Wouldn't mass stay the same and acceleration increase or am I mistaken?
Answer:
rpm
Explanation:
Given that rotational kinetic energy = ![4.66*10^9J](https://tex.z-dn.net/?f=4.66%2A10%5E9J)
Mass of the fly wheel (m) = 19.7 kg
Radius of the fly wheel (r) = 0.351 m
Moment of inertia (I) = ![\frac{1}{2} mr ^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20mr%20%5E2)
Rotational K.E is illustrated as ![(K.E)_{rt} = \frac{1}{2} I \omega^2](https://tex.z-dn.net/?f=%28K.E%29_%7Brt%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20I%20%5Comega%5E2)
![\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%5Cfrac%7B2%28K.E%29_%7Brt%7D%7D%7BI%7D%20%7D)
![\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%5Cfrac%7B2%28KE%29_%7Brt%7D%7D%7B1%2F2%20mr%5E2%7D%20%7D)
![\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%5Cfrac%7B4%28K.E%29_%7Brt%7D%7D%7Bmr%5E2%7D%20%7D)
![\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%5Cfrac%7B4%2A4.66%2A10%5E9J%7D%7B19.7kg%2A%280.351%29%5E2%7D%20%7D)
![\omega = 87636.04](https://tex.z-dn.net/?f=%5Comega%20%3D%2087636.04)
Since 1 rpm = ![\frac{2 \pi}{60} rad/s](https://tex.z-dn.net/?f=%5Cfrac%7B2%20%5Cpi%7D%7B60%7D%20%20rad%2Fs)
![\omega = 8.76*10^4(\frac{60}{2 \pi})](https://tex.z-dn.net/?f=%5Comega%20%3D%208.76%2A10%5E4%28%5Cfrac%7B60%7D%7B2%20%5Cpi%7D%29)
![\omega = 836518.38](https://tex.z-dn.net/?f=%5Comega%20%3D%20836518.38)
![\omega = 8.37 *10^5 rpm](https://tex.z-dn.net/?f=%5Comega%20%3D%208.37%20%2A10%5E5%20rpm)