A flywheel has a radius of $3 \mathrm{~m}$ and a mass of $6000 \mathrm{~kg}$. This mass is mainly distributed along the circumference of the rim. For approximate calculations, we will neglect the moment of inertia of the spokes and the shaft. The rotational speed of the wheel is $n=800$ revolutions per minute. A driving engine can bring this wheel from rest to the full rotational speed in 3 minutes of operation. What is the horsepower of the driving engine (its power), assuming it does no other work during this time?
The work of the machine for 3 minutes is equal to the kinetic energy of the wheel. In this case, this energy can be calculated without knowing the moment of inertia, because the approximate speed of each point of the rotating mass is $v=\frac{2 \cdot 3 \cdot \pi \cdot 800}{60} \frac{\mathrm{m}}{\mathrm{sec}}$.

If we want to express the kinetic energy in ergs, we need to convert the speed to $\frac{\mathrm{cm}}{\sec }$ and the mass to grams. The energy thus obtained should then be divided by $98,1 \cdot 10^{6}$ to get the energy in kilogram meters. However, without this conversion, we can immediately express the energy in kgm (i.e., practical units) if we leave the speed in $\frac{\mathrm{m}}{\sec }$ and express the mass in practical units as well, which is done by dividing the mass in kilograms by the acceleration due to gravity, $9,81-$ times. The energy is therefore

$$
E=\frac{1}{2} \cdot \frac{6000}{9,81} \cdot\left(\frac{2 \cdot 3 \cdot \pi \cdot 800}{60}\right)^{2} \mathrm{kgm}
$$

Since this energy was created by the machine in 3 minutes $=3 \cdot 60 \mathrm{sec}$, the power of the machine is

$$
\begin{gathered}
H=\frac{1}{2} \cdot \frac{6000}{9,81} \cdot\left(\frac{2 \cdot 3 \cdot \pi \cdot 800}{60}\right)^{2} \frac{1}{3 \cdot 60 \cdot 75} \mathrm{HP} \\
H=1431 \mathrm{HP}
\end{gathered}
$$[^0]


[^0]:    ${ }^{1}$ We would like to alert our readers who deal with solving such problems to this note. Whenever there is a calculation of kinetic energy, it is always necessary to think in advance in which units we want to obtain it. In practical life, kgm is most commonly used. Therefore, in the expression $\frac{1}{2} m v^{2}$, the mass and the speed should also be expressed in practical units. The practical unit of mass is $\frac{\mathrm{kg} \text { weight }}{9,81 \cdot \frac{\mathrm{m}}{\mathrm{sec}}^{2}}$, and the speed is $\frac{\mathrm{m}}{\mathrm{sec}}$.