Problem 92
Question
BIO All birds, independent of their size, must maintain a power output of \(10-25\) watts per kilogram of body mass in order to fly by flapping their wings. (a) The Andean giant hummingbird (Patagona gigas) has mass 70 \(\mathrm{g}\) and flaps its wings 10 times per second while hovering. Estimate the amount of work done by such a hummingbird in each wingbeat. (b) A 70 -kg athlete can maintain a power output of 1.4 \(\mathrm{kW}\) for no more than a few seconds; the steady power output of a typical athlete is only 500 \(\mathrm{W}\) or so. Is it possible for a human-powered aircraft to fly for extended periods by flapping its wings? Explain.
Step-by-Step Solution
Verified Answer
(a) The hummingbird does 0.1225 J of work per wingbeat. (b) Due to insufficient sustainable power, humans cannot fly by flapping wings.
1Step 1: Determine Power Output
First, we need to determine the power output that the hummingbird must maintain. The hummingbird's mass is 70 g, which is 0.07 kg. We know the power output requirement is between 10 and 25 watts per kilogram. Let's find the power range: \[ 0.07 \times 10 = 0.7 \, \text{W} \] \[ 0.07 \times 25 = 1.75 \, \text{W} \] Thus, the hummingbird's power output is between 0.7 to 1.75 W.
2Step 2: Determine Work Done per Wingbeat
Assuming the constant power output is roughly the median 1.225 W (since it's a continuous process instead of fluctuating to the extreme ends), the work done per wingbeat can be found by considering the time per wingbeat:The hummingbird flaps 10 times per second, so the time per wingbeat is:\[ \frac{1 \text{ second}}{10 \text{ flaps/second}} = 0.1 \text{ seconds} \]The work done per wingbeat is the power output multiplied by the time:\[ 1.225 \, \text{W} \times 0.1 \, \text{s} = 0.1225 \, \text{J} \]Therefore, the work done per wingbeat is approximately 0.1225 J.
3Step 3: Analyze Human Power Output
A human weighing 70 kg can exert a maximum power of 1.4 kW for a few seconds. The sustainable power output of a typical athlete is about 500 W. Flapping suitable wings to fly demands consistent high power output similar to birds, which is much higher than 500 W for extended periods.
4Step 4: Determine Feasibility of Human Flight by Flapping
Considering the human power output, it's challenging to sustain the necessary power output continually, which is less than what's required for bird flight. Therefore, it is not feasible for a human-powered aircraft to fly by simply flapping wings like a bird.
Key Concepts
Bird Flight MechanicsHuman-Powered FlightWork and Energy in PhysicsPhysics of Flapping Wings
Bird Flight Mechanics
Understanding bird flight mechanics is key to comprehending how birds sustain flight. Birds, regardless of their size, maintain a power output ranging from 10 to 25 watts per kilogram of body mass. This is crucial for flight, as power needs to be sufficient for the wings to generate lift and overcome drag forces. The Andean giant hummingbird, for example, has a mass of 70 grams, which converts to 0.07 kilograms. During flight, particularly hovering, it flaps its wings 10 times per second. Given its weight, the bird must produce a power output within the specified range. This involves using energy efficiently to convert through its wing flaps into lift, all balanced by the power it generates.
Human-Powered Flight
Human-powered flight is an intriguing and challenging concept. While humans possess significant muscle strength, it becomes apparent that maintaining prolonged flight by purely flapping wings, akin to birds, is not feasible. A 70 kg athlete can generate a peak power of 1.4 kW, but only for a short duration. More sustainably, an athlete's steady power output is around 500 W. Comparatively, birds require a higher power output to maintain flight over time. The difference lies in the mechanics and efficiency of human muscles versus bird anatomy specialized for flight. Efforts in human-powered flight often incorporate mechanisms like pedaling to harness leg muscles more efficiently, distinct from the bird's natural wing flapping method.
Work and Energy in Physics
Understanding work and energy in the physics of flight involves breaking down actions like flapping wings into mechanical energy. The equation for work is given by \( W = P \times t \), where \( W \) is work in joules, \( P \) is power in watts, and \( t \) is time in seconds. For a hummingbird flapping its wings 10 times per second, each flap takes 0.1 seconds. With a consistent power of approximately 1.225 watts during hovering, the work done per wingbeat totals about 0.1225 joules. This example shows how birds distribute their energy efficiently over time to maintain flight, exemplifying the practical application of these physics principles.
Physics of Flapping Wings
The physics of flapping wings involves converting stored energy into motion through intricate biomechanical processes. Birds have evolved to maximize efficiency, with wings acting as both propellers and lift generators. This involves a complex interaction of forces including thrust, drag, and lift, which dictates their flight capabilities. When a bird flaps its wings, it creates a pressure difference, producing lift and thrust concurrently. This is constant even in small birds like the hummingbird. Conversely, a human's muscular arrangement and biomechanical constraints make it improbable to achieve the same with just arm movements. Nonetheless, studying this mode of propulsion provides insights, aiding in technological advancements such as simple mechanical flight apparatuses for study and recreation.
Other exercises in this chapter
Problem 90
Rescue. Your friend (mass 65.0 \(\mathrm{kg} )\) is standing on the ice in the middle of a frozen pond. There is very litle friction between her feet and the ic
View solution Problem 91
A pump is required to lift 800 kg of water (about 210 gallons) per minute from a well 14.0 \(\mathrm{m}\) deep and eject it with a speed of 18.0 \(\mathrm{m} /
View solution Problem 93
A physics student spends part of her day walking between classes or for recreation, during which time she expends energy at an average rate of 280 \(\mathrm{W}\
View solution Problem 94
The Grand Coulee Dam is 1270 \(\mathrm{m}\) long and 170 \(\mathrm{m}\) high. The electrical power output from generators at its base is approximately 2000 \(\m
View solution