EXAMPLE 10.9Calculating Helicopter EnergiesA typical small rescue helicopter, similar to the one in Figure 10.18, has four blades, each is 4.00 m longand has a mass of 50.0 kg. The blades can be approximated as thin rods that rotate about one end of anaxis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. (a) Calculate therotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kineticenergy 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 liftit?Solution for (a)The rotational kinetic energy is1KErot = lu.10.73We must convert the angular velocity to radians per second and calculate the moment of inertia before wecan find KErot - The angular velocity w israd= 31.4300 rev 27 rad 1.00 minW =10.741.00 min1 rev60.0 sThe 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,(50.0 kg)(4.00 m)²I = 4-= 4 x3= 1067 kg · m?.10.753Entering w and I into the expression for rotational kinetic energy givesKErot = 0.5(1067 kg - m²)(31.4 rad/s)²10.76= 5.26 x 105 JSolution for (b)Translational kinetic energy was defined in Uniform Circular Motion and Gravitation. Entering the givenvalues of mass and velocity, we obtain1KEtrans =mv² =(0.5)(1000 kg)(20.0 m/s)² = 2.00 × 105 J.10.77To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy.This ratio is2.00 x 105 J= 0.380.10.785.26 x 105 JSolution for (c)At the maximum height, all rotational kinetic energy will have been converted to gravitational energy. Tofind this height, we equate those two energies:KErot = PEgrav10.79orlu = mgh.- Iw?10.80We now solve for h and substitute known values into the resulting equation5.26 x 105 Jh =mg= 53.7 m.10.81(1000 kg) (9.80 m/s²) 5. In Section 10.4 (Rotational Kinetic Energy) of the OpenStax College Physics textbook, go toExample 10.9 (Calculating Helicopter Energies) and change the mass of the helicopter bladesfrom 50.0 kg to Y kg. Also, change the rotation rate from 300 rpm to Z rpm. Then write theanswers (a), (b), and (c) to the question below.Y = 46:83Z = 294.83t boilgge

Part (5): Given:The mass of the helicopter is Y = 46.83 kg. The anguar speed is Z = 294.83 rpm. The…

5. In Section 10.4 (Rotational Kinetic Energy) of the OpenStax College Physics textbook, go to Example 10.9 (Calculating Helicopter Energies) and change the mass of the helicopter blades from 50.0 kg to Y kg. Also, change the rotation rate from 300 rpm to Z rpm. Then write the answers (a), (b), and (c) to the question below. Y = 46:83 Z = 294.83

5. In Section 10.4 (Rotational Kinetic Energy) of the OpenStax College Physics textbook, go to Example 10.9 (Calculating Helicopter Energies) and change the mass of the helicopter blades from 50.0 kg to Y kg. Also, change the rotation rate from 300 rpm to Z rpm. Then write the answers (a), (b), and (c) to the question below. Y = 46:83 Z = 294.83

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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Concept explainers

Angular speed, acceleration and displacement

Angular acceleration is defined as the rate of change in angular velocity with respect to time. It has both magnitude and direction. So, it is a vector quantity.

Angular Position

Before diving into angular position, one should understand the basics of position and its importance along with usage in day-to-day life. When one talks of position, it’s always relative with respect to some other object. For example, position of earth with respect to sun, position of school with respect to house, etc. Angular position is the rotational analogue of linear position.

Question

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?
Solution for (a)
The rotational kinetic energy is
1
KErot = lu.
10.73
We must convert the angular velocity to radians per second and calculate the moment of inertia before we
can find KErot - The angular velocity w is
rad
= 31.4
300 rev 27 rad 1.00 min
W =
10.74
1.00 min
1 rev
60.0 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,
(50.0 kg)(4.00 m)²
I = 4-
= 4 x
3
= 1067 kg · m?.
10.75
3
Entering w and I into the expression for rotational kinetic energy gives
KErot = 0.5(1067 kg - m²)(31.4 rad/s)²
10.76
= 5.26 x 105 J
Solution for (b)
Translational kinetic energy was defined in Uniform Circular Motion and Gravitation. Entering the given
values of mass and velocity, we obtain
1
KEtrans =
mv² =(0.5)(1000 kg)(20.0 m/s)² = 2.00 × 105 J.
10.77
To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy.
This ratio is
2.00 x 105 J
= 0.380.
10.78
5.26 x 105 J
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:
KErot = PEgrav
10.79
or
lu = mgh.
- Iw?
10.80
We now solve for h and substitute known values into the resulting equation
5.26 x 105 J
h =
mg
= 53.7 m.
10.81
(1000 kg) (9.80 m/s²)

5. In Section 10.4 (Rotational Kinetic Energy) of the OpenStax College Physics textbook, go to
Example 10.9 (Calculating Helicopter Energies) and change the mass of the helicopter blades
from 50.0 kg to Y kg. Also, change the rotation rate from 300 rpm to Z rpm. Then write the
answers (a), (b), and (c) to the question below.
Y = 46:83
Z = 294.83
t boilgge

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