Aaron Boyd's Assignment 8 - Revision history

Aaron.boyd at 18:16, 19 November 2010

2010-11-19T18:16:39Z

← Older revision		Revision as of 11:16, 19 November 2010
Line 58:		Line 58:
	\\		\\
	\\		\\
	\theta(t) = \~~theta_0cosh~~(t\sqrt(\frac{g}{L}))\\		\theta(t) = \theta_0cos(t\sqrt(\frac{g}{L}))\\
	\end{align}		\end{align}
	</math>		</math>

Aaron.boyd at 20:27, 1 November 2010

2010-11-01T20:27:40Z

← Older revision		Revision as of 13:27, 1 November 2010
Line 58:		Line 58:
	\\		\\
	\\		\\
	\theta(t) = ~~cosh~~(t\sqrt(\frac{g}{L}))\\		\theta(t) = \theta_0cosh(t\sqrt(\frac{g}{L}))\\
	\end{align}		\end{align}
	</math>		</math>

Aaron.boyd at 20:24, 1 November 2010

2010-11-01T20:24:35Z

← Older revision		Revision as of 13:24, 1 November 2010
Line 5:		Line 5:
	F_y &= T\cos(\theta)-mg = 0		F_y &= T\cos(\theta)-mg = 0
	\end{align}</math>		\end{align}</math>


	Polar coordinates may be easier to use, lets try that.		Polar coordinates may be easier to use, lets try that.
	~~now:~~


	<math>\begin{align}		<math>\begin{align}
	F_r &= T - mg\cos(\theta) = 0\\		F_r &= T - mg\cos(\theta) = 0\\

Aaron.boyd at 19:02, 1 November 2010

2010-11-01T19:02:38Z

← Older revision		Revision as of 12:02, 1 November 2010
Line 1:		Line 1:
	I decided to use laplace transforms to solve a pendulum equation. A pendulum with a weight of mass m and a massless rod length L is released from an initial angle \theta<sub>0</sub>. Find a function to determine the angle at any time t.		I decided to use laplace transforms to solve a pendulum equation. A pendulum with a weight of mass m and a massless rod length L is released from an initial angle <math>\theta<sub>0</sub></math>. Find a function to determine the angle at any time t.
	The summation of forces yields		The summation of forces yields
	<math>\begin{align}		<math>\begin{align}
Line 5:		Line 5:
	F_y &= T\cos(\theta)-mg = 0		F_y &= T\cos(\theta)-mg = 0
	\end{align}</math>		\end{align}</math>



	Polar coordinates may be easier to use, lets try that.		Polar coordinates may be easier to use, lets try that.

	now:		now:


	<math>\begin{align}		<math>\begin{align}
	F_r &= T - mg\cos(\theta) = 0\\		F_r &= T - mg\cos(\theta) = 0\\
Line 31:		Line 25:
	canceling the common mass term and rearranging a bit we get.		canceling the common mass term and rearranging a bit we get.

	<math>\begin{align}		<math> \begin{align}
	\ddot\theta - (g/L)\sin(\theta) = 0\\		\ddot\theta - (g/L)\sin(\theta) = 0\\
	\\		\\
Line 61:		Line 55:
	\\		\\
	\theta(t) = cosh(t\sqrt(\frac{g}{L}))\\		\theta(t) = cosh(t\sqrt(\frac{g}{L}))\\
	\end{align}</math>		\end{align}
			</math>

Aaron.boyd at 18:51, 1 November 2010

2010-11-01T18:51:27Z

← Older revision		Revision as of 11:51, 1 November 2010
Line 62:		Line 62:
	\theta(t) = cosh(t\sqrt(\frac{g}{L}))\\		\theta(t) = cosh(t\sqrt(\frac{g}{L}))\\
	\end{align}</math>		\end{align}</math>


			You can solve for the same thing from the cartesian coordinates. Taking:


