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Chapter 8: Problem 54

The penny-farthing is a bicycle that was popular between 1870 and 1890 . As the drawing shows, this type of bicycle has a large front wheel and a small rear wheel. On a Sunday ride in the park the front wheel (radius \(=1.20 \mathrm{~m}\) ) makes 276 revolutions. How many revolutions does the rear wheel (radius \(=0.340 \mathrm{~m}\) ) make?

Short Answer

Expert verified
The rear wheel makes approximately 973 revolutions.

Step by step solution

01

Calculate the circumference of the front wheel

To find the distance traveled in one revolution, calculate the circumference of the front wheel using the formula \( C = 2\pi r \), where \( r \) is the radius. Here, \( r = 1.20 \text{ m} \).Substitute the given value: \( C = 2\pi (1.20) \approx 7.54 \text{ m} \).
02

Find the total distance traveled by the front wheel

Multiply the number of revolutions made by the front wheel by its circumference to find the total distance.Total distance \( = 276 \times 7.54 \approx 2,081.04 \text{ m} \).
03

Calculate the circumference of the rear wheel

Use the same circumference formula for the rear wheel, which has a radius \( r = 0.340 \text{ m} \).\( C = 2\pi (0.340) \approx 2.14 \text{ m} \).
04

Determine the number of revolutions for the rear wheel

To find the number of revolutions the rear wheel makes, divide the total distance traveled by the circumference of the rear wheel.Revolutions of rear wheel \( = \frac{2,081.04}{2.14} \approx 973 \).

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Key Concepts

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

Penny-Farthing Bicycle
The penny-farthing bicycle, a remarkable invention from the late 19th century, is known for its distinctive large front wheel and small rear wheel. This unique design was both a product of innovation and the technology available during that era.
Unlike today's bicycles, where gears enable smoother rides and speed variation, the penny-farthing relied heavily on direct pedal power.
  • Front Wheel: The large front wheel allowed cyclists to cover more ground with each pedal stroke, contributing to greater speeds compared to other bicycles of the time.
  • Rear Wheel: The smaller rear wheel provided stability and supported the rider's weight.
Both wheels played crucial roles in the penny-farthing's operation, and understanding their dimensions is key in solving related physics problems.
Wheel Circumference Calculation
Calculating the wheel circumference is essential for understanding how far the bicycle travels with each revolution. The formula for circumference is given by:\[ C = 2\pi r \]where \( C \) is the circumference, \( \pi \) is a mathematical constant approximately equal to 3.14159, and \( r \) is the radius of the wheel. This simple yet powerful formula helps determine how much distance a wheel covers per complete turn.
  • For the front wheel with a radius of 1.20 m, the circumference becomes:\[ C = 2\pi(1.20) \approx 7.54 \text{ m} \]
  • The rear wheel, with a radius of 0.340 m, computes to:\[ C = 2\pi(0.340) \approx 2.14 \text{ m} \]
Knowing the circumference allows for calculations related to distance and speed, linking geometry with real-world applications in physics.
Revolutions Calculation
Understanding revolutions is crucial when analyzing a bicycle's movement. It involves determining how many times a wheel turns to cover a certain distance. To find the total distance traveled by the large front wheel, you multiply the number of revolutions by the wheel's circumference.Given:
  • 276 revolutions for the front wheel
  • The front wheel circumference as 7.54 m
The total distance traveled is:\[ \text{Total Distance} = 276 \times 7.54 \approx 2,081.04 \text{ m} \]For the rear wheel, to determine how many revolutions it makes to cover the same distance:
  • Use the total distance of 2,081.04 m
  • The rear wheel circumference of 2.14 m
The number of revolutions is:\[ \text{Revolutions of rear wheel} = \frac{2,081.04}{2.14} \approx 973 \]This calculation demonstrates how differences in wheel sizes affect their rotation and movement.
Mathematical Physics Problem-Solving
Solving physics problems using mathematics involves understanding and applying formulas to real-world scenarios. In this penny-farthing problem, students use measurements, apply concepts of distance and rotation, and consolidate their calculation skills.
  • Identify the Givens: Know the wheel radii and number of revolutions. This collects the relevant physical information.
  • Use the Formula: Attribute the correct formula; here, the circumference formula is central.
  • Calculate Step-by-Step: Break down the problem into manageable steps 鈥� compute wheel circumferences, total distance, and then determine the rear wheel's revolutions.
  • Verify Units and Answers: Ensure all calculations are in proper units for coherence and that the final answers make logical sense.
Engaging in such exercises not only aids in understanding mathematical physics but also fosters critical thinking and problem-solving skills.

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Most popular questions from this chapter

During a time \(t,\) a wheel has a constant angular acceleration, so its angular velocity increases from an initial value of \(\omega_{0}\) to a final value of \(\omega\). During this time, is the angular displacement less than, greater than, or equal to (a) \(\omega_{0} t\) or (b) \(\omega t\) ? In each case, justify your answer. (c) If you conclude that the angular displacement does not equal \(\omega_{0} t\) or \(\omega t,\) then how does it depend on \(\omega_{0}, \omega,\) and \(t ?\) Problem A car is traveling along a road, and the engine is turning over with an angular velocity of \(+220 \mathrm{rad} / \mathrm{s}\). The driver steps on the accelerator, and in a time of \(10.0 \mathrm{~s}\) the angular velocity increases to \(+280 \mathrm{rad} / \mathrm{s}\). (a) What would have been the angular displacement of the engine if its angular velocity had remained constant at the initial value of \(+220 \mathrm{rad} / \mathrm{s}\) during the \(10.0 \mathrm{~s} ?\) (b) What would have been the angular displacement if the angular velocity had been equal to its final value of \(+280 \mathrm{rad} / \mathrm{s}\) during the \(10.0 \mathrm{~s} ?\) (c) Determine the actual value of the angular displacement during this period. Check your answers to see that they are consistent with those to the Concept Questions.

A ball of radius \(0.200 \mathrm{~m}\) rolls along a horizontal table top with a constant linear speed of \(3.60 \mathrm{~m} / \mathrm{s} .\) The ball rolls off the edge and falls a vertical distance of \(2.10 \mathrm{~m}\) before hitting the floor. What is the angular displacement of the ball while the ball is in the air?

The sun appears to move across the sky, because the earth spins on its axis. To a person standing on the earth, the sun subtends an angle of \(\theta_{\mathrm{sun}}=9.28 \times 10^{-3} \mathrm{rad}\) (see Conceptual Example 2). How much time (in seconds) does it take for the sun to move a distance equal to its own diameter?

A compact disc (CD) contains music on a spiral track. Music is put onto a CD with the assumption that, during playback, the music will be detected at a constant tangential speed at any point. Since \(\mathcal{V}_{\mathrm{T}}=r \omega,\) a CD rotates at a smaller angular speed for music near the outer edge and a larger angular speed for music near the inner part of the disc. For music at the outer edge \((r=0.0568 \mathrm{~m})\), the angular speed is \(3.50 \mathrm{rev} / \mathrm{s}\). Find (a) the constant tangential speed at which music is detected and (b) the angular speed (in rev/s) for music at a distance of \(0.0249 \mathrm{~m}\) from the center of a \(\mathrm{CD}\).

A person lowers a bucket into a well by turning the hand crank, as the drawing illustrates. The crank handle moves with a constant tangential speed of \(1.20\mathrm{~m} / \mathrm{s}\) on its circular path. Find the linear speed with which the bucket moves down the well.
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