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A racing car travels with a constant tangential speed of \(75.0 \mathrm{m} / \mathrm{s}\) around a circular track of radius \(625 \mathrm{m}\). Find (a) the magnitude of the car's total acceleration and (b) the direction of its total acceleration relative to the radial direction.

Short Answer

Expert verified
Total acceleration is 9.0 m/s², directed radially towards the center.

Step by step solution

01

Determine the Tangential and Radial Components of Acceleration

The car travels at a constant tangential speed of \( 75.0 \mathrm{m/s} \), meaning there is no change in speed along the path, so the tangential acceleration is zero. Next, calculate the centripetal (radial) acceleration as the car is moving in a circular path. Centripetal acceleration is given by the formula \( a_r = \frac{v^2}{r} \), where \( v = 75.0 \mathrm{m/s} \) and \( r = 625 \mathrm{m} \). Calculate this to find \( a_r \).
02

Calculate Centripetal Acceleration

Use the formula \( a_r = \frac{v^2}{r} \) to find the centripetal acceleration. Substitute \( v = 75.0 \mathrm{m/s} \) and \( r = 625 \mathrm{m} \): \[a_r = \frac{(75.0 \mathrm{m/s})^2}{625 \mathrm{m}} = \frac{5625 \mathrm{m^2/s^2}}{625 \mathrm{m}} = 9.0 \mathrm{m/s^2}.\]
03

Find Total Acceleration

Since the tangential acceleration is zero (constant speed), the total acceleration is purely radial. Thus, the magnitude of total acceleration is the same as the radial (centripetal) acceleration: \( 9.0 \mathrm{m/s^2} \).
04

Determine the Direction of Total Acceleration

The total acceleration direction is the same as the radial (centripetal) acceleration direction, which is always towards the center of the circle. Therefore, relative to the radial direction, the total acceleration direction is purely radial without any tangential component.

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Key Concepts

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

Centripetal Acceleration
Centripetal acceleration is a key concept when dealing with objects in circular motion. This acceleration is always directed towards the center of the circular path the object is following. Although the object is traveling in a path that may seem constant, the continuous change in direction results in this type of acceleration.Since the car in the example travels at a constant speed, its speed doesn't change, but its direction does. The formula to calculate centripetal acceleration is \(a_r = \frac{v^2}{r}\), where \(v\) is the tangential speed, and \(r\) is the radius of the circle.
  • For the car, \(v = 75.0 \, \mathrm{m/s}\) and \(r = 625 \, \mathrm{m}\).
  • Substitute these values into the formula: \(a_r = \frac{(75.0 \, \mathrm{m/s})^2}{625 \, \mathrm{m}} = 9.0 \, \mathrm{m/s^2}\).
This centripetal acceleration ensures that the car continues its circular trajectory by constantly steering it towards the center of the track. If this force were absent, the car would move off in a straight line.
Tangential Speed
Tangential speed, sometimes referred to as linear speed in the context of a circular path, determines how fast an object is moving along its curved path, measured in meters per second (m/s). In the provided example, the car maintains a constant tangential speed of \(75.0 \, \mathrm{m/s}\).The key point is that while the speed remains constant, what differentiates tangential speed from linear speed on a straight path is the direction of movement. Since it's following a circular path, this direction is continuously changing, influencing how we perceive motion.
  • A constant tangential speed means there's no tangential acceleration within the motion.
  • The absence of tangential acceleration indicates that the object's speed along the path does not increase or decrease.
Thus, in this exercise, because the speed is constant, the tangential acceleration remains zero, and the car's total acceleration is entirely centripetal, keeping the car on its circular path.
Radial Direction
The radial direction is the line that points directly from the perimeter of the circle towards its center. In contexts involving circular motion, understanding radial direction is crucial. It is this direction that the centripetal force and acceleration are aligned with. In the context of this problem, the radial direction defines where the car's total acceleration is headed—straight towards the circle's middle.
  • For any object moving in a circle, the radial direction changes as the object travels along the curve.
  • In our car example, total acceleration is radially inward, meaning it's always pulling the car towards the track's core.
No tangential components exist in the case presented, as there is no speed change, emphasizing that radial direction and centripetal forces are significant in maintaining the car's circular motion.

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Most popular questions from this chapter

A pitcher throws a curveball that reaches the catcher in 0.60 s. The ball curves because it is spinning at an average angular velocity of 330 rev/min (assumed constant) on its way to the catcher’s mitt. What is the angular displacement of the baseball (in radians) as it travels from the pitcher to the catcher?

Conceptual Example 2 provides some relevant background for this problem. A jet is circling an airport control tower at a distance of \(18.0 \mathrm{km}\) An observer in the tower watches the jet cross in front of the moon. As seen from the tower, the moon subtends an angle of \(9.04 \times 10^{-3}\) radians. Find the distance traveled (in meters) by the jet as the observer watches the nose of the jet cross from one side of the moon to the other.

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 \(v_{\mathrm{T}}=r \omega, \mathrm{a} \mathrm{CD}\) 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 rev/s. Find \((\mathrm{a})\) 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 CD.

The drive propeller of a ship starts from rest and accelerates at \(2.90 \times 10^{-3} \mathrm{rad} / \mathrm{s}^{2}\) for \(2.10 \times 10^{3} \mathrm{s} .\) For the next \(1.40 \times 10^{3} \mathrm{s}\) the propeller rotates at a constant angular speed. Then it decelerates at \(2.30 \times 10^{-3} \mathrm{rad} / \mathrm{s}^{2}\) until it slows (without reversing direction) to an angular speed of \(4.00 \mathrm{rad} / \mathrm{s}\). Find the total angular displacement of the propeller.

?wo people start at the same place and walk around a circular lake in opposite directions. One walks with an angular speed of \(1.7 \times 10^{-3}\) \(\mathrm{rad} / \mathrm{s},\) while the other has an angular speed of \(3.4 \times 10^{-3} \mathrm{rad} / \mathrm{s} .\) How long will it be before they meet?

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