Question

what will be the breaking strength of the material if a rod of density 'p' and length unity which is rotating about a vertical axis passing through its centre , ruptures at 10 rps.

Answer 1

Dear Student,
Breaking strength = F/A
F = force due to rotation
A = Area of cross section of Rod
dm = small mass taken

$$\begin{gathered} l=1 \\ \omega =2\pi \times 10 \\ \frac{F}{A}=? \\ dm=\rho Ax \\ F={\int }_0^{\frac{l}{2}}m{\omega }^{2}dx \\ F=\rho A{\omega }^2{x}^2÷2 \\ F=\rho A{\omega }^{2}l÷8 \\ \frac{F}{A}=\rho {\omega }^2\frac{l}{8} \\ \frac{F}{A}=\rho \times 20\times 20\times 1÷8=50\rho Ans \end{gathered}$$​​
Regards