – x/2) ( x /2 + 1/4 ) dx

= 2 pw^{2} / E ( t^{2} / 16 – x^{2} / 4 ) dx

= 2 pw^{2} / E [ t^{2} / 16 x ½ – 2 pw^{2} / E x ¼ [ x^{3} / 3]_{ 0} ½

= 2 pw^{2} [ t^{3} / 32 – 2 pw^{2} / 4E [ t^{3} / 24]

= pw^{2} / E [ t^{3} / 16 – t^{3} / 48]

= pw^{2} / E [ 3t3 – t3 / 48] = p w2 t3 / 24 E

Total extension of the bar = 2δ = p w^{2} t^{3} / 12 E