Question

A steel cylinder of 500mm outside diameter and 200mm inside diameter is set in rotation about its axis. If the cylinder is 900mm long, of density 7800kg/m^3, calculate the torque required to give an angular acceleration of 0.5rad/s^2.

Answer 1

Since the angular acceleration is equal to the ratio of torque to moment of inertia, then $\varepsilon =\frac {M} {J}$

$M =\varepsilon\times J$ , since $J=\frac{m\times R^2} {2}$ , a $m=\rho\times V$ , then

$M=\varepsilon\times\frac{\rho\times V \times R^2}{2}$ $=0.5\times\frac{7800\times(900\times\frac{\pi}{4}(0.5^2-0.2^2))\times (0.5\times 0.5)^2}{2}\approx18091.16$ N$\times$m.