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A motorcycle accelerates uniformly from rest and reaches a linear speed of 22.0 m/s in a time of 9.00 s. The radius of each tire is 0.280 m. What is the magnitude of the angular acceleration of each tire?

Short Answer

Expert verified
The angular acceleration is 8.71 rad/s².

Step by step solution

01

Understanding the Problem

The motorcycle starts from rest and reaches a final speed of 22.0 m/s in 9.00 seconds. We are asked to find the angular acceleration of the tires, given their radius is 0.280 m. Angular acceleration is the change in angular velocity over time, which we can find using the relationship between linear and angular quantities.
02

Calculating Linear Acceleration

First, calculate the linear acceleration of the motorcycle using the formula for acceleration:\[ a = \frac{v_f - v_i}{t} \]where \( v_f = 22.0 \ \text{m/s} \) (final velocity), \( v_i = 0 \ \text{m/s} \) (initial velocity), and \( t = 9.00 \ \text{s} \). So,\[ a = \frac{22.0 - 0}{9.00} = 2.44 \ \text{m/s}^2 \]
03

Relating Linear and Angular Acceleration

Use the relationship between linear acceleration \( a \) and angular acceleration \( \alpha \):\[ a = \alpha \cdot r \]Here, \( a = 2.44 \ \text{m/s}^2 \) and \( r = 0.280 \ \text{m} \). Solve for \( \alpha \):
04

Calculating Angular Acceleration

Rearrange the formula to solve for angular acceleration \( \alpha \):\[ \alpha = \frac{a}{r} = \frac{2.44}{0.280} \]Perform the division to find the angular acceleration:\[ \alpha = 8.71 \ \text{rad/s}^2 \]
05

Conclusion

The magnitude of the angular acceleration of each tire is \( 8.71 \ \text{rad/s}^2 \).

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Key Concepts

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

Linear Acceleration
Linear acceleration is how quickly the speed of an object changes over time along a straight path. In the context of our problem, the motorcycle's linear acceleration can be determined using the formula for acceleration:
\[ a = \frac{v_f - v_i}{t} \]
where:
  • \( v_f = 22.0 \ \text{m/s} \) is the final speed,
  • \( v_i = 0 \ \text{m/s} \) is the starting speed (since it begins from rest),
  • and \( t = 9.00 \ \text{s} \) is the time over which this change happens.
Using these values, the motorcycle accelerates linearly at \( 2.44 \ \text{m/s}^2 \). This value reveals how rapidly the motorcycle increases its speed, and is crucial for finding other related quantities, such as the angular acceleration of the tires.
Relationship Between Linear and Angular Quantities
Understanding the relationship between linear and angular quantities is essential when dealing with rotational systems. One key aspect of this relationship is connecting linear acceleration (\( a \)) with angular acceleration (\( \alpha \)). This is done using the formula:
\[ a = \alpha \cdot r \]
where \( r \) is the radius of rotation. In our exercise:
  • \( a = 2.44 \ \text{m/s}^2 \) is the linear acceleration,
  • and \( r = 0.280 \ \text{m} \) is the radius of the tire.
This formula implies that the linear acceleration experienced at the edge of a rotating object is directly proportional to both its angular acceleration and its radius.
Therefore, by rearranging the formula to solve for \( \alpha \):
\[ \alpha = \frac{a}{r} \],
we can compute the angular acceleration, helping to visualize how rotational dynamics relate to everyday linear experiences.
Physics Problem-Solving
Solving physics problems effectively often involves breaking down complex situations into simpler parts. Here's a general approach you can follow when tackling similar exercises:
  • **Understand the Problem**: Clearly identify what quantities are given and what you need to find.
  • **List Relevant Formulas**: Write down formulas that connect the given quantities to the ones you need to find. In this problem, you need both linear and angular acceleration formulas.
  • **Calculate Step-by-Step**: Work through the problem systematically, one step at a time, as we did by finding the linear acceleration first and then using it to find the angular acceleration.
  • **Check Your Work**: Ensure your calculations make sense and your final answer has the correct units, reflecting the magnitude of a physical situation.
By applying systematic problem-solving techniques, you increase your chances of arriving at the correct answer while deepening your understanding of physical concepts.
Uniform Acceleration
Uniform acceleration refers to a constant acceleration, meaning the speed of the object increases by the same amount each second. In our exercise, both the linear and angular accelerations are uniform, which simplifies the calculations.
For uniform linear acceleration, constant formulas like:
\[ v_f = v_i + a t \] and\[ a = \frac{v_f - v_i}{t} \]
are particularly useful. It's the constancy of the acceleration that allows us to use these basic kinematic equations effectively without needing more complex calculus-based methods.
When acceleration is uniform, all derived quantities, such as velocity and displacement, follow predictable patterns. This predictability not only simplifies calculations but also provides insight into the motion's nature, making physics problems more intuitive to handle.

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Most popular questions from this chapter

A car is traveling with a speed of \(20.0 \mathrm{~m} / \mathrm{s}\) along a straight horizontal road. The wheels have a radius of \(0.300 \mathrm{~m} .\) If the car speeds up with a linear acceleration of \(1.50 \mathrm{~m} / \mathrm{s}^{2}\) for \(8.00 \mathrm{~s},\) find the angular displacement of each wheel during this period.

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