Forest UCM NLM Oscilations - Revision history

Foretony: /* The phi-hat direction */

2014-08-31T12:38:08Z

The phi-hat direction

2014-08-31T12:34:27Z

Skate boarder in Half pipe

2014-08-31T12:33:14Z

The phi-hat direction

2014-08-31T12:30:55Z

The phi-hat direction

2014-08-31T12:30:35Z

The phi-hat direction

2014-08-31T12:22:11Z

The phi-hat direction

2014-08-31T12:19:18Z

The phi-hat direction

2014-08-31T12:18:00Z

The phi-hat direction

2014-08-31T12:17:06Z

The r-hat direction

2014-08-31T12:16:15Z

The r-hat direction

← Older revision		Revision as of 12:38, 31 August 2014
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	:<math>\phi(t) = \phi_0 \cos ( \omega t)</math>		:<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]]		[[Forest_UCM_NLM#Oscillatiions]]

← Older revision		Revision as of 12:34, 31 August 2014
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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.
		+
		+	Determine its equation of motion

	==Step 1: System==		==Step 1: System==

@@ Line 72: / Line 72: @@
 In this problem we can say that at time t=0
-:<math>\theta(t=0) = \theta_0</math>
+:<math>\theta(t=0) = \theta_0= A \cos (0) + B \sin (0) = A</math>
 and
-:<math>\dot \theta (t=0) = 0</math>
+:<math>\dot \theta (t=0) = 0 = - A \omega \sin(0) + B \cos(0) =B</math>
 [[Forest_UCM_NLM#Oscillatiions]]

@@ Line 72: / Line 72: @@
 In this problem we can say that at time t=0
-\theta(t=0) = \theta_0
+:<math>\theta(t=0) = \theta_0</math>
 and
-\dot \theta (t=0) = 0
+:<math>\dot \theta (t=0) = 0</math>
 [[Forest_UCM_NLM#Oscillatiions]]

← Older revision		Revision as of 12:30, 31 August 2014
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	: <math>\theta = A \cos (\omega t) + B \sin (\omega t)</math>		: <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
		+
		+	\theta(t=0) = \theta_0
		+	and
		+	\dot \theta (t=0) = 0


	[[Forest_UCM_NLM#Oscillatiions]]		[[Forest_UCM_NLM#Oscillatiions]]

@@ Line 51: / Line 51: @@
 Then
 :<math>\sin \theta \approx \theta</math>
 [[Forest_UCM_NLM#Oscillatiions]]

@@ Line 42: / Line 42: @@
 === The phi-hat direction===
-:<math>-mg \sin \theta = mR ddot \phi</math>
+:<math>-mg \sin \theta = mR \ddot \phi</math>
 : <math>\ddot \phi = - \frac{g}{R} \sin \theta</math>
 [[Forest_UCM_NLM#Oscillatiions]]

← Older revision		Revision as of 12:17, 31 August 2014
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	::<math>a_c = \frac{v^2}{R} = R\dot{\phi}^2 =</math> centripetal acceleration		::<math>a_c = \frac{v^2}{R} = R\dot{\phi}^2 =</math> centripetal acceleration
		+
		+	=== The phi-hat direction===
		+
		+	:-mg \sin \theta = mR ddot \phit

	[[Forest_UCM_NLM#Oscillatiions]]		[[Forest_UCM_NLM#Oscillatiions]]

@@ Line 35: / Line 35: @@
 :<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 = m \left ( g \cos \theta +  R\dot{\phi}^2 \right )</math>
 [[Forest_UCM_NLM#Oscillatiions]]