Rotational Kinetic Energy

Example 10.9 Calculating Helicopter Energies

A typical small rescue helicopter, similar to the one in Figure 10.18, has four blades, each is 4.00 m long and has a mass of 50.0 kg. The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length.

The helicopter has a total loaded mass of 1000 kg. (a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter when it flies at 20.0 m/s, and compare it with the rotational energy in the blades. (c) To what height could the helicopter be raised if all of the rotational kinetic energy could be used to lift it?

Strategy

Rotational and translational kinetic energies can be calculated from their definitions. The last part of the problem relates to the idea that energy can change form, in this case from rotational kinetic energy to gravitational potential energy.

Solution for (a)

The rotational kinetic energy is

$\mathrm{KE}_{\mathrm{rot}}=\frac{1}{2} I \omega^{2}$.

We must convert the angular velocity to radians per second and calculate the moment of inertia before we can find $\mathrm{KE}_{\text {rot }}$. The angular velocity$\omega$ is

$\omega=\frac{300 \mathrm{rev}}{1.00 \mathrm{~min}} \cdot \frac{2 \pi \mathrm{rad}}{1 \mathrm{rev}} \cdot \frac{1.00 \mathrm{~min}}{60.0 \mathrm{~s}}=31.4 \frac{\mathrm{rad}}{\mathrm{s}}$

The moment of inertia of one blade will be that of a thin rod rotated about its end, found in Figure 10.12. The total $I$ is four times this moment of inertia, because there are four blades. Thus,

$I=4 \frac{M \ell^{2}}{3}=4 \times \frac{(50.0 \mathrm{~kg})(4.00 \mathrm{~m})^{2}}{3}=1067 \mathrm{~kg} \cdot \mathrm{m}^{2}$.

Entering $\omega$ and $I$ into the expression for rotational kinetic energy gives

$\begin{aligned} \mathrm{KE}_{\mathrm{rot}} &=0.5\left(1067 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(31.4 \mathrm{rad} / \mathrm{s})^{2} \\ &=5.26 \times 10^{5} \mathrm{~J} \end{aligned}$

Solution for (b)

Translational kinetic energy was defined in Uniform Circular Motion and Gravitation. Entering the given values of mass and velocity, we obtain

$\mathrm{KE}_{\text {trans }}=\frac{1}{2} m v^{2}=(0.5)(1000 \mathrm{~kg})(20.0 \mathrm{~m} / \mathrm{s})^{2}=2.00 \times 10^{5} \mathrm{~J}$.

To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy. This ratio is

$\frac{2.00 \times 10^{5} \mathrm{~J}}{5.26 \times 10^{5} \mathrm{~J}}=0.380$.

Solution for (c)

At the maximum height, all rotational kinetic energy will have been converted to gravitational energy. To find this height, we equate those two energies:

$\mathrm{KE}_{\mathrm{rot}}=\mathrm{PE}_{\text {grav }}$.

or

$\frac{1}{2} I \omega^{2}=m g h$.

We now solve for $h$ and substitute known values into the resulting equation

$h=\frac{\frac{1}{2} I \omega^{2}}{m g}=\frac{5.26 \times 10^{5} \mathrm{~J}}{(1000 \mathrm{~kg})\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)}=53.7 \mathrm{~m}$.

Discussion

The ratio of translational energy to rotational kinetic energy is only 0.380. This ratio tells us that most of the kinetic energy of the helicopter is in its spinning blades – something you probably would not suspect. The 53.7 m height to which the helicopter could be raised with the rotational kinetic energy is also impressive, again emphasizing the amount of rotational kinetic energy in the blades.

Figure 10.18 The first image shows how helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades. The second image shows a helicopter from the Auckland Westpac Rescue Helicopter Service. Over 50,000 lives have been saved since its operations beginning in 1973. Here, a water rescue operation is shown.