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A uniform metal rod of length 2L, cross sectional area A and mass M is set rotating with an angular speed ω in a horizontal plane. The axis of rotation passes through centre of the rod and normal to its length. If Y is young’s modulus of material of the rod, the extension in the length of the rod will be

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a

ML2ω26AY

b

ML2w23AY

c

ML2ω22AY

d

2ML2ω23AY

answer is B.

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Detailed Solution

Centripetal force on the element is dF=(M2L)xω2dx Tension on the half length of the rod at a distance x from the axis of rotation is F=∫xLdF=Mω22L∫xLxdx =Mω24L(L2−x2) If de is extension of element dx dx=FdxAY e=∫0Lde=∫0LFAdx=Mω24LYA∫0L(L2−x2)dx = Mω24LAY[L3−L33]=Mω2L26AY Total extension = 2e=Mω2L23AY

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