The flywheel of a steam engine begins to rotate from rest with a constant angular acceleration of 1.23 rad/s². It accelerates for 31.9 s, then maintains a constant angular velocity. Calculate the total angle through which the wheel has turned 65.5 s after it begins rotating.

Physics for Scientists and Engineers: Foundations and Connections
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Author:Katz, Debora M.
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Chapter12: Rotation I: Kinematics And Dynamics
Section: Chapter Questions
Problem 63PQ
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**Problem Statement:**

The flywheel of a steam engine begins to rotate from rest with a constant angular acceleration of 1.23 rad/s². It accelerates for 31.9 seconds, then maintains a constant angular velocity. Calculate the total angle through which the wheel has turned 65.5 seconds after it begins rotating.

**Solution Approach:**

1. **Calculate Angular Displacement During Acceleration Phase:**
   - Use the formula for angular displacement during uniform acceleration:
     \[
     \theta_1 = \omega_0 t + \frac{1}{2} \alpha t^2
     \]
   - Where \(\omega_0 = 0\) (initial angular velocity), \(\alpha = 1.23 \, \text{rad/s}^2\), and \(t = 31.9 \, \text{s}\).

2. **Calculate Angular Velocity at the End of Acceleration Phase:**
   - Angular velocity (\(\omega\)) can be calculated using:
     \[
     \omega = \omega_0 + \alpha t
     \]

3. **Calculate Angular Displacement During Constant Velocity Phase:**
   - After 31.9 seconds, the wheel travels with constant angular velocity for \(65.5 - 31.9 = 33.6 \, \text{s}\).
   - Angular displacement during this phase (\(\theta_2\)):
     \[
     \theta_2 = \omega \times 33.6 \, \text{s}
     \]

4. **Calculate Total Angular Displacement:**
   - Total angle \(\theta\) is the sum of \(\theta_1\) and \(\theta_2\):
     \[
     \theta = \theta_1 + \theta_2
     \]

**Visual Aid:**

- No graphs or diagrams are present in the original problem statement. A detailed step-by-step explanation accompanies the problem for clarity.
Transcribed Image Text:**Problem Statement:** The flywheel of a steam engine begins to rotate from rest with a constant angular acceleration of 1.23 rad/s². It accelerates for 31.9 seconds, then maintains a constant angular velocity. Calculate the total angle through which the wheel has turned 65.5 seconds after it begins rotating. **Solution Approach:** 1. **Calculate Angular Displacement During Acceleration Phase:** - Use the formula for angular displacement during uniform acceleration: \[ \theta_1 = \omega_0 t + \frac{1}{2} \alpha t^2 \] - Where \(\omega_0 = 0\) (initial angular velocity), \(\alpha = 1.23 \, \text{rad/s}^2\), and \(t = 31.9 \, \text{s}\). 2. **Calculate Angular Velocity at the End of Acceleration Phase:** - Angular velocity (\(\omega\)) can be calculated using: \[ \omega = \omega_0 + \alpha t \] 3. **Calculate Angular Displacement During Constant Velocity Phase:** - After 31.9 seconds, the wheel travels with constant angular velocity for \(65.5 - 31.9 = 33.6 \, \text{s}\). - Angular displacement during this phase (\(\theta_2\)): \[ \theta_2 = \omega \times 33.6 \, \text{s} \] 4. **Calculate Total Angular Displacement:** - Total angle \(\theta\) is the sum of \(\theta_1\) and \(\theta_2\): \[ \theta = \theta_1 + \theta_2 \] **Visual Aid:** - No graphs or diagrams are present in the original problem statement. A detailed step-by-step explanation accompanies the problem for clarity.
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