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A rod of mass M, length l and crosssectional area A, made of material of young's modulus Y is rotated about its one end in horizontal plane with constant angular speed \omega.
Its extension is ______
 

Option: 1

\frac{M \omega^2 l^2}{3 A Y}


Option: 2

\frac{M \omega^2 l^2}{A Y}


Option: 3

\frac{M \omega^2 l^3}{A Y}


Option: 4

\frac{M \omega^2 l^2}{4 A Y}


Answers (1)

best_answer

For the element of mass dM

\begin{aligned} -(d F) & =(d M) \omega^2 x \\ F & =-\int_l^x\left(\frac{M}{l} d x\right) \omega^2 x \\ F & =\frac{M}{2 l} \omega^2\left(l^2-x^2\right) \end{aligned}

Now, using d l=\frac{F(d x)}{A Y}
\begin{aligned} & d L=\frac{M \omega^2}{2 l} \frac{\left(l^2-x^2\right)}{A Y} d x \\ & \Delta l=\frac{M \omega^2}{2Y A l} \int_0^l\left(l^2-x^2\right) d x \\ & \Delta l=\frac{M \omega^2 l^2}{3 A Y} \end{aligned}

Posted by

manish painkra

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