### Video Transcript

A small helicopter is propelled
horizontally at a speed of 20.0 metres per second by the rotation at 300 rpm of four
rotor blades. Each blade is 50.0 kilograms in
mass, 4.00 metres in length, and can be modelled as a thin rod that is joined at one
end to an axis of rotation that is perpendicular to its length. The mass of the helicopter
including its rotor blades is 1000 kilograms. Find the rotational kinetic energy
of the blades. Find the ratio of the rotational
kinetic energy of the blades to the translational kinetic energy of the
helicopter.

In this two-part exercise, we first
want to solve for the rotational kinetic energy of the blades of the helicopter and
then we want to solve for the ratio of that rotational kinetic energy to the
translational kinetic energy of the helicopter overall. Let’s start out with a diagram of
the helicopter in flight.

If we consider a top down view of
the helicopter while it’s flying, we know that its main rotor is spinning with an
angular speed — we’ve called capital Ω — given as 300 revolutions per minute. And while this is happening, the
helicopter is translating along forward at a linear speed — we’ve called 𝑣 — given
as 20.0 metres per second. We’re also told that each of the
four blades in the main rotor has a length of 4.00 metres and a mass of 50.0
kilograms.

Knowing all this, we want to
calculate the rotational kinetic energy of the blades in the rotor. We can recall that an object’s
rotational kinetic energy is equal to one-half its moment of inertia times its
angular speed in radians per second squared. The statement tells us that each
one of the blades in the main rotor can be modelled as a thin rod rotating
perpendicularly about one of its ends.

When we look up the moment of
inertia of an object like this rotating about an axis like this, we find that it’s
equal to one-third the object’s mass times its length squared. This all means that the rotational
kinetic energy of one of the four rotor blades is equal to one-half times a third
times the mass of the blade times its length squared times its angular speed 𝜔
squared.

We aren’t given that angular speed
𝜔 in units of radians per second. But we are given that speed in
units of revolutions per minute. To convert 300 revolutions per
minute into units of radians per second, we can first multiply this fraction by one
minute every 60 seconds, which will cancel out the units of minutes and give us a
time unit of seconds. Then since there are two 𝜋 radians
in every one complete circular revolution, we multiply by this ratio which leads us
to cancelling out the units for revolutions.

We find then that the angular speed
of these rotor blades in units of radians per second is 300 times two 𝜋 over
60. We now have an expression for the
rotational kinetic energy of one of our rotor blades in terms of values that are all
known. In order to find the total
rotational kinetic energy of the entire main rotor, we multiply this term by four
which creates a leading factor of two in place of our one-half previously.

We’re now ready to plug in for 𝑚,
𝐿, and 𝜔 to solve for KE sub 𝑟. When we do and then enter this
expression on our calculator, we find that to three significant figures KE sub 𝑟 is
5.26 times 10 to the fifth joules. That’s the rotational kinetic
energy of the blades in the main rotor of this helicopter.

Next, we want to solve for the
ratio of the rotational to translational kinetic energy of this helicopter. Recalling that translational
kinetic energy is one-half an object’s mass times its speed squared, we can say that
the translational kinetic energy of this helicopter as it flies is one-half the mass
of the helicopter 𝑚 sub ℎ times its speed 𝑣 squared.

In the problem statement, we’re
told that the overall mass of the helicopter 𝑚 sub ℎ is 1000 kilograms. Since we already know 𝑣, the
linear speed of the helicopter, we’re ready to plug in and solve for KE sub 𝑡. When we plug in for 𝑚 sub ℎ and 𝑣
and enter this expression on our calculator, we find a result for the overall
translational kinetic energy of the helicopter. We divide this result into the
rotational kinetic energy of the blades in the helicopter’s rotor to solve for the
ratio we’re interested in. To three significant figures, this
ratio is 2.63. So the blades of the main rotor of
the helicopter have 2.63 times as much kinetic energy rotationally as the helicopter
has overall translationally.