Math / Calculus

QuestionAs you will learn in Chapter 13, the angular acceleration of a simple pendulum is given by $\ddot{\theta}=-(g / L) \sin \theta$ , where $g$ is the acceleration of gravity and $L$ is the length of the pendulum cord.
Problem 12.74. Derive the expression of the angular velocity $\ddot{\theta}$ as a function of the angular coordinate $\theta$ . The initial conditions are $\theta(0)=\theta_{0}$ and $\dot{\theta}(0)=\dot{\theta}_{0}$
Answer Problem 12.75 Let the length of the pendulum cord be $L=1.5 \mathrm{~m}$ . If $\dot{\theta}=3.7 \mathrm{rad} / \mathrm{s}$ when $\theta=14^{\circ}$ , determine the maximum value of $\theta$ achieved by the pendulum.
Problem 12.76 The given angular acceleration remains valid even if the pendulum cord is replaced by a massless rigid bar. For this case, let $L=5.3 \mathrm{ft}$ and assume that the pendulum is placed in motion at $\theta=0^{\circ}$ . What is the minimum angular velocity at this position for the pendulum to swing through a full circle?
Answer $\dot{\theta}_{\min }=4.930 \mathrm{rad} / \mathrm{s}$

Studdy Solution

STEP 1

1. The angular acceleration of a simple pendulum is given by $\ddot{\theta}=-(g / L) \sin \theta$ .
2. The initial conditions are $\theta(0)=\theta_{0}$ and $\dot{\theta}(0)=\dot{\theta}_{0}$ .
3. The length of the pendulum cord is $L=1.5 \mathrm{~m}$ .
4. When $\theta=14^{\circ}$ , $\dot{\theta}=3.7 \mathrm{rad} / \mathrm{s}$ .

STEP 2

1. Derive the expression for angular velocity $\dot{\theta}$ as a function of angular coordinate $\theta$ .
2. Determine the maximum value of $\theta$ achieved by the pendulum using the initial conditions and conservation of energy.

STEP 3

Start with the given differential equation for angular acceleration:
$\ddot{\theta} = -\frac{g}{L} \sin \theta$

STEP 4

Multiply both sides by $\dot{\theta}$ to facilitate integration:
$\ddot{\theta} \dot{\theta} = -\frac{g}{L} \sin \theta \dot{\theta}$

STEP 5

Recognize that the left side is the derivative of $\frac{1}{2} \dot{\theta}^2$ and the right side can be integrated with respect to $\theta$ :
$\frac{d}{dt} \left( \frac{1}{2} \dot{\theta}^2 \right) = -\frac{g}{L} \frac{d}{dt} (\cos \theta)$

STEP 6

Integrate both sides with respect to $t$ :
$\frac{1}{2} \dot{\theta}^2 = \frac{g}{L} \cos \theta + C$

STEP 7

Determine the constant of integration $C$ using the initial conditions $\theta(0)=\theta_{0}$ and $\dot{\theta}(0)=\dot{\theta}_{0}$ :
$\frac{1}{2} \dot{\theta}_{0}^2 = \frac{g}{L} \cos \theta_{0} + C$ $C = \frac{1}{2} \dot{\theta}_{0}^2 - \frac{g}{L} \cos \theta_{0}$

STEP 8

Substitute $C$ back into the expression:
$\frac{1}{2} \dot{\theta}^2 = \frac{g}{L} \cos \theta + \left( \frac{1}{2} \dot{\theta}_{0}^2 - \frac{g}{L} \cos \theta_{0} \right)$

STEP 9

Solve for $\dot{\theta}$ :
$\dot{\theta}^2 = \dot{\theta}_{0}^2 - \frac{2g}{L} (\cos \theta - \cos \theta_{0})$ $\dot{\theta} = \sqrt{\dot{\theta}_{0}^2 - \frac{2g}{L} (\cos \theta - \cos \theta_{0})}$

