Question

A steel cylinder of 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/m3
, calculate the torque required to
give an angular acceleration of 0.5rad/s2
.

Answer 1

Solution

Inner redius(b)=200 mm=0.2 m

Outer redius(a)=500 mm=0.5 m

Height of cylender(h)=900 mm=0.9 m

Dencity of cylender($\rho$ )=7800 kg/m^3

Moment of inertia can be

$I=\pi h\rho \frac{(a^4-b^4)}{2}$

So by putting the value

$I=3.14\times 7800\times0.9\times \frac{(0.5^4-0.2^4)}{2}$

$I=671.2 kg m^2$

Torque can be given as

$\tau=I\alpha$

$\tau=671.2\times0.5=335.6 Nm$