			<math>\begin{align}
			F_x &= T\sin(\theta) = 0\\
			\text { and recognizing } T = mg\\
			\end{align}</math>


			you can arrive at the same answer

Aaron.boyd at 18:41, 1 November 2010

2010-11-01T18:41:39Z

← Older revision		Revision as of 11:41, 1 November 2010
Line 60:		Line 60:
	\\		\\
	\\		\\
	\theta(t) = cosh(t\sqrt(\frac{g}{L})\\		\theta(t) = cosh(t\sqrt(\frac{g}{L}))\\
	\end{align}</math>		\end{align}</math>

Aaron.boyd at 18:40, 1 November 2010

2010-11-01T18:40:23Z

← Older revision		Revision as of 11:40, 1 November 2010
Line 20:		Line 20:

	<math>\begin{align}		<math>\begin{align}
	Now since F_r &= 0 we can ignore it and look only at F_\theta.\\		\text{Now since } F_r &= 0 \text{ we can ignore it and look only at } F_\theta.\\
	Since we know F_\theta &= maL and ma &= mLa". We can conclude		\text{ Since we know } F_\theta &= maL \text{ and } ma = mLa. \text{ We can conclude}
	\end{align}</math>		\end{align}</math>

			<math>\begin{align}

	sin(\theta)*mg = mL\theta"		\sin(\theta)*mg = mL\ddot\theta
			\end{align}</math>

	canceling the common mass term and rearranging a bit we get.		canceling the common mass term and rearranging a bit we get.

			<math>\begin{align}
	\theta" - sin(\theta~~)(g/L~~) = 0		\ddot\theta - (g/L)\sin(\theta) = 0\\
			\\
			\text{Now we take the laplace transform of this.}\\
	~~now~~ we take the laplace transform of this. Unfortunately the laplace transform of that is horrible. Long~~, complicated~~, and ~~nearly~~ impossible to solve. So we use the approximation sin(\theta) = \theta where \theta is small.		\text{Unfortunately the laplace transform of that is horrible. Long, and impossible to solve.}\\
	(I tried to leave sin(\theta) in the equation. After 4 hours and many wolframalpha.com timeouts I gave up)		\text{So we use the approximation } \\
			\\
	with this new equation we get:		\sin(\theta) = \theta
			\\
			\text{ where } \theta \text{ is small. }\\
	g\theta(t)/L +s^2\theta(t) - s*\theta(0) - \theta'(0) = 0		\\
			\text{(I tried to leave } sin(\theta) \text{ in the equation.}\\
	we know that \theta(0) = \~~theta0~~ and \theta'(0) = 0		\text{After 4 hours and many wolframalpha.com timeouts I gave up) }\\
			\\
			\text{with this new equation we get:}\\
	solving for \theta(t) we get		\\
			g\frac{\theta(t)}{L +s^2}\theta(t) - s\theta(0) - \dot\theta(0) = 0\\
			\\
	\theta(t) = s*\~~theta<sub>0</sub>/(~~(-g/L)+S^2)		\text{we know that } \theta(0) = \theta_0 \text{ and } \dot\theta(0) = 0\\
			\\
			\\
	now we take the inverse laplace transform of that which yields		\text{solving for } \theta(t) \text{ we get} \\
			\\
			\theta(t) = s*\frac{\theta_0}{(\frac{-g}{L}+S^2)}\\
	\theta(t) = cosh(t*(g/L)~~^(1~~/~~2))~~		\\
			\\
			\text{now we take the inverse laplace transform of that which yields }\\
			\\
			\\
			\theta(t) = cosh(t\sqrt(\frac{g}{L})\\
			\end{align}</math>

Aaron.boyd at 18:11, 1 November 2010

2010-11-01T18:11:34Z

← Older revision		Revision as of 11:11, 1 November 2010
Line 1:		Line 1:
	I decided to use laplace transforms to solve a pendulum equation. A pendulum with a weight of mass m and a massless rod length L is released from an initial angle A<sub>0</sub>. Find a function to determine the angle at any time t.		I decided to use laplace transforms to solve a pendulum equation. A pendulum with a weight of mass m and a massless rod length L is released from an initial angle \theta<sub>0</sub>. Find a function to determine the angle at any time t.
	The summation of forces yields		The summation of forces yields
			<math>\begin{align}
			F_x &= T\sin(\theta)\\
			F_y &= T\cos(\theta)-mg = 0
			\end{align}</math>


