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The diameters of the main rotor and tail rotor of a single-engine helicopter are \(7.60 \mathrm{~m}\) and \(1.02 \mathrm{~m}\), respectively. The respective rotational speeds are \(450 \mathrm{rev} / \mathrm{min}\) and \(4138 \mathrm{rev} / \mathrm{min}\). Calculate the speeds of the tips of both rotors. Compare these speeds with the speed of sound, \(343 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
The speed of the main rotor's tip is \(179.26 \, \mathrm{m/s}\) and the tail rotor's tip is \(220.59 \, \mathrm{m/s}\). Both these speeds are lower than the speed of sound, which is \(343 \, \mathrm{m/s}\)

Step by step solution

01

Conversion of Rotational Speed

Convert the rotational speeds from revolutions per minute to radians per second. This can be done using the fact that 1 revolution is \(2\pi\) radians and 1 minute is 60 seconds. The conversions are: For main rotor, \(\omega_1 = 450 \mathrm{rev/min} \cdot \frac {2\pi \mathrm{rad}} {1 \mathrm{rev}} \cdot \frac {1\mathrm{min}} {60 \mathrm{s}} = 47.12 \, \mathrm{rad/s}\) For tail rotor, \(\omega_2 = 4138 \mathrm{rev/min} \cdot \frac {2\pi \mathrm{rad}} {1 \mathrm{rev}} \cdot \frac {1 \mathrm{min}} {60 \mathrm{s}} = 433.11 \, \mathrm{rad/s}\)
02

Calculation of Linear Speeds

Apply the formula of linear velocity, \(v = r \cdot \omega\), to calculate the speeds of the tips of both rotors. Remember to convert the diameter to radius by halving it. For main rotor, \(v_1 = r_1 \cdot \omega_1\), substituting the values we have \(v_1 = \frac{7.60 m}{2} \times 47.12 \mathrm{rad/s} = 179.26 \, \mathrm{m/s}\) For tail rotor, \(v_2 = r_2 \cdot \omega_2\), substituting the values we have \(v_2 = \frac{1.02 m}{2} \times 433.11 \mathrm{rad/s} = 220.59 \, \mathrm{m/s}\)
03

Comparision with Speed of Sound

Compare the calculated speeds with the speed of sound (343 m/s). The speeds of the main rotor and tail rotor tips are lower than the speed of sound.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Helicopter Physics
Understanding the physics behind a helicopter is crucial in grasping how these incredible machines operate. A helicopter utilizes rotary wings or rotors to provide lift, which allows it to hover, take off, and land vertically. The main rotor, typically the larger one, is responsible for lifting the helicopter vertically by generating a low-pressure region above it. The smaller tail rotor counteracts the torque effect created by the main rotor's spin, allowing for directional control.

Both these rotors spin at significant speeds; their rotational speeds are measured in revolutions per minute (rev/min). While the main rotor generates lift, the tail rotor provides stability, making it an indispensable part of helicopter design. Without it, the helicopter would uncontrollably spin in the opposite direction of the main rotor. Understanding the dynamics and mechanics of these rotors helps explain how helicopters can perform their unique aerial maneuvers.
Speed of Sound Comparison
When we talk about the speed of sound, we refer to how quickly sound waves travel through a medium, typically air, at standard atmospheric conditions. This speed is approximately 343 meters per second (m/s) at sea level and room temperature (20°C or 68°F). The speed of sound changes with the medium and temperature; it increases in warmer air and decreases in colder conditions due to variations in air density.

Comparing rotor speeds with the speed of sound provides insight into a helicopter's performance and safety. For instance, if the tip speed of a helicopter's rotor exceeds the speed of sound, it can create supersonic shock waves. This phenomenon can cause excessive noise, increased aerodynamic drag, and potential structural damage to the rotor blades. In our example exercise, both the main and tail rotor speeds are below the speed of sound, which keeps the helicopter in a safe and efficient operational range.
Linear Velocity Calculation
Linear velocity refers to the speed at which a point on a rotating object travels along a straight path. In the context of helicopter rotors, this is the speed of the rotor tips as they cut through the air. Calculating this involves determining the product of the rotor's angular velocity (in radians per second) and the radius of the rotor.

To find the linear velocity, you need to first convert the rotational speed from revolutions per minute (rev/min) to radians per second (rad/s) using the conversion factor \(1 \, \text{rev} = 2\pi \, \text{radians}\) and \(1 \, \text{minute} = 60 \, \text{seconds}\). Once you have the rotor's angular velocity, you calculate the linear velocity using \(v = r \cdot \omega\), where \(v\) is the linear velocity, \(r\) is the radius of the rotor, and \(\omega\) is the angular velocity.

In the solved example, the main rotor's tips travel at approximately 179.26 m/s, and the tail rotor's tips reach 220.59 m/s, illustrating the high-speed mechanics involved in helicopter flight. Converting and understanding these values not only aids in mechanical calculations but also enhances safety and performance assessments.

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Most popular questions from this chapter

An air puck of mass \(m_{1}=0.25 \mathrm{~kg}\) is tied to a string and allowed to revolve in a circle of radius \(R=1.0 \mathrm{~m}\) on a frictionless horizontal table. The other end of the string passes through a hole in the center of the table, and a mass of \(m_{2}=1.0 \mathrm{~kg}\) is tied to it (Fig. P7.27). The suspended mass remains in equilibrium while the puck on the tabletop revolves. (a) What is the tension in the string? (b) What is the horizontal force acting on the puck? (c) What is the speed of the puck? the horizontal force acting on the puck? (c) What is the the horizontal force acting on the puck? (c) What is the speed of the puck?

A stuntman whose mass is \(70 \mathrm{~kg}\) swings from the end of a \(4.0-\mathrm{m}-\mathrm{long}\) rope along the arc of a vertical circle. Assuming he starts from rest when the rope is horizontal, find the tensions in the rope that are required to make him follow his circular path (a) at the beginning of his motion, (b) at a height of \(1.5 \mathrm{~m}\) above the bottom of the circular arc, and (c) at the bottom of the arc.

A car moves at speed \(v\) across a bridge made in the shape of a circular arc of radius \(r\). (a) Find an expression for the normal force acting on the car when it is at the top of the arc. (b) At what minimum speed will the normal force become zero (causing the occupants of the car to seem weightless) if \(r=30.0 \mathrm{~m}\) ?

(a) One of the moons of Jupiter, named Io, has an orbital radius of \(4.22 \times 10^{8} \mathrm{~m}\) and a period of \(1.77\) days. Assuming the orbit is circular, calculate the mass of Jupiter. (b) The largest moon of Jupiter, named Ganymede, has an orbital radius of \(1.07 \times 10^{9} \mathrm{~m}\) and a period of \(7.16\) days. Calculate the mass of Jupiter from this data. (c) Are your results to parts (a) and (b) consistent? Explain.

Q|C A bicycle is turned upside down while its owner repairs a flat tire. A friend spins the other wheel and observes that drops of water fly off tan- gentially. She measures the heights reached by drops moving vertically (Fig. P7.8). A drop that breaks loose from the tire on one turn rises vertically \(54.0\) cm above the tangent point. A drop that breaks loose on the next turn rises \(51.0 \mathrm{~cm}\) above the tan- gent point. The radius of the wheel is \(0.381 \mathrm{~m}\). (a) Why does the first drop rise higher than the second drop? (b) Neglecting air friction and using only the observed heights and the radius of the wheel, find the wheel's angular acceleration (assuming it to be constant).

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