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A steel uniform rod of length 2L cross sectional area A and mass Mis set rotating in a horizontal plane about an axis passing through the centre. If Y is the Young's modulus for steel, find the extension in the length of the rod. |
Answer» Solution : Let us CONSIDER an element of width dr at a distance r from the given AXIS of rotation as showin in the diagram. Suppose tention in the rod at r and dr distance are `T(r ) and T (r + dr)` Centripetal force acting on small length element `F = dm r omega ^(2) ""[because F _(C) = mr omega ^(2) ]` `therefore F = mu r omega ^(2) dr ""...(1)` `[because mu = (dm )/( dr) implies dm = mu dr ]` `therefore T (r) -T ( r + dr) =mu r omega ^(2) dr ` `[ because T(r) -T (r + dr) =` resultant force F] `therefore -dT = mu r omega ^(2) dr` Centripetal force and tension force are opposite to each other hence negative SIGN present. Integrating on both side. `int _(T=T) ^(T=0) dT = mu omega ^(2) int _(r ) ^(l) r dr` ` -[T] _(T) ^(0) = mu omega ^(2) [ (r ^(2))/( 2 )] _(r ) ^(l)` `therefore -[0-T] = (mu omega ^(2))/( 2) [ l ^(2) - r ^(2) ]` `therefore T = ( mu omega ^(2))/( 2) (l ^(2) -r ^(2)) ""...(2)` Now if `Delta r` is the extension in the elemente of length dr `therefore` Young.s modulus `Y = (T (r))/(A) xx (dr )/(Delta r)` `therefore (Delta r )/( dr ) = (T (r))/(YA) = (T)/(YA)` `therefore (Delta r )/(dr) = (mu omega ^(2))/( 2 YA) (l ^(2) - r ^(2))` `therefore Delta r = ( mu omega ^(2))/( 2 YA) int (l ^(2) -r ^(2)) dr` Integrating on both side, `int Delta r = (mu omega ^(2))/( 2 YA) int ( l ^(2) -r ^(2)) dr` `therefore Delta r = (mu omega ^(2))/( 2 YA) [ l ^(2) r - (r ^(3))/(3)]_(0) ^(l)` `therefore Delta r = (mu omega ^(2))/( 2 YA) [[ l ^(3) - (l ^(3))/(3)]` `therefore Deltar= (mu omega ^(2))/(2YA)xx(2L^(3))/(3)``therefore Delta r = ( mu omega ^(2) l ^(3))/( 3YA)` `therefore ` Extension of `Delta r` in both side length, then tota extension `=2 Delta r` `= ( 2 mu omega ^(2) l ^(3))/(3YA)` |
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