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the angular speed of an automobile engine is increased at a constant rate from \(1200 \mathrm{rev} / \mathrm{min}\) to \(3200 \mathrm{rev} / \mathrm{min}\) in \(12 \mathrm{~s}\). (a) What is its angular acceleration in revolutions per minute-squared? (b) How many revolutions does the engine make during this \(12 \mathrm{~s}\) interval?

Short Answer

Expert verified
Angular acceleration is 10000 rev/min²; total revolutions are 440.

Step by step solution

01

Convert Angular Speeds to Consistent Units

First, we need to convert the initial and final angular speeds from revolutions per minute to revolutions per second to work with time in seconds. \[ \omega_i = 1200 \, \text{rev/min} = \frac{1200}{60} \, \text{rev/s} = 20 \, \text{rev/s} \]\[ \omega_f = 3200 \, \text{rev/min} = \frac{3200}{60} \, \text{rev/s} = 53.33 \, \text{rev/s} \]
02

Calculate Angular Acceleration

The angular acceleration \( \alpha \) can be found using the formula:\[ \alpha = \frac{\omega_f - \omega_i}{t} \]Substituting the values, we have:\[ \alpha = \frac{53.33 \, \text{rev/s} - 20 \, \text{rev/s}}{12 \, \text{s}} = 2.777 \, \text{rev/s}^2 \]Now, convert \( \alpha \) to revolutions per minute-squared:\[ \alpha = 2.777 \, \text{rev/s}^2 \times (60 \, \text{seconds/minute})^2 = 10000 \, \text{rev/min}^2 \]
03

Calculate the Total Number of Revolutions

The total number of revolutions \( \theta \) can be calculated using the formula:\[ \theta = \omega_i \times t + \frac{1}{2} \times \alpha \times t^2 \]Replace \( \alpha \) with \( 2.777 \, \text{rev/s}^2 \) for calculations in seconds:\[ \theta = 20 \, \text{rev/s} \times 12 \, \text{s} + \frac{1}{2} \times 2.777 \, \text{rev/s}^2 \times (12 \, \text{s})^2 \]\[ \theta = 240 \, \text{rev} + \frac{1}{2} \times 2.777 \, \text{rev/s}^2 \times 144 \, \text{s}^2 \]\[ \theta = 240 \, \text{rev} + 199.94 \, \text{rev} \approx 439.94 \, \text{rev} \]
04

Conclusion

The engine's angular acceleration is \( 10000 \, \text{rev/min}^2 \), and the total number of revolutions during the 12 seconds interval is approximately 440.

