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

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Chapter12: Rotation I: Kinematics And Dynamics
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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)
Transcribed Image Text: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
Transcribed Image Text: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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