Question

A thin rim with a mean diameter of 1.2 m and cross section of 15 mm ×200mm rotates at 800 r/min. Calculate change in diameter due to rotation if E=200GPa and density is 7800kg/m^3

Answer 1

Radius, r $=\frac{D}{2}=\frac{1.2}{2}=0.6 m$

Area, A$= 15 \times 2000=3000 mm^2= \frac{3000}{1000000}=0.003 m^2$

$\omega= \frac{2 \pi N}{60}=\frac{2 \times \pi \times800}{60}=83.775 rad/s$

Velocity $,v=r\omega=0.6 \times 83.775 =50.2655 m/s$

Magnitude of force, F=mv=$\rho A v^2=7800 \times 0.003 \times (50.2655)^2=59122.919 N$

Stress, $\sigma =\frac{F}{A}=\frac{59122.919}{0.003}=19.708 MPa$

Change in diameter , $\triangle d=\frac{Fd}{AE}=\frac{59122.919 \times1.2}{0.003 \times 200 \times 10^9}=0.000118245=0.118245 mm$