	~~F<sub>x</sub> = T*sin(A)~~

	~~F<sub>y</sub> = T*cos(A) - mg = 0~~


Line 13:		Line 13:


	F<~~sub>r</sub~~> = T- cos(A)*mg = 0		<math>\begin{align}
			F_r &= T - mg\cos(\theta) = 0\\
	~~F<sub>A</sub>~~ = sin(A)*mg = maL		F_\theta &= \sin(\theta)mg = maL
			\end{align}</math>

	~~now since F<sub>r~~</~~sub~~> ~~= 0 we can ignore it and look only at FA.~~

	~~Since we know F<sub>A</sub> = maL and ma = mLa". We can conclude~~

			<math>\begin{align}
			Now since F_r &= 0 we can ignore it and look only at F_\theta.\\
			Since we know F_\theta &= maL and ma &= mLa". We can conclude
			\end{align}</math>


	sin(A)*mg = ~~mLA~~"		sin(\theta)*mg = mL\theta"


Line 30:		Line 31:


	A" - sin(A)(g/L) = 0		\theta" - sin(\theta)(g/L) = 0


	now we take the laplace transform of this. Unfortunately the laplace transform of that is horrible. Long, complicated, and nearly impossible to solve. So we use the approximation sin(A) = A where A is small.		now we take the laplace transform of this. Unfortunately the laplace transform of that is horrible. Long, complicated, and nearly impossible to solve. So we use the approximation sin(\theta) = \theta where \theta is small.
	(I tried to leave sin(A) in the equation. After 4 hours and many wolframalpha.com timeouts I gave up)		(I tried to leave sin(\theta) in the equation. After 4 hours and many wolframalpha.com timeouts I gave up)

	with this new equation we get:		with this new equation we get:


	gA(t)/L +s^2A(t) - s*A(0) - A'(0) = 0		g\theta(t)/L +s^2\theta(t) - s*\theta(0) - \theta'(0) = 0

	we know that A(0) = A0 and A'(0) = 0		we know that \theta(0) = \theta0 and \theta'(0) = 0


	solving for A(t) we get		solving for \theta(t) we get


	A(t) = s*A0/((-g/L)+S^2)		\theta(t) = s*\theta<sub>0</sub>/((-g/L)+S^2)


Line 53:		Line 54:


	A(t) = cosh(t*(g/L)^(1/2))		\theta(t) = cosh(t*(g/L)^(1/2))

Aaron.boyd: Created page with 'I decided to use laplace transforms to solve a pendulum equation. A pendulum with a weight of mass m and a massless rod length L is released from an initial angle A₀. …'

2010-11-01T17:45:22Z

Created page with 'I decided to use laplace transforms to solve a pendulum equation. A pendulum with a weight of mass m and a massless rod length L is released from an initial angle A0. …'

New page

I decided to use laplace transforms to solve a pendulum equation. A pendulum with a weight of mass m and a massless rod length L is released from an initial angle A0. Find a function to determine the angle at any time t.
The summation of forces yields

Fx = T*sin(A)

Fy = T*cos(A) - mg = 0

Polar coordinates may be easier to use, lets try that.

now:

Fr = T- cos(A)*mg = 0

FA = sin(A)*mg = maL

now since Fr = 0 we can ignore it and look only at FA.

Since we know FA = maL and ma = mLa". We can conclude

sin(A)*mg = mLA"

canceling the common mass term and rearranging a bit we get.

A" - sin(A)(g/L) = 0

now we take the laplace transform of this. Unfortunately the laplace transform of that is horrible. Long, complicated, and nearly impossible to solve. So we use the approximation sin(A) = A where A is small.
(I tried to leave sin(A) in the equation. After 4 hours and many wolframalpha.com timeouts I gave up)

with this new equation we get:

g*A(t)/L +s^2*A(t) - s*A(0) - A'(0) = 0

we know that A(0) = A0 and A'(0) = 0

solving for A(t) we get

A(t) = s*A0/((-g/L)+S^2)

now we take the inverse laplace transform of that which yields

A(t) = cosh(t*(g/L)^(1/2))