A solid cast iron disk, 1 m in diameter and 0.2 m thick, is used as a flywheel. It is rotating at 350 rpm. It is brought to rest in 1.5 s by means of a brake. Calculate
(i) the energy absorbed by the brake; and
(ii) the torque capacity of the brake.

Question

Accepted Answer

[latex] ext { Given } D=1 m \quad t=0.2 m \quad n_{1}=350 rpm [/latex].

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 [latex] (d / \sqrt{8}) [/latex]. Therefore,

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

[latex] k^{2}=\frac{D^{2}}{8}=\frac{1}{8} m ^{2} [/latex].

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

[latex] ext { and } \quad \omega_{2}=0 [/latex].

[latex] E=\frac{1}{2} m k^{2}\left(\omega_{1}^{2}-\omega_{2}^{2} ight) [/latex].

[latex] =\frac{1}{2}(1130.97)\left(\frac{1}{8} ight)(36.65)^{2}=94946.52 J ( i ) [/latex].

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

[latex] \left(\omega_{1}+\omega_{2} ight) / 2 ext { or }\left(\omega_{1} / 2 ight) ext {. Therefore, } [/latex].

[latex] heta=\left(\frac{\omega_{1}}{2} ight) t=\left(\frac{36.65}{2} ight)(1.5)=27.49 rad [/latex].

From Eq. (12.5),

[latex] E=M_{t} heta [/latex] (12.5).

[latex] M_{t}=\frac{E}{ heta}=\frac{94946.52}{27.49}=3453.86 N - m [/latex] (ii).

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

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