STEP 10

Use the expression for $\dot{\theta}$ to determine the conditions for maximum $\theta$ . At the maximum value of $\theta$ , $\dot{\theta} = 0$ :
$0 = \dot{\theta}_{0}^2 - \frac{2g}{L} (\cos \theta_{\max} - \cos \theta_{0})$

STEP 11

Solve for $\cos \theta_{\max}$ :
$\cos \theta_{\max} = \cos \theta_{0} + \frac{L \dot{\theta}_{0}^2}{2g}$

STEP 12

Convert the given values to the appropriate units and substitute: $L = 1.5 \mathrm{~m}$ , $\dot{\theta}_{0} = 3.7 \mathrm{rad} / \mathrm{s}$ , and $\theta_{0} = 14^{\circ}$ .
$\cos \theta_{\max} = \cos 14^{\circ} + \frac{1.5 \cdot (3.7)^2}{2 \cdot 9.81}$

STEP 13

Evaluate $\cos 14^{\circ}$ and the other terms:
$\cos 14^{\circ} \approx 0.9703$ $\frac{1.5 \cdot (3.7)^2}{2 \cdot 9.81} \approx 1.047$ $\cos \theta_{\max} = 0.9703 + 1.047 \approx 2.0173$

STEP 14

Since $\cos \theta$ cannot be greater than 1, this indicates that the pendulum would theoretically achieve a full swing, but we need to verify the physical feasibility:
Given the calculated value exceeds 1, the physical maximum $\theta$ would be when $\dot{\theta} = 0$ at the highest point of the swing, but practically the pendulum reaches closer to $\theta = \pi$ radians if it has enough energy. Therefore, the maximum value of $\theta$ would theoretically be close to $180^{\circ}$ or $\pi$ radians.

Was this helpful?

Studdy Solution

STEP 1

STEP 2

1. Derive the expression for angular velocity $\dot{\theta}$ as a function of angular coordinate $\theta$ .
2. Determine the maximum value of $\theta$ achieved by the pendulum using the initial conditions and conservation of energy.

STEP 3

Start with the given differential equation for angular acceleration:
$\ddot{\theta} = -\frac{g}{L} \sin \theta$

STEP 4

Multiply both sides by $\dot{\theta}$ to facilitate integration:
$\ddot{\theta} \dot{\theta} = -\frac{g}{L} \sin \theta \dot{\theta}$

STEP 5

Recognize that the left side is the derivative of $\frac{1}{2} \dot{\theta}^2$ and the right side can be integrated with respect to $\theta$ :
$\frac{d}{dt} \left( \frac{1}{2} \dot{\theta}^2 \right) = -\frac{g}{L} \frac{d}{dt} (\cos \theta)$

STEP 6

Integrate both sides with respect to $t$ :
$\frac{1}{2} \dot{\theta}^2 = \frac{g}{L} \cos \theta + C$

STEP 7

Determine the constant of integration $C$ using the initial conditions $\theta(0)=\theta_{0}$ and $\dot{\theta}(0)=\dot{\theta}_{0}$ :
$\frac{1}{2} \dot{\theta}_{0}^2 = \frac{g}{L} \cos \theta_{0} + C$ $C = \frac{1}{2} \dot{\theta}_{0}^2 - \frac{g}{L} \cos \theta_{0}$

STEP 8

Substitute $C$ back into the expression:
$\frac{1}{2} \dot{\theta}^2 = \frac{g}{L} \cos \theta + \left( \frac{1}{2} \dot{\theta}_{0}^2 - \frac{g}{L} \cos \theta_{0} \right)$

STEP 9

Solve for $\dot{\theta}$ :
$\dot{\theta}^2 = \dot{\theta}_{0}^2 - \frac{2g}{L} (\cos \theta - \cos \theta_{0})$ $\dot{\theta} = \sqrt{\dot{\theta}_{0}^2 - \frac{2g}{L} (\cos \theta - \cos \theta_{0})}$

STEP 10

Use the expression for $\dot{\theta}$ to determine the conditions for maximum $\theta$ . At the maximum value of $\theta$ , $\dot{\theta} = 0$ :
$0 = \dot{\theta}_{0}^2 - \frac{2g}{L} (\cos \theta_{\max} - \cos \theta_{0})$

