Question 12.1: A solid cast iron disk, 1 m in diameter and 0.2 m thick, is ...

\text { Given } D=1 m \quad t=0.2 m \quad n_{1}=350 rpm .

t = 1.5 s
Step I Energy absorbed by brake
The brake absorbs the kinetic energy of the rotating flywheel. The mass density of cast iron is taken as 7200 kg/m³. The radius of gyration of a solid disk about its axis of rotation is (d / \sqrt{8}) . Therefore,

m=\frac{\pi}{4}(1)^{2}(0.2)(7200)=1130.97 kg .

k^{2}=\frac{D^{2}}{8}=\frac{1}{8} m ^{2} .

\omega_{1}=\frac{2 \pi n_{1}}{60}=\frac{2 \pi(350)}{60}=36.65 rad / s .

\text { and } \quad \omega_{2}=0 .

E=\frac{1}{2} m k^{2}\left(\omega_{1}^{2}-\omega_{2}^{2}\right) .

=\frac{1}{2}(1130.97)\left(\frac{1}{8}\right)(36.65)^{2}=94946.52 J ( i ) .

Step II Torque capacity of brake
The average velocity during the braking period is

\left(\omega_{1}+\omega_{2}\right) / 2 \text { or }\left(\omega_{1} / 2\right) \text {. Therefore, } .

\theta=\left(\frac{\omega_{1}}{2}\right) t=\left(\frac{36.65}{2}\right)(1.5)=27.49 rad .

From Eq. (12.5),

E=M_{t} \theta             (12.5).

M_{t}=\frac{E}{\theta}=\frac{94946.52}{27.49}=3453.86 N – m         (ii).