Question 10.8: A flywheel of mass 18 kg and radius of gyration 250 mm is mo...

Angular speed of flywheel, ω = 120 rpm = (2π )(120)/60 rad/s = 4π rad/s.
If the flywheel is suddenly stopped, the change of kinetic energy of flywheel =(1 / 2) \bar{I} \omega^2 ; \text { where } \bar{I} is the mass moment of inertia of flywheel about axis of rotation:

\bar{I}=m \bar{K}^2

where \bar{K} is the centroidal radius of gyration of the flywheel. Therefore, change in rotational kinetic energy of the flywheel is

\frac{1}{2} m \bar{K}^2 \omega^2=\frac{1}{2}(18)(0.25)^2(4 \pi)^2 J =88.83  J

If the flywheel is symmetrically placed at the shaft middle point, assuming no energy loss, we can say that

2\left\lgroup\begin{array}{l} \text { Torsional strain energy } \\ \text { of one-half of shaft } \end{array} \right\rgroup =\text { Change in rotational kinetic energy of flywheel }

or            2 \frac{T^2 L}{4 G J}=88.83 \Rightarrow \frac{T^2 L}{G J}=177.66                 (1)

But τ = 16T/π d³ for solid shaft, therefore

T_{\max }=\frac{\pi d^3}{16} \tau_{\max }=\frac{\pi}{16}(0.05)^3 80\left(10^6\right)  N m

or              T_{\max }=1963.5  Nm               (2)

From Eqs. (1) and (2), we get

L_{\min }=\frac{(82)\left(10^9\right) \frac{\pi}{32}(0.05)^4(177.66)}{(1963.5)^2}=2.32  m

The required minimum shaft length is 2.32 m.