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Chapter 14: Problem 67

CP SHM in a Car Engine. The motion of the piston of an automobile engine is approximately simple harmonic. (a) If the stroke of an engine (twice the amplitude) is 0.100 \(\mathrm{m}\) and the engine runs at 4500 \(\mathrm{rev} / \mathrm{min}\) , compute the acceleration of the piston at the endpoint of its stroke. (b) If the piston has mass \(0.450 \mathrm{kg},\) what net force must be exerted on it at this point? (c) What are the speed and kinetic energy of the piston at the mid- point of its stroke? (d) What average power is required to accelerate the piston from rest to the speed found in part (c)? (e) If the engine runs at 7000 rev/min, what are the answers to parts (b), (c), and (d)?

Short Answer

Expert verified
Acceleration: 11,105 m/s^2, Force: 4997.25 N, Speed: 23.562 m/s, Kinetic Energy: 124.94 J, Power: 18,722 W at 4500 rev/min; at 7000 rev/min: Force: 12,080 N, Speed: 36.652 m/s, K.E: 302.5 J, Power: 70,648 W.

Step by step solution

01

Understand Stroke and Amplitude

The stroke of an engine is given as twice the amplitude of oscillation. Thus, if the stroke is 0.100 m, then the amplitude \( A \) is 0.050 m.
02

Calculate Angular Frequency for 4500 rev/min

First, convert revolutions per minute to radians per second. The engine speed is 4500 rev/min. Use the formula: \[\omega = 2\pi \times \frac{4500}{60} \approx 471.24 \, \text{rad/s}.\]
03

Compute Acceleration at Endpoint for 4500 rev/min

The maximum acceleration of the piston occurs at the endpoints and is given by:\[a_{ ext{max}} = \omega^2 \times A.\]Substituting the values, we have:\[a_{ ext{max}} = (471.24)^2 \times 0.050 \approx 11,105 \, \text{m/s}^2.\]
04

Calculate Net Force at Endpoint for Mass 0.450 kg

Use the formula for force, \( F = m \times a \). Given mass \( m = 0.450 \, \text{kg} \) and \( a = 11,105 \, \text{m/s}^2 \):\[F = 0.450 \times 11,105 \approx 4997.25 \, \text{N}. \]
05

Determine Speed at Midpoint for 4500 rev/min

At the midpoint, the maximum speed of the piston can be calculated using:\[v_{ ext{max}} = \omega \times A.\]Substitute \( \omega = 471.24 \, \text{rad/s} \) and \( A = 0.050 \, \text{m} \):\[v_{ ext{max}} = 471.24 \times 0.050 \approx 23.562 \, \text{m/s}.\]
06

Calculate Kinetic Energy at Midpoint

The kinetic energy \( K \) at the midpoint is given by:\[K = \frac{1}{2} m v^2.\]Substitute \( m = 0.450 \, \text{kg} \) and \( v = 23.562 \, \text{m/s} \):\[K = \frac{1}{2} \times 0.450 \times (23.562)^2 \approx 124.94 \, \text{J}.\]
07

Find Average Power Required

The average power \( P \) required to accelerate the piston can be calculated as:\[P = \frac{K}{T/2},\]where \( T \) is the period \(T = \frac{1}{f} = \frac{2\pi}{\omega}\). For \( \omega = 471.24 \, \text{rad/s}\):\[T = \frac{2\pi}{471.24} \approx 0.01334 \, \text{s}.\]Thus:\[P = \frac{124.94}{0.01334/2} \approx 18,722 \, \text{W}.\]
08

Repeat Previous Steps for 7000 rev/min

Convert 7000 rev/min to radians per second to get \(\omega\). \[\omega = 2\pi \times \frac{7000}{60} \approx 733.04 \, \text{rad/s}.\]For acceleration, force, speed, kinetic energy, and power, substitute \(\omega = 733.04\) in previous formulas. - Maximum acceleration: \[a_{\text{max}} = (733.04)^2 \times 0.050 \approx 26,844.5 \, \text{m/s}^2.\]- Force at endpoint:\[F = 0.450 \times 26,844.5 \approx 12,080 \, \text{N}. \]- Speed at midpoint:\[v_{\text{max}} = 733.04 \times 0.050 \approx 36.652 \, \text{m/s}.\]- Kinetic energy:\[K = \frac{1}{2} \times 0.450 \times (36.652)^2 \approx 302.50 \, \text{J}.\]- Average power:\[T = \frac{2\pi}{733.04} \approx 0.00857 \, \text{s},\]\[P = \frac{302.50}{0.00857/2} \approx 70,648 \, \text{W}.\]

