Answer:
[tex]8.37*10^5[/tex] rpm
Explanation:
Given that rotational kinetic energy = [tex]4.66*10^9J[/tex]
Mass of the fly wheel (m) = 19.7 kg
Radius of the fly wheel (r) = 0.351 m
Moment of inertia (I) = [tex]\frac{1}{2} mr ^2[/tex]
Rotational K.E is illustrated as [tex](K.E)_{rt} = \frac{1}{2} I \omega^2[/tex]
[tex]\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }[/tex]
[tex]\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }[/tex]
[tex]\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }[/tex]
[tex]\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }[/tex]
[tex]\omega = 87636.04[/tex]
[tex]\omega = 8.76*10^4 rad/s[/tex]
Since 1 rpm = [tex]\frac{2 \pi}{60} rad/s[/tex]
[tex]\omega = 8.76*10^4(\frac{60}{2 \pi})[/tex]
[tex]\omega = 836518.38[/tex]
[tex]\omega = 8.37 *10^5 rpm[/tex]
