Problem 67 - University Physics with Modern Physics

Question

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)?

Step-by-Step Solution

Answer

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 1: 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.

Step 2: 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: \[\omega = 2\pi \times \frac{4500}{60} \approx 471.24 \, \text{rad/s}.\]

Step 3: 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:\[a_{ ext{max}} = (471.24)^2 \times 0.050 \approx 11,105 \, \text{m/s}^2.\]

Step 4: 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 \( a = 11,105 \, \text{m/s}^2 \):\[F = 0.450 \times 11,105 \approx 4997.25 \, \text{N}. \]

Step 5: 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 \( A = 0.050 \, \text{m} \):\[v_{ ext{max}} = 471.24 \times 0.050 \approx 23.562 \, \text{m/s}.\]

Step 6: 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 \( v = 23.562 \, \text{m/s} \):\[K = \frac{1}{2} \times 0.450 \times (23.562)^2 \approx 124.94 \, \text{J}.\]

Step 7: 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:\[P = \frac{124.94}{0.01334/2} \approx 18,722 \, \text{W}.\]

Step 8: 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:\[T = \frac{2\pi}{733.04} \approx 0.00857 \, \text{s},\]\[P = \frac{302.50}{0.00857/2} \approx 70,648 \, \text{W}.\]

Key Concepts

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.