Revolutions 81906

How much energy was supplied by the flywheel of the device if its number of revolutions decreased from 1200 to 720 revolutions per minute if it has a moment of inertia of 71.3 kg/m²?

Correct answer:

W =  360295.8304 J

Step-by-step explanation:

$$\begin{gathered} r_1=1200\text{rpm} \\ {r}_2=720\text{rpm} \\ {\omega }_1=2\pi \cdot \frac{r_1}{60}=2\cdot 3.1416\cdot \frac{1200}{60}\doteq 125.6637\text{rad/s} \\ {\omega }_2=2\pi \cdot \frac{r_2}{60}=2\cdot 3.1416\cdot \frac{720}{60}\doteq 75.3982\text{rad/s} \\ I=71.3{\text{kg/m}}^2 \\ {E}_1=\frac{1}{2}\cdot I\cdot {\omega }_1^2=\frac{1}{2}\cdot 71.3\cdot 125.6637^2\doteq 562962.235\text{J} \\ {E}_2=\frac{1}{2}\cdot I\cdot {\omega }_2^2=\frac{1}{2}\cdot 71.3\cdot 75.3982^2\doteq 202666.4046\text{J} \\ W=E_1-E_2=562962.235-202666.4046=360295.8304\text{J}\doteq 3.603\cdot 10^5\text{J} \end{gathered}$$

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