Motion in a circle of helicopter rotor

In summary, a model of a helicopter rotor with four blades, each 3.20m in length, is rotated at 600rev/min in a wind tunnel. The linear speed of the blade is 32.0m/s and the radial acceleration is 320m/s^2. To express the radial acceleration as a multiple of g, divide by 9.8 m/s^2.
  • #1
Edwardo_Elric
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0

Homework Statement


A model of a helicopter rotor has four blades, each 3.20m in length from the central shaft to the blade tip. The model is rotated in a wind tunnel at 600rev/min. a.) What is the linear speed of the blade in m/s?
b.) What is the radial acceleration of the blade expressed as a multiple of the acceleration due to gravity g?

Homework Equations


[tex] a_{rad} = \frac{V^2}{R}[/tex]
[tex] a_{rad} = \frac{4{\pi}^2R}{T^2}[/tex]


The Attempt at a Solution


a.) convert: 600rev / min ( 1 min / 60secs) = 10 rev / s
Multiplied 3.20m by 10 rev / s = 32.0m/s?

b.) [tex] a_{rad} = \frac{V^2}{R}[/tex]
a_{rad} = (32.0m/s)^2 / (3.20m) = 320m/s^2

i don't understand the radial acceleration of blade expressed as a multiple of g...
is my answers correct?
 
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  • #2
Edwardo_Elric said:
a.) convert: 600rev / min ( 1 min / 60secs) = 10 rev / s
Multiplied 3.20m by 10 rev / s = 32.0m/s?
Careful. To convert between linear and angular speed, use [itex]v = \omega r[/itex], where [itex]\omega[/itex] is in radians/sec, not rev/s.

Alternatively, realize that 1 revolution covers 1 circumference, which equals [itex]2 \pi r[/itex].

b.) [tex] a_{rad} = \frac{V^2}{R}[/tex]
a_{rad} = (32.0m/s)^2 / (3.20m) = 320m/s^2
Redo this with the correct speed.

i don't understand the radial acceleration of blade expressed as a multiple of g...
Since g = 9.8 m/s^2, just divide your answer by that value to get an answer as a multiple of g.
 
  • #3


Your calculation for the linear speed of the blade is correct. To calculate the radial acceleration of the blade, you can use the equation a_{rad} = \frac{V^2}{R}, where V is the linear speed and R is the radius of the circle. In this case, the radius is 3.20m. Plugging in the value for V (32.0m/s), we get a_{rad} = \frac{(32.0m/s)^2}{3.20m} = 320m/s^2. This means that the radial acceleration of the blade is 320 times the acceleration due to gravity (g), which is approximately 9.8m/s^2. So, the radial acceleration of the blade is 320g. Your answer is correct.
 

1. What is the purpose of the helicopter rotor?

The helicopter rotor is responsible for generating lift and thrust, allowing the helicopter to fly and maneuver in the air.

2. How does the helicopter rotor create motion in a circular path?

The helicopter rotor blades are shaped like airfoils, similar to airplane wings. As the rotor spins, the airfoils create lift by pushing air downwards. This lift force causes the helicopter to move in a circular path.

3. What factors affect the motion of the helicopter rotor?

The speed and angle of the rotor blades, as well as the weight and balance of the helicopter, can affect the motion of the rotor. Other external factors such as wind and air density can also impact the motion.

4. How does the helicopter pilot control the motion of the rotor?

The pilot can adjust the speed and angle of the rotor blades using the helicopter's control system. This allows them to control the lift and thrust generated by the rotor, and ultimately control the direction and speed of the helicopter's motion.

5. Can the motion of the helicopter rotor be affected by changing the shape of the blades?

Yes, the shape and design of the rotor blades can greatly impact the performance and motion of the helicopter. Different shapes and designs can affect the lift and thrust generated, as well as the efficiency and stability of the rotor system.

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