Difference between revisions of "Forest UCM NLM Oscilations"

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Note: because the skateboard is frictionless, its wheels are not going to turn.
 
Note: because the skateboard is frictionless, its wheels are not going to turn.
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Determine its equation of motion
  
 
==Step 1: System==
 
==Step 1: System==
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The skateboard of mass <math>m</math> is the system.
 
The skateboard of mass <math>m</math> is the system.
  
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== Step 1: Coordinate system==
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Polar coordinate may be a good coordinate system to use since the skateboard's motion will be along the half circle.
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[[File:Forest_UCM_NLM_Osc_Fig1.png | 300 px]]
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==Step 3: Free Body Diagram==
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==Step 4: External Force vectors==
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:<math>\vec{F}_g = -mg \cos \theta \hat{r} - mg \sin \theta \hat{\phi}</math>
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:<math>\vec{N} = N \hat{r}</math>
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==Step 5: apply Netwon's 2nd Law==
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: <math>\sum \vec{F}_{ext} = \vec{F}_g + \vec{N} = m  \left (  \ddot{r}  -r\dot{\phi}^2 \right) \hat{r}  + \left ( 2\dot{r} \dot{\phi} +r \ddot{\phi} \right ) \hat{\phi}</math>
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For the case of circular motion at constant <math> r=R, \dot{r} = 0</math>
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:<math>\vec{F}_g + \vec{N}  =  m \left ( -R\dot{\phi}^2 \hat{r}  + R \ddot{\phi} \hat{\phi} \right ) </math>
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=== The r-hat direction===
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:<math>mg \cos \theta - N = -m R\dot{\phi}^2</math>
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:<math>N = m \left ( g \cos \theta +  R\dot{\phi}^2 \right )</math>
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:<math>N = mg \cos \theta + ma_c</math>
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::<math>a_c = \frac{v^2}{R} = R\dot{\phi}^2 =</math> centripetal acceleration
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=== The phi-hat direction===
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:<math>-mg \sin \theta = mR \ddot \phi</math>
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: <math>\ddot \phi = - \frac{g}{R} \sin \theta</math>
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; small angle approximation
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If we release the skateboard close to the bottom
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Then
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:<math>\sin \theta \approx \theta</math>
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This approximation make the differential equation easier to solve as it looks more like the simple harmonic motion typical of a spring.
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: <math>\ddot \phi = - \frac{g}{R} \theta</math>
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where
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:<math>\omega = \sqrt{\frac{g}{R}} =</math> oscillation frequency
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The general solution for this second order differential equation may be written as
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: <math>\theta = A \cos (\omega t) + B \sin (\omega t)</math>
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;Theorem: the solutions of an nth order equation contain n independent constants
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Once you have a general solution to a differential equation, your next step is to evaluate the constant A and B using constriant given in the problem such as the initial conditions ( t=0) or boundary conditions.
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In this problem we can say that at time t=0
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:<math>\theta(t=0) = \theta_0= A \cos (0) + B \sin (0) = A</math>
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and
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:<math>\dot \theta (t=0) = 0 = - A \omega \sin(0) + B \cos(0) =B</math>
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thus the specific solution is
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:<math>\phi(t) = \phi_0 \cos ( \omega t)</math>
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=Time for skateboard to go down and come back up=
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The time the skateboard takes to go down the semi-circle, rise to the opposite side from its relese point, and then return to its release point is just the period of oscillation.
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: <math>\tau = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{g}{R}}</math>
  
 
[[Forest_UCM_NLM#Oscillatiions]]
 
[[Forest_UCM_NLM#Oscillatiions]]

Latest revision as of 12:38, 31 August 2014

Skate boarder in Half pipe

Consider a frictionless skateboard released from the top of a semi-circle (half pipe) and oriented to fall directly towards the bottom. The semi-circle has a radius [math]R[/math] and the skateboard has a mass [math]m[/math].

Note: because the skateboard is frictionless, its wheels are not going to turn.

Determine its equation of motion

Step 1: System

The skateboard of mass [math]m[/math] is the system.

Step 1: Coordinate system

Polar coordinate may be a good coordinate system to use since the skateboard's motion will be along the half circle.

300 px

Step 3: Free Body Diagram

Step 4: External Force vectors

[math]\vec{F}_g = -mg \cos \theta \hat{r} - mg \sin \theta \hat{\phi}[/math]
[math]\vec{N} = N \hat{r}[/math]

Step 5: apply Netwon's 2nd Law

[math]\sum \vec{F}_{ext} = \vec{F}_g + \vec{N} = m \left ( \ddot{r} -r\dot{\phi}^2 \right) \hat{r} + \left ( 2\dot{r} \dot{\phi} +r \ddot{\phi} \right ) \hat{\phi}[/math]

For the case of circular motion at constant [math] r=R, \dot{r} = 0[/math]

[math]\vec{F}_g + \vec{N} = m \left ( -R\dot{\phi}^2 \hat{r} + R \ddot{\phi} \hat{\phi} \right ) [/math]

The r-hat direction

[math]mg \cos \theta - N = -m R\dot{\phi}^2[/math]
[math]N = m \left ( g \cos \theta + R\dot{\phi}^2 \right )[/math]
[math]N = mg \cos \theta + ma_c[/math]
[math]a_c = \frac{v^2}{R} = R\dot{\phi}^2 =[/math] centripetal acceleration

The phi-hat direction

[math]-mg \sin \theta = mR \ddot \phi[/math]
[math]\ddot \phi = - \frac{g}{R} \sin \theta[/math]
small angle approximation

If we release the skateboard close to the bottom

Then

[math]\sin \theta \approx \theta[/math]

This approximation make the differential equation easier to solve as it looks more like the simple harmonic motion typical of a spring.

[math]\ddot \phi = - \frac{g}{R} \theta[/math]

where

[math]\omega = \sqrt{\frac{g}{R}} =[/math] oscillation frequency

The general solution for this second order differential equation may be written as

[math]\theta = A \cos (\omega t) + B \sin (\omega t)[/math]
Theorem
the solutions of an nth order equation contain n independent constants


Once you have a general solution to a differential equation, your next step is to evaluate the constant A and B using constriant given in the problem such as the initial conditions ( t=0) or boundary conditions.


In this problem we can say that at time t=0

[math]\theta(t=0) = \theta_0= A \cos (0) + B \sin (0) = A[/math]

and

[math]\dot \theta (t=0) = 0 = - A \omega \sin(0) + B \cos(0) =B[/math]

thus the specific solution is

[math]\phi(t) = \phi_0 \cos ( \omega t)[/math]

Time for skateboard to go down and come back up

The time the skateboard takes to go down the semi-circle, rise to the opposite side from its relese point, and then return to its release point is just the period of oscillation.

[math]\tau = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{g}{R}}[/math]

Forest_UCM_NLM#Oscillatiions