Answer to Question #142437 in Mechanics | Relativity for Tafadzwa Mubariri

Question

A flywheel, together with its shaft , has a total mass of 300kg, and its radius of gyration is 900mm. If the effect of bearing friction is equivalent to a couple of 70Nm, calculate the breaking torque required to bring the flywheel to rest from a speed of 12rev/s in 8s.

Answer 1

Total mass $M=300kg$

Radius of gyration $k=900mm$

Coupled form by friction of bearing $C=70Nm$

Moment of inertia $I=MK^2$

$I=300/times (900)^2\\=243kgm^2$

Speed $\omega=12rev/s$

Work done by shaft $W=\dfrac{1}{2}I\omega^2$

$\dfrac{1}{2}\times 243\times (12)^2\\=243\times 72=17496Nm$

Hence the breaking torques required to brake the shaft=$17496-70=17426Nm$