Q5.6-8E A double pendulum swinging in a ... [FREE SOLUTION]

Chapter 5: Q5.6-8E (page 288)

A double pendulum swinging in a vertical plane under the influence of gravity (see Figure5.35) satisfies the system
$(m_{1} + m_{2}) l_{1}^{2} 胃_{1}^{}'' + m_{2} l_{1} l_{2} 胃_{2}^{}'' + (m_{1} + m_{2}) l_{1} {驳胃}_{1} = 0 m_{2} l_{2}^{2} 胃_{2}^{}'' + m_{2} l_{1} l_{2} 胃_{1}^{}'' + m_{2} l_{2} {驳胃}_{2} = 0$
When $胃_{1}$ and $胃_{2}$ are small angles. Solve the system when
$m_{1} = 3 kg, m_{2} = 2 kg, l_{1} = l_{2} = 5 m, 胃_{1} (0) = 蟺 / 6, 胃_{2} {(0) = 胃}_{1}^{} {' (0) = 胃}_{2}^{}' (0) = 0$ .

Short Answer

Expert verified

The solutions for $胃_{1}$ and $胃_{2}$ are:

$胃_{1} (t) = \frac{蟺}{12} \cos 1.1 t + \frac{蟺}{12} \cos 2.31 t 胃_{2} (t) = \frac{5}{12} \cos 1.1 t - \frac{5}{12} \cos 2.31 t$

Step by step solution

Substituting the values of m1,m2,l1,l2

Substituting the given values for $m_{1}, m_{2}, l_{1}, l_{2}$ into the given system and taking $g = 9.8 m / s^{2}$ one will get:

$(3 + 2) 5^{2} 胃_{1}^{} {'' + 2 脳 5 脳 5 胃}_{2}^{}'' + (3 + 2) 5 脳 9 . 8 胃_{1} = 0 2 脳 5^{2} 胃_{2}^{} {'' + 2 脳 5 脳 5 胃}_{1}^{}'' + 2 脳 5 脳 9.8 胃_{2} = 0$

Dividing the first equation by 25 and the second by 10 one will get:

${5 胃}_{1}^{} {'' + 2 胃}_{2}^{}'' + 9.8 胃_{1} = 0 {5 胃}_{2}^{} {'' + 5 胃}_{1}^{}'' + 9.8 胃_{2} = 0$

One will rewrite this system in operator form:

$(5 D^{2} + 9.8) [胃_{1}] + 2 D^{2} [胃_{2}] = 0 5 D^{2} [胃_{1}] + (5 D^{2} + 9.8) [胃_{2}] = 0$

Finding θ1

One will eliminate $胃_{2}$ from the system. To do so one will multiply the first equation by $(D^{2} + 9.8)$ and the second by $- 2 D^{2}$ and then add those two equations together.

$\begin{array}{rcl} {(5 D^{2} + 9.8)}^{2} [胃_{1}] + 2 D^{2} (5 D^{2} + 9.8) [胃_{2}] & = & 0 \\ - 10 D^{4} - 2 D^{2} (5 D^{2} + 9.8) [胃_{2}] & = & 0 \\ (25 D^{4} + 98 D^{2} + 96.04 - 10 D^{4}) [胃_{1}] & = & 0 \\ (15 D^{4} + 98 D^{2} + 96.04) [胃_{1}] & = & 0 \end{array}$

Now, one can find a general solution for $胃_{1}$ . The auxiliary equation is:

$15 r^{4} + 98 r^{2} + 96.04 = 0$ , and its roots are:

$r^{2} = \frac{- 98 卤 \sqrt{9604 - 5762.4}}{30} = \frac{- 98 卤 61.98}{30} r_{1, 2}^{2} = - 1.2, r_{3, 4}^{2} = - 5.33 r_{1, 2} = 卤 \sqrt{1.2} i, r_{3, 4} = 卤 \sqrt{5.33} i r_{1, 2} = 卤 1.1 i, r_{3, 4} = 卤 2.31 i$

So, the general solution for $胃_{1}$ is $胃_{1} (t) = c_{1} \cos 1.1 t + c_{2} \sin 1.1 t + c_{3} \cos 2.31 t + c_{4} \sin 2.31 t$