STEP 11

Solve for $\cos \theta_{\max}$ :
$\cos \theta_{\max} = \cos \theta_{0} + \frac{L \dot{\theta}_{0}^2}{2g}$

STEP 12

Convert the given values to the appropriate units and substitute: $L = 1.5 \mathrm{~m}$ , $\dot{\theta}_{0} = 3.7 \mathrm{rad} / \mathrm{s}$ , and $\theta_{0} = 14^{\circ}$ .
$\cos \theta_{\max} = \cos 14^{\circ} + \frac{1.5 \cdot (3.7)^2}{2 \cdot 9.81}$

STEP 13

Evaluate $\cos 14^{\circ}$ and the other terms:
$\cos 14^{\circ} \approx 0.9703$ $\frac{1.5 \cdot (3.7)^2}{2 \cdot 9.81} \approx 1.047$ $\cos \theta_{\max} = 0.9703 + 1.047 \approx 2.0173$

STEP 14

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

STEP 2

1. Derive the expression for angular velocity θ˙\dot{\theta}θ˙ as a function of angular coordinate θ\thetaθ.2. Determine the maximum value of θ\thetaθ achieved by the pendulum using the initial conditions and conservation of energy.

STEP 3

Start with the given differential equation for angular acceleration:θ¨=−gLsin⁡θ \ddot{\theta} = -\frac{g}{L} \sin \theta θ¨=−Lg​sinθ

STEP 4

Multiply both sides by θ˙\dot{\theta}θ˙ to facilitate integration:θ¨θ˙=−gLsin⁡θθ˙ \ddot{\theta} \dot{\theta} = -\frac{g}{L} \sin \theta \dot{\theta} θ¨θ˙=−Lg​sinθθ˙

STEP 5

STEP 6

Integrate both sides with respect to ttt:12θ˙2=gLcos⁡θ+C \frac{1}{2} \dot{\theta}^2 = \frac{g}{L} \cos \theta + C 21​θ˙2=Lg​cosθ+C

STEP 7

STEP 8

Substitute CCC back into the expression:12θ˙2=gLcos⁡θ+(12θ˙02−gLcos⁡θ0) \frac{1}{2} \dot{\theta}^2 = \frac{g}{L} \cos \theta + \left( \frac{1}{2} \dot{\theta}_{0}^2 - \frac{g}{L} \cos \theta_{0} \right) 21​θ˙2=Lg​cosθ+(21​θ˙02​−Lg​cosθ0​)

STEP 9

STEP 10

STEP 11

Solve for cos⁡θmax⁡\cos \theta_{\max}cosθmax​:cos⁡θmax⁡=cos⁡θ0+Lθ˙022g \cos \theta_{\max} = \cos \theta_{0} + \frac{L \dot{\theta}_{0}^2}{2g} cosθmax​=cosθ0​+2gLθ˙02​​

STEP 12

STEP 13

STEP 14

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Derive the expression for angular velocity $\dot{\theta}$ as a function of angular coordinate $\theta$ .
2. Determine the maximum value of $\theta$ achieved by the pendulum using the initial conditions and conservation of energy.

Start with the given differential equation for angular acceleration:
$\ddot{\theta} = -\frac{g}{L} \sin \theta$

Multiply both sides by $\dot{\theta}$ to facilitate integration:
$\ddot{\theta} \dot{\theta} = -\frac{g}{L} \sin \theta \dot{\theta}$

Integrate both sides with respect to $t$ :
$\frac{1}{2} \dot{\theta}^2 = \frac{g}{L} \cos \theta + C$

Substitute $C$ back into the expression:
$\frac{1}{2} \dot{\theta}^2 = \frac{g}{L} \cos \theta + \left( \frac{1}{2} \dot{\theta}_{0}^2 - \frac{g}{L} \cos \theta_{0} \right)$

Solve for $\cos \theta_{\max}$ :
$\cos \theta_{\max} = \cos \theta_{0} + \frac{L \dot{\theta}_{0}^2}{2g}$