Question

A thin uniform metallic rod of mass $M$ and length $L$ is rotated with a angular velocity $\omega$ in a horizontal plane about a vertical axis passing through one of its ends. The tension in the middle of the rod is :

• A. $\frac{1}{2}ML{\omega }^2$
• B. $\frac{1}{4}ML{\omega }^2$
• C. $\frac{1}{8}ML{\omega }^2$
• D. $\frac{3}{8}ML{\omega }^2$

Answer 1

The correct option is D $\frac{3}{8}ML{\omega }^2$

$T\left(x\right)-T(x+dx)=\frac{M}{L}{w}^{2}xdx$ (by taylors series formula)

$-dT=\frac{M}{L}{w}^{2}xdx$ (formulating the differential equation)

${\int }_{T^x}^{T^0}dT=-{\int }_x^0\frac{M}{L}{w}^{2}xdx$

$T_{\left(x\right)}-T_{\left(0\right)}=-[\frac{M}{2L}{w}^2{x}^2{]}_0^x$

$Now,whenx=L,T\left(L\right)=0$

$Thus,T\left(0\right)=\frac{M}{2L}{w}^2{L}^2$

$Hence,T\left(x\right)=\frac{M}{2L}{w}^2(L^2-x^2)$

$\therefore T\left(\frac{L}{2}\right)=\frac{M}{2L}{w}^2(L^2-\frac{L^2}{4})$

$Hence,T\left(\frac{L}{2}\right)=\frac{3}{8}ML{w}^2$