Findingθ2

Now, one can find a solution for $胃_{2}$ from the first equation of the system

$\begin{array}{rcl} 2 D^{2} [胃_{2}] & = & - (5 D^{2} + 9.8) [胃_{1}] \\ D [胃_{1}] & = & - 1.1 c_{1} \sin 1.1 t + 1.1 c_{2} \cos 1.1 t - 2.31 c_{3} \sin 2.31 t + 2.31 c_{4} \cos 2.31 t \\ D^{2} [胃_{1}] & = & - 1.2 c_{1} \cos 1.1 t - 1.2 c_{2} \sin 1.1 t - 5.33 c_{3} \cos 2.31 t - 5.33 c_{4} \sin 2.31 t 2 \\ D^{2} [胃_{2}] & = & - 5 D^{2} [胃_{1}] - 9.8 胃_{1} \\ = & 6 c_{1} \cos 1.1 t + 6 c_{2} \sin 1.1 t + 26.65 c_{3} \cos 2.31 t + 26.65 c_{4} \sin 2.31 t - 9.8 c_{1} \cos 1.1 t \\ - 9.8 c_{2} \sin 1.1 t - 9.8 c_{3} \cos 2.31 t - 9.8 c_{4} \sin 2.31 t \\ = & - 3.8 c_{1} \cos 1.1 t - 3.8 c_{2} \sin 1.1 t + 16.85 c_{3} \cos 2.31 t + 16.85 c_{4} \sin 2.31 t \\ D^{2} [胃_{2}] & = & - 1.9 c_{1} \cos 1.1 t - 1.9 c_{2} \sin 1.1 t + 8.42 c_{3} \cos 2.31 t + 8.42 c_{4} \sin 2 \end{array}$

Integration

Integrating the previous equation twice one will get:

$\begin{array}{rcl} D [胃_{2}] & = & 鈭� (- 1.9 c_{1} \cos 1.1 t - 1.9 c_{2} \sin 1.1 t + 8.42 c_{3} \cos 2.31 t + 8.42 c_{4} \sin 2.31 t) dt \\ = & - 1.9 c_{1} 鈭� \cos 1.1 tdt - 1.9 c_{2} 鈭� \sin 1.1 tdt + 8.42 c_{3} 鈭� \cos 2.31 tdt + 8.42 c_{4} 鈭� \sin 2.31 tdt \\ = & - 1.73 c_{1} \sin 1.1 t + 1.73 c_{2} \cos 1.1 t + 3.65 c_{3} \sin 2.31 t - 3.65 c_{4} \cos 2.31 t + A \\ 胃_{2} & = & 鈭� (- 1.73 c_{1} \sin 1.1 t + 1.73 c_{2} \cos 1.1 t + 3.65 c_{3} \sin 2.31 t - 3.65 c_{4} \cos 2.31 t + A) dt \\ = & - 1.73 c_{1} 鈭� \sin 1.1 tdt + 1.73 c_{2} 鈭� \cos 1.1 tdt + 3.65 c_{3} 鈭� \sin 2.31 tdt - 3.65 c_{4} 鈭� \cos 2.31 tdt \\ + A 鈭� dt \\ = & 1.57 c_{1} \cos 1.1 t + 1.57 c_{2} \sin 1.1 t - 1.57 c_{3} \cos 2.31 t - 1.57 \sin 2.31 t + At + B \end{array}$

But this equation must satisfy the second equation of the system, which is:

$5 D^{2} [胃_{1}] + 5 D^{2} [胃_{2}] + 9.8 [胃_{2}] = 0$

Finding the solution for θ1 and θ2

Since one does not have any constant nor a term multiplying t in $D^{2} [胃_{1}]$ and $D^{2} [胃_{2}]$ , one cannot have it in $胃_{1}$ , and therefore $A = B = 0$ , so the solution for $胃_{2}$ is $胃_{2} (t) = 1 . 57 c_{1} \cos 1.1 t + 1.57 c_{2} \sin 1.1 t - 1.57 c_{3} \cos 2.31 t - 1.57 \sin 2.31 t$

The initial conditions give us:

$胃_{1} (0) = c_{1} + c_{3} = \frac{蟺}{6}, 胃_{2} (0) = 1 . 57 c_{1} - 1.57 c_{3} = 0 c_{1} = c_{3}, 2 c_{1} = \frac{蟺}{6}, c_{1} = c_{3} = \frac{蟺}{12} 胃_{1}^{}' (0) = 1 . 1 c_{2} + 2.31 c_{4} = 0, 胃_{2}^{}' (0) = 1 . 73 c_{2} - 3.65 c_{4} = 0 c_{2} = 2.11 c_{4}, 4.62 c_{4} = 0, c_{2} = c_{4} = 0$

Finally, the solutions for $胃_{1}$ and $胃_{2}$ are

$\begin{array}{rcl} 胃_{1} (t) = \frac{蟺}{12} \cos 1.1 t + \frac{蟺}{12} \cos 2.31 t \\ 胃_{2} (t) = 1 . 57 \frac{蟺}{12} \cos 1.1 t - 1.57 \frac{蟺}{12} \cos 2.31 t \\ 胃_{1} (t) = \frac{蟺}{12} \cos 1.1 t + \frac{蟺}{12} \cos 2.31 t \\ 胃_{2} (t) = \frac{5}{12} \cos 1.1 t - \frac{5}{12} \cos 2.31 t \end{array}$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

91影视

Short Answer

Step by step solution

Substituting the values of m1,m2,l1,l2

Finding θ1

Findingθ2

Integration

Finding the solution for θ1 and θ2

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Theoretical and Mathematical Physics

Statistics

Decision Maths

Discrete Mathematics

Applied Mathematics

Study anywhere. Anytime. Across all devices.