Question

Consider the pendulum given in the figure above. The motion of the object with mass M is governed by the following Ordinary Differential Equation. This equation also considers the air drag of the mass due to motion in air. \frac{d^2\theta}{dt^2} = -\frac{g}{L}sin(\theta) - \frac{\alpha}{M}\frac{d\theta}{dt} g = 9.81 m/s^2 M = 0.1 kg L = 1 m \alpha = 0.1 kg s^{-1} (Drag Coefficient) Initially (at t = 0), the angular position of the object is set to $\theta$ = 0 and it's angular velocity is $\omega = \frac{d\theta}{dt} = 0.1/s$ [5] a) Express this initial value problem with a 1st order ODE system and define the initial conditions accordingly as well. [10] b) Write a Matlab script that would solve for the motion of the pendulum within the time window of $0 \le t \le 2$ using ode solver 'ode45'. Finally, the script should also plot t vs. $\omega$ of the object. 

          Consider the pendulum given in the figure above.

The motion of the object with mass M is governed by the following Ordinary Differential Equation. This equation also considers the air drag of the mass due to motion in air.
\frac{d^2\theta}{dt^2} = -\frac{g}{L}sin(\theta) - \frac{\alpha}{M}\frac{d\theta}{dt}
g = 9.81 m/s^2
M = 0.1 kg
L = 1 m
\alpha = 0.1 kg s^{-1} (Drag Coefficient)
Initially (at t = 0), the angular position of the object is set to $\theta$ = 0 and it's angular velocity is $\omega = \frac{d\theta}{dt} = 0.1/s$
[5] a) Express this initial value problem with a 1st order ODE system and define the initial conditions accordingly as well.
[10] b) Write a Matlab script that would solve for the motion of the pendulum within the time window of $0 \le t \le 2$ using ode solver 'ode45'.
Finally, the script should also plot t vs. $\omega$ of the object.
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Consider the pendulum given in the figure above. The motion of the object with mass M is governed by the following Ordinary Differential Equation. This equation also considers the air drag of the mass due to motion in air. (d^2θ)/(dt^2) = -(g)/(L)sin(θ) - (α)/(M)(dθ)/(dt) g = 9.81 m/s^2 M = 0.1 kg L = 1 m α= 0.1 kg s^-1 (Drag Coefficient) Initially (at t = 0), the angular position of the object is set to θ = 0 and it's angular velocity is ω = (dθ)/(dt) = 0.1/s [5] a) Express this initial value problem with a 1st order ODE system and define the initial conditions accordingly as well. [10] b) Write a Matlab script that would solve for the motion of the pendulum within the time window of 0 ≤ t ≤ 2 using ode solver 'ode45'. Finally, the script should also plot t vs. ω of the object.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Consider the pendulum given in the figure above from Encyclopedia Britannica, Inc. The motion of the object with mass M is governed by the following Ordinary Differential Equation. This equation also considers the air drag of the mass due to motion in air. d^2θ/dt^2 + (a/M) * dθ/dt + (g/L) * sin(θ) = 0 where: g = 9.81 m/s^2 M = 0.1 kg L = 1 m a = 0.1 kg/s (Drag Coefficient) a) Express this initial value problem with a 1st order ODE system and define the initial conditions accordingly as well. b) Write a Matlab script that would solve for the motion of the pendulum within the time window of 0 < t < 2 using the ode solver 'ode45'. Finally, the script should also plot t vs. θ of the object.
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Transcript

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00:01 So in this question, we have a pendulum of length l, where this point here has a position x -tilda of t equals, or sorry, we're calling this psi of t equals a -cos omega -t.
00:27 And then let's say we have this angle here, theta, and this point here has a mass of m.
00:34 Then there's going to be a force here, m g.
00:43 And this is going to cause a restoring force.
00:48 This is the restoring force here, minus m g sine theta.
00:57 Now the position here, x, x is going to be l sine theta plus sigh of t.
01:14 So for small angles theta, we have x is approximately l theta plus xxai of t.
01:26 So l theta plus a cos omega -t.
01:33 Now there's also a drag force, fd, is minus b, dx by dt.
01:43 So that's the x components of that force.
01:47 And we're also, what else are we going to have? so, yeah.
01:56 So now we can say that, well, we've got this restoring force here.
02:06 So f restoring is minus mg sine theta, which is approximately minus mg theta.
02:15 But theta equals x over l minus a over l cos omega t.
02:26 So this restoring force here is minus mg times this stuff.
02:34 So we've got the restoring force in terms of x is going to be minus mg theta of x.
02:43 So minus m g over l x plus mg a over l cos omega t.
02:52 So now we've got these two restoring, well we've got these two forces which are acting on the object.
02:58 So we can write newton's second law, m, d2x by dt squared, is the sum of these two forces.
03:07 So minus b, the x by dt.
03:11 Now we're going to use g over l is omega -naught squared, because we're told that in the question.
03:15 So minus m omega -naut squared x plus m -a, omega -naut -squared, cos omega -t.
03:22 So now this is the equation m d2x by d t squared plus b d x by d t plus m omega nought squared x is equal to m omega nought squared a cos omega t.
03:44 Now let's say that x is the real part of a complex variable z equals x plus i y...
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