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Key Concepts

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

Angular Frequency
Angular frequency is a fundamental concept in simple harmonic motion (SHM). It represents how rapidly something oscillates in a circular motion, which in this case means how quickly the engine's piston moves back and forth. To calculate angular frequency, you convert revolutions per minute (rev/min) into radians per second (rad/s).
In this exercise, the engine runs at 4500 rev/min. Use the formula \( \omega = 2\pi \times \frac{\text{RPM}}{60} \) to convert RPM to \( \omega \). This results in an angular frequency of approximately 471.24 radians per second. This math helps understand the cyclical speed of the system, affecting energy calculations and motion predictions.
Understanding angular frequency in these scenarios is key because it determines several attributes of the piston's motion like acceleration and speed.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. For the piston's simple harmonic motion, the kinetic energy is highest at the midpoint of its stroke.
To calculate it, use the formula \( K = \frac{1}{2} m v^2 \), where \( m \) is the mass of the piston, and \( v \) is the velocity at the midpoint. In this exercise, the maximum speed of the piston at the midpoint can be derived from \( v_{\text{max}} = \omega \times A \), which results in 23.562 meters per second at 4500 rev/min. The kinetic energy at this speed is approximately 124.94 Joules.
The kinetic energy gives insights into how much energy is needed to keep the piston moving or bring it to a stop from a given speed.
Average Power
Average power in physics quantifies how much energy is used or converted over time. It tells us how much work needs to be done per unit of time to accelerate or decelerate the piston's motion in the engine.
To calculate average power, you need the kinetic energy and the time over which this energy change happens. The formula \( P = \frac{K}{T/2} \) helps find the average power, where \( T \) is the period of oscillation, \( T = \frac{2\pi}{\omega} \). At 4500 rev/min, the period \( T \) is approximately 0.01334 seconds, leading to an average power usage of about 18,722 Watts.
This concept helps understand the engine's energy requirements during operation, ensuring efficient and effective functioning.

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Most popular questions from this chapter

An object with mass 0.200 \(\mathrm{kg}\) is acted on by an elastic restoring force with force constant 10.0 \(\mathrm{N} / \mathrm{m}\) . (a) Graph elastic potential energy \(U\) as a function of displacement \(x\) over a range of \(x\) from \(-0.300 \mathrm{m}\) to \(+0.300 \mathrm{m}\) . On your graph, let \(1 \mathrm{cm}=0.05 \mathrm{J}\) vertically and \(1 \mathrm{cm}=0.05 \mathrm{m}\) horizontally. The object is set into oscillation with an initial potential energy of 0.140 \(\mathrm{J}\) and an initial kinetic energy of 0.060 \(\mathrm{J}\) . Answer the following questions by referring to the graph. (b) What is the amplitude of oscillation? (c) What is the potential energy when the displacement is one half the amplitude? (d) At what displacement are the kinetic and potential energies equal? (e) What is the value of the phase angle \(\phi\) if the initial velocity is positive and the initial displacement is negative?

An object is undergoing SHM with period 1.200 s and amplitude 0.600 \(\mathrm{m}\) . At \(t=0\) the object is at \(x=0\) and is moving in the negative \(x\) -direction. How far is the object from the equilibrium position when \(t=0.480 \mathrm{s} ?\)

CP A holiday ornament in the shape of a hollow sphere with mass \(M=0.015 \mathrm{kg}\) and radius \(R=0.050 \mathrm{m}\) is hung from a tree limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance and released, it swings back and forth as a physical pendulum with negligible friction. Calculate its period. (Hint: Use the parallel-axis theorem to find the moment of inertia of the sphere about the pivot at the tree limb.)

A Pendulum on Mars. A certain simple pendulum has a period on the earth of 1.60 s. What is period on the surface of Mars, where \(g=3.71 \mathrm{m} / \mathrm{s}^{2} ?\)

BID (a) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note that is sung. If someone sings the note \(B\) flat, which has a fre- quency of 466 Hz, how much time does it take the person's vocal cords to vibrate through one complete cycle, and what is the angu- lar frequency of the cords? (b) Hearing. When sound waves strike the eardrum, this membrane vibrates with the same frequency as the sound. The highest pitch that typical humans can hear has a period of 50.0\(\mu \mathrm{s} .\) What are the frequency and angular frequency of the vibrating eardrum for this sound? (c) Vision. When light having vibrations with angular frequency ranging from \(2.7 \times 10^{15} \mathrm{rad} / \mathrm{s}\) to \(4.7 \times 10^{15} \mathrm{rad} / \mathrm{s}\) strikes the retina of the eye, it stimulates the receptor cells there and is perceived as visible light. What are the limits of the period and frequency of this light? (d) Ultrasound. High-frequency sound waves (ultrasound) are used to probe the interior of the body, much as x rays do. To detect small objects such as tumors, a frequency of around 5.0 \(\mathrm{MHz}\) is used. What are the period and angular frequency of the molecular vibrations caused by this pulse of sound?
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