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A racing car travels on a circular track of radius \(250 \mathrm{m} .\) If the car moves with a constant linear speed of 45.0 \(\mathrm{m} / \mathrm{s}\) , find (a) its angular speed and (b) the magnitude and direction of its acceleration.

Short Answer

Expert verified
The angular speed of the car is 0.18 rad/s. The magnitude of the car's acceleration is 8.1 m/s^2, and it is directed towards the center of the circular track.

Step by step solution

01

Find the angular speed

The angular speed, often denoted as \(\omega\), can be calculated using the formula \(\omega = \frac{v}{r}\), where \(v\) is the linear speed and \(r\) is the radius of the circular path. Knowing that \(v = 45.0 \text{ m/s}\) and \(r = 250 \text{ m}\), substitute these values to get \(\omega = \frac{45.0}{250}\).
02

Calculate the angular speed

Perform the division to find \(\omega\) which is \(\omega = \frac{45.0}{250} = 0.18 \text{ rad/s}\). This is the constant angular speed of the car.
03

Identify the type of acceleration

For an object moving in a circular path at a constant speed, the acceleration is centripetal (toward the center of the circular path). The magnitude of this acceleration, \(a_c\), is given by the formula \(a_c = \frac{v^2}{r}\).
04

Calculate the magnitude of acceleration

Using the values \(v = 45.0 \text{ m/s}\) and \(r = 250 \text{ m}\), substitute into the formula to get \(a_c = \frac{45.0^2}{250} = \frac{2025}{250}\).
05

Find the final value of the acceleration

By performing the division, the magnitude of the acceleration is \(a_c = \frac{2025}{250} = 8.1 \text{ m/s}^2\).
06

Determine the direction of the acceleration

Since the car is moving in a circle and its speed is constant, the acceleration is centripetal, which means it points towards the center of the circular path.

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Key Concepts

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

Angular Speed
When an object moves along a circular path, its angular speed, denoted by the Greek letter \(\omega\), is a measure of how quickly it rotates around the center of the circle. Specifically, it represents the angle covered per unit time and is usually measured in radians per second (rad/s). To calculate angular speed, the equation \(\omega = \frac{v}{r}\) is used, where \(v\) is the object's linear speed and \(r\) is the radius of the circular path.

In the context of a racing car on a track, knowing the angular speed is crucial for understanding the car's performance and for making precise turns. It allows engineers and drivers to coordinate speed and timing, ensuring the car navigates the curve efficiently without skidding out of control.
Centripetal Acceleration
Centripetal acceleration is the acceleration experienced by an object moving in a circular path at a constant speed. It's directed towards the center of the circle and maintains the object's circular motion. The magnitude of centripetal acceleration \(a_c\) can be found using the formula \(a_c = \frac{v^2}{r}\), with \(v\) being the linear speed and \(r\) the radius of the path.

Centripetal acceleration is essential for understanding how forces act on objects in circular motion. For example, a race car driver feels the force pushing them into the seat while navigating a turn. This force is due to centripetal acceleration and is necessary to change the car's direction.
Linear Speed
Linear speed refers to the distance an object covers over time along a linear or straight trajectory and is often measured in meters per second (m/s). In circular motion, while the speed may remain constant, the velocity (which includes both speed and direction) does not, as the direction is constantly changing.

Understanding the linear speed in a circular track is crucial for calculating other aspects of motion, such as angular speed and centripetal acceleration. It also plays a significant role in ensuring safety and stability, particularly in high-speed settings like car racing or amusement park rides, where the correct speed is key to preventing accidents.
Circular Track Physics
Circular track physics encompasses all the physical principles that come into play when an object moves along a circular path. This includes concepts like angular speed, centripetal force, centripetal acceleration, and the relationship between linear and angular velocities. Understanding these concepts is vital in many real-world applications, such as designing race tracks, roller coasters, and analyzing planetary orbits.

In applications such as car racing, engineers and drivers must account for forces and motions on the track to optimize performance and safety. A deep understanding of circular track physics allows for better predictions and control over the behavior of objects in circular motion.

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Most popular questions from this chapter

A shaft is turning at 65.0 \(\mathrm{rad} / \mathrm{s}\) at time \(t=0 .\) Thereafter, its angular acceleration is given by $$ \alpha=-10.0 \mathrm{rad} / \mathrm{s}^{2}-5.00 t \mathrm{rad} / \mathrm{s}^{3} $$ where \(t\) is the elapsed time. (a) Find its angular speed at \(t=3.00 \mathrm{s}\) . (b) How far does it turn in these 3 \(\mathrm{s} ?\)

A disk 8.00 \(\mathrm{cm}\) in radius rotates at a constant rate of 1 200 rev/min about its central axis. Determine (a) its angular speed, (b) the tangential speed at a point 3.00 \(\mathrm{cm}\) from its center, (c) the radial acceleration of a point on the rim, and (d) the total distance a point on the rim moves in 2.00 s.

A grinding wheel is in the form of a uniform solid disk of radius 7.00 \(\mathrm{cm}\) and mass 2.00 \(\mathrm{kg}\) . It starts from rest and accelerates uniformly under the action of the constant torque of 0.600 \(\mathrm{N} \cdot \mathrm{m}\) that the motor exerts on the wheel. (a) How long does the wheel take to reach its final operating speed of 1200 \(\mathrm{rev} / \mathrm{min}\) ? (b) Through how many revolutions does it turn while accelerating?

A uniform solid disk and a uniform hoop are placed side by side at the top of an incline of height \(h\) . If they are leased from rest and roll without slipping, which object reaches the bottom first? Verify your answer by calculating their speeds when they reach the bottom in terms of \(h .\)

During a certain period of time, the angular position of a swinging door is described by \(\theta=5.00+10.0 t+2.00 t^{2}\) , where \(\theta\) is in radians and \(t\) is in seconds. Determine the angular position, angular speed, and angular acceleration of the door (a) at \(t=0\) (b) at \(t=3.00 \mathrm{s}\) .

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