Math Problem Statement

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?

Solution

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

  1. Initial Phase (t = 0 to t = 3.00 s):

    • Initial angular velocity (ω0\omega_0) = 27.0 rad/s
    • Angular acceleration (α1\alpha_1) = 31.0 rad/s²
    • Time (t1t_1) = 3.00 s

    Angular velocity at t = 3.00 s:

    ω1=ω0+α1t\omega_1 = \omega_0 + \alpha_1 t ω1=27.0+(31.0)(3.00)=27.0+93.0=120.0rad/s\omega_1 = 27.0 + (31.0)(3.00) = 27.0 + 93.0 = 120.0 \, \text{rad/s}

    Angular displacement during this phase:

    θ1=ω0t+12α1t2\theta_1 = \omega_0 t + \frac{1}{2} \alpha_1 t^2 θ1=(27.0)(3.00)+12(31.0)(3.00)2\theta_1 = (27.0)(3.00) + \frac{1}{2}(31.0)(3.00)^2 θ1=81.0+12(31.0)(9.00)=81.0+139.5=220.5rad\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 = ω1=120.0rad/s\omega_1 = 120.0 \, \text{rad/s}
    • Final angular velocity = 0 (stopped)
    • Angular displacement (θ2\theta_2) = 441 rad
    • Use the equation: ω2=ω02+2αθ\omega^2 = \omega_0^2 + 2\alpha \theta Solve for α2\alpha_2: 0=(120.0)2+2α2(441)0 = (120.0)^2 + 2\alpha_2 (441) 14400=2α2(441)14400 = -2\alpha_2 (441) α2=144002(441)=14400882=16.33rad/s2\alpha_2 = -\frac{14400}{2(441)} = -\frac{14400}{882} = -16.33 \, \text{rad/s}^2

    Time taken to decelerate (t2t_2):

    Using the equation: ω=ω0+α2t\omega = \omega_0 + \alpha_2 t 0=120.0+(16.33)t20 = 120.0 + (-16.33)t_2 t2=120.016.33=7.35st_2 = \frac{120.0}{16.33} = 7.35 \, \text{s}

  3. Total Results: a) Total angle rotated (θtotal\theta_{\text{total}}): θtotal=θ1+θ2\theta_{\text{total}} = \theta_1 + \theta_2 θtotal=220.5+441=661.5rad\theta_{\text{total}} = 220.5 + 441 = 661.5 \, \text{rad}

    b) Total time (ttotalt_{\text{total}}): ttotal=t1+t2t_{\text{total}} = t_1 + t_2 ttotal=3.00+7.35=10.35st_{\text{total}} = 3.00 + 7.35 = 10.35 \, \text{s}

    c) Deceleration (α2\alpha_2): α2=16.33rad/s2\alpha_2 = -16.33 \, \text{rad/s}^2

Final Answers:

a) Total angle rotated: 661.5rad661.5 \, \text{rad}
b) Total time: 10.35s10.35 \, \text{s}
c) Deceleration: 16.33rad/s2-16.33 \, \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 II?
  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.

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Math Problem Analysis

Mathematical Concepts

Rotational Kinematics
Angular Displacement
Angular Velocity
Angular Acceleration

Formulas

ω₁ = ω₀ + α₁t
θ₁ = ω₀t + (1/2)α₁t²
ω² = ω₀² + 2αθ
t = (ω - ω₀)/α

Theorems

Equations of Motion for Rotational Systems

Suitable Grade Level

Grades 10-12