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Key Concepts

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

Angular Speed
Angular speed is an essential concept when discussing rotational motion. Simply put, it measures how fast something spins or rotates. For example, in our exercise, we have an automobile engine that increases its rotational speed from \(1200 \text{ rev/min} \) to \(3200 \text{ rev/min} \). When dealing with problems involving time in seconds, it's crucial to convert angular speed to revolutions per second (rev/s). This conversion aids in simplifying calculations, as seen when we changed \(1200 \text{ rev/min} \) to \(20 \text{ rev/s} \), and \(3200 \text{ rev/min} \) to \(53.33 \text{ rev/s} \). Remember, to convert from rev/min to rev/s, simply divide by 60, since there are 60 seconds in a minute. Hence, understanding angular speed and effectively converting it to the appropriate units will enable smooth progress in solving angular motion problems.
Angular Acceleration
Angular acceleration refers to the rate at which angular speed changes over time. It represents how quickly a spinning object speeds up or slows down. In our exercise, the engine's angular speed goes from \(20 \text{ rev/s} \) to \(53.33 \text{ rev/s} \) over 12 seconds. Therefore, the angular acceleration is calculated using the formula:\[ \alpha = \frac{\omega_f - \omega_i}{t} \]Where \( \omega_f \) is the final angular speed, \( \omega_i \) is the initial angular speed, and \( t \) represents time. Substituting the given values results in:\[ \alpha = \frac{53.33 \text{ rev/s} - 20 \text{ rev/s}}{12 \text{ s}} = 2.777 \text{ rev/s}^2 \]To convert angular acceleration to revolutions per minute-squared, you multiply by \((60 \text{ seconds/minute})^2\). Thus, the angular acceleration becomes \(10000 \text{ rev/min}^2\). The process of calculating and converting angular acceleration is crucial for effectively analyzing and understanding rotational motion dynamics.
Revolutions
Revolutions are a way of expressing the total angle an object has turned through its motion. For rotational problems, such as the one in the exercise, the number of revolutions is calculated using the initial angular speed, angular acceleration, and the time over which the motion occurs.The formula used in calculations is:\[ \theta = \omega_i \times t + \frac{1}{2} \times \alpha \times t^2 \]Let's break it down:
  • \( \omega_i \times t \) represents the revolutions made by the object due to its initial speed.
  • \(\frac{1}{2} \times \alpha \times t^2 \) describes the extra revolutions from the object speeding up.
Inserting the given values:\[ \theta = 20 \text{ rev/s} \times 12 \text{ s} + \frac{1}{2} \times 2.777 \text{ rev/s}^2 \times (12 \text{ s})^2 \]Calculating gives a total of approximately \(440 \text{ rev} \). Understanding revolutions and the ability to compute them helps in thoroughly grasping the change in position during an object's rotational motion.

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Most popular questions from this chapter

Disks \(A\) and \(B\) each have a rotational inertia of \(0.300 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about the central axis and a radius of \(20.0 \mathrm{~cm}\) and are free to rotate on a central rod through both of them. To set them spinning around the rod in the same direction, each is wrapped with a string that is then pulled for \(10.0 \mathrm{~s}\) (the string detaches at the end). The magnitudes of the forces pulling the strings are \(30.0 \mathrm{~N}\) for disk \(A\) and \(20.0 \mathrm{~N}\) for disk \(B\). After the strings detach, the disks happen to collide and the frictional force between them brings them to the same final angular speed in \(6.00 \mathrm{~s}\). What are (a) magnitude of the average frictional torque that brings them to the final angular speed and (b) the loss in kinetic energy as that torque acts on them? (c) Where did the "lost energy" go?

A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at \(10 \mathrm{rev} / \mathrm{s} ; 60\) revolutions later, its angular speed is \(15 \mathrm{rev} / \mathrm{s}\). Calculate (a) the angular acceleration, (b) the time required to complete the 60 revolutions, (c) the time required to reach the 10 rev/s angular speed, and (d) the number of revolutions from rest until the time the disk reaches the \(10 \mathrm{rev} / \mathrm{s}\) angular speed.

A vinyl record is played by rotating the record so that an approximately circular groove in the vinyl slides under a stylus. Bumps in the groove run into the stylus, causing it to oscillate. The equipment converts those oscillations to electrical signals and then to sound. Suppose that a record turns at the rate of \(33 \frac{1}{3} \mathrm{rev} / \mathrm{min}\), the groove being played is at a radius of \(10.0 \mathrm{~cm}\), and the bumps in the groove are uniformly separated by \(1.85 \mathrm{~mm}\). At what rate (hits per second) do the bumps hit the stylus?

If a \(42.0 \mathrm{~N} \cdot \mathrm{m}\) torque on a wheel causes angular acceleration \(25.0 \mathrm{rad} / \mathrm{s}^{2}\), what is the wheel's rotational inertia?

An open jar of water moves in a vertical circle of radius \(0.50 \mathrm{~m}\) with a frequency that is small enough to put the water on the verge of falling out of the jar at the top of the circle. If the same demonstration were repeated on Mars, where the gravitational acceleration is only \(3.7 \mathrm{~m} / \mathrm{s}^{2}\), what is the change in the circling frequency to again put the water on the verge of falling out at the top point?

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