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Chapter 7: Problem 29

In a noninjury, noncontact skid on icy pavement on an empty road, a car spins 1.75 revolutions while it skids to a halt. It was initially moving at \(15.0 \mathrm{~m} / \mathrm{s}\), and because of the ice it was able to decelerate at a rate of only \(1.50 \mathrm{~m} / \mathrm{s}^{2}\). Viewed from above, the car spun clockwise. Determine its average angular velocity as it spun and slid to a halt.

Short Answer

Expert verified
The average angular velocity is \(0.35\pi\) radians per second.

Step by step solution

01

Convert Revolutions to Radians

First, we need to convert the number of revolutions the car spins into radians. Since one revolution is equal to \(2\pi\) radians, 1.75 revolutions is:\[1.75 \times 2\pi = 3.5\pi \text{ radians.}\]
02

Calculate Time to Stop

Next, we will find the time it takes for the car to come to a complete stop. We use the formula for uniform acceleration: \[v_f = v_i + a \cdot t.\] Here, \(v_f = 0\) m/s, \(v_i = 15.0\) m/s, and \(a = -1.50\) m/s虏. Solving for \(t\):\[0 = 15 - 1.5t\implies t = \frac{15}{1.5} = 10 \text{ seconds.}\]
03

Calculate Average Angular Velocity

The average angular velocity \(\omega_{avg}\) is given by the formula \(\omega_{avg} = \frac{\Delta \theta}{\Delta t}\), where \(\Delta \theta\) is the angular displacement and \(\Delta t\) is the time. Using the values from previous steps: \[\omega_{avg} = \frac{3.5\pi}{10} = 0.35\pi \text{ radians/second.}\]

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Key Concepts

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

Revolutions to Radians
Angular motion involves converting revolutions into radians because radians are the standard unit for angular measurements in physics. To convert the number of revolutions to radians, we use the relationship that one complete revolution corresponds to an angle of \(2\pi\) radians. So, if you spin through 1.75 revolutions, you would calculate the radians by multiplying:
  • \(1.75 \times 2\pi\).
This gives us \(3.5\pi\) radians. Radians provide a more natural measure for angles, especially when dealing with circular motion, as they relate directly to the arc length of a circle and its radius.
Uniform Acceleration
Uniform acceleration refers to the constant rate at which an object speeds up or slows down. In the context of this problem, the car is decelerating uniformly, meaning its speed decreases at a constant rate due to the force applied in opposition to its motion (in this case, the friction on ice). The formula for uniform acceleration is expressed as:
  • \(v_f = v_i + a \cdot t\)
where:
  • \(v_f\) is the final velocity. For the car stopping, this is 0 m/s.
  • \(v_i\) is the initial velocity, which here is 15.0 m/s.
  • \(a\) is the acceleration (negative for deceleration), given as \(-1.50\, \text{m/s}^2\).
  • \(t\) is the time taken to stop.
By rearranging to solve for \(t\), we determine how long it takes for the car to stop, which helps in calculating other parameters of motion.
Stopping Time Calculation
To calculate stopping time under uniform acceleration, we again use the equation:
  • \(v_f = v_i + a \cdot t\)
Plug in the values:
  • Final velocity \(v_f = 0\, \text{m/s}\)
  • Initial velocity \(v_i = 15.0\, \text{m/s}\)
  • Acceleration \(a = -1.50\, \text{m/s}^2\)
This comes to:
  • \(0 = 15.0 - 1.5 \cdot t\)
Solving for \(t\), we find:
  • \(t = \frac{15.0}{1.5} = 10\) seconds
This means it takes 10 seconds for the car to slide to a halt. Knowing the stopping time is crucial for estimating the angular velocity during this period. It tells us how long the car remains in motion under the measured conditions.

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Most popular questions from this chapter

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