At time t = 0 the angular velocity of the grinding wheel is 27.0 rad/s. The wheel speed increases at a constant acceleration of
31.0 rad/s2 until the circuit breaker trips at time t = 3.00 s. After this, the wheel rotates an additional
angle of 441 rad before it comes to a complete stop (the speed slows down at a constant acceleration).
a) What is the total angle the wheel rotates from time t = 0 to the moment the wheel stops?
b) At what time did the wheel stop?
c) What was the acceleration of the wheel as it slowed down?

### To solve the problem, we will use equations of rotational motion:

1. **Initial Phase (t = 0 to t = 3.00 s):**
   - Initial angular velocity (\( \omega_0 \)) = 27.0 rad/s
   - Angular acceleration (\( \alpha_1 \)) = 31.0 rad/s²
   - Time (\( t_1 \)) = 3.00 s

   #### Angular velocity at t = 3.00 s:
   \[
   \omega_1 = \omega_0 + \alpha_1 t
   \]
   \[
   \omega_1 = 27.0 + (31.0)(3.00) = 27.0 + 93.0 = 120.0 \, \text{rad/s}
   \]

   #### Angular displacement during this phase:
   \[
   \theta_1 = \omega_0 t + \frac{1}{2} \alpha_1 t^2
   \]
   \[
   \theta_1 = (27.0)(3.00) + \frac{1}{2}(31.0)(3.00)^2
   \]
   \[
   \theta_1 = 81.0 + \frac{1}{2}(31.0)(9.00) = 81.0 + 139.5 = 220.5 \, \text{rad}
   \]

2. **Second Phase (Deceleration to a stop):**
   - Final angular displacement = 441 rad
   - Initial angular velocity for this phase = \( \omega_1 = 120.0 \, \text{rad/s} \)
   - Final angular velocity = 0 (stopped)
   - Angular displacement (\( \theta_2 \)) = 441 rad
   - Use the equation:
     \[
     \omega^2 = \omega_0^2 + 2\alpha \theta
     \]
     Solve for \( \alpha_2 \):
     \[
     0 = (120.0)^2 + 2\alpha_2 (441)
     \]
     \[
     14400 = -2\alpha_2 (441)
     \]
     \[
     \alpha_2 = -\frac{14400}{2(441)} = -\frac{14400}{882} = -16.33 \, \text{rad/s}^2
     \]

   #### Time taken to decelerate (\( t_2 \)):
   Using the equation:
   \[
   \omega = \omega_0 + \alpha_2 t
   \]
   \[
   0 = 120.0 + (-16.33)t_2
   \]
   \[
   t_2 = \frac{120.0}{16.33} = 7.35 \, \text{s}
   \]

3. **Total Results:**
   a) **Total angle rotated (\( \theta_{\text{total}} \)):**
   \[
   \theta_{\text{total}} = \theta_1 + \theta_2
   \]
   \[
   \theta_{\text{total}} = 220.5 + 441 = 661.5 \, \text{rad}
   \]

   b) **Total time (\( t_{\text{total}} \)):**
   \[
   t_{\text{total}} = t_1 + t_2
   \]
   \[
   t_{\text{total}} = 3.00 + 7.35 = 10.35 \, \text{s}
   \]

   c) **Deceleration (\( \alpha_2 \)):**
   \[
   \alpha_2 = -16.33 \, \text{rad/s}^2
   \]

### Final Answers:
a) Total angle rotated: \( 661.5 \, \text{rad} \)  
b) Total time: \( 10.35 \, \text{s} \)  
c) Deceleration: \( -16.33 \, \text{rad/s}^2 \)  

Let me know if you'd like further clarifications or detailed explanations.

---

**5 Related Questions:**
1. What is the average angular velocity of the wheel during its entire motion?
2. How much energy is dissipated during the deceleration phase, assuming the wheel has a moment of inertia \( I \)?
3. What is the total distance traveled by a point at the rim of the wheel (if the radius is known)?
4. How would the results change if the acceleration were not constant?
5. How can these equations be adapted for non-rotational (linear) motion?

**Tip:** Always verify your kinematic equations by double-checking unit consistency and physical interpretations of the results.

At time t = 0 the angular velocity of the grinding wheel is 27.0 rad/s. The wheel speed increases at a constant acceleration of 31.0 rad/s² until the circuit breaker trips at time t = 3.00 s. After this, the wheel rotates an additional angle of 441 rad before it comes to a complete stop (the speed slows down at a constant acceleration). a) What is the total angle the wheel rotates from time t = 0 to the moment the wheel stops? b) At what time did the wheel stop? c) What was the acceleration of the wheel as it slowed down?

Discover how to solve a rotational kinematics problem involving angular velocity, acceleration, and displacement. This example explores the total angle rotated, time to stop, and deceleration of a grinding wheel. Suitable for Grades 10-12 physics students.

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