009C Sample Final 3, Problem 9 - Revision history

MathAdmin at 19:52, 19 March 2017

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MathAdmin at 19:05, 19 March 2017

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MathAdmin: Created page with "350px A wheel of radius 1 rolls along a straight line, say the $x$ -axis. A point
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Created page with "right|350px A wheel of radius 1 rolls along a straight line, say the <math>x</math>-axis. A point <math style="vertica..."

New page
[[File:9CSF3_9a_GP.png|right|350px]]

A wheel of radius 1 rolls along a straight line, say the  <math>x</math>-axis. A point  <math style="vertical-align: 0px">P</math>  is located halfway between the center of the wheel and the rim. As the wheel rolls,  <math style="vertical-align: 0px">P</math>  traces a curve. Find parametric equations for the curve.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
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|Many concepts in physics involve the notion of a relative frame. For example, if I'm in a box dropped from an airplane, I won't be moving relative to the box. However, I'm still heading towards the ground with acceleration <math style="vertical-align: -5px">10\,\textrm{m/sec}^{2}. </math>  Say it drops for 5 seconds, so the box is going <math style="vertical-align: -5px">50\,\textrm{m/sec} </math> when it hits the ground. Even if I jump with all my might and pull off something like <math style="vertical-align: -5px">5\,\textrm{m/sec} </math> of upward velocity, I'll still feel the impact of hitting the ground at

     <math>50-5\,\textrm{m/sec}=45\,\textrm{m/sec}. </math>

Essentially, equations of motion can often be broken into parts, and
then added up.
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'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|If a wheel of radius one is resting at the origin, its axis will be at the point <math style="vertical-align: -5px">(1,0). </math> For this solution, we will assume the point <math style="vertical-align: 0px">P </math> is below the axle, although the problem does not state the position of <math style="vertical-align: 0px">P </math>. When the wheel rotates clockwise, it will move to the right. Since the length of the arc defined by an angle <math style="vertical-align: 0px">\theta </math> on a circle of radius <math style="vertical-align: 0px">R </math> is <math style="vertical-align: -4px">L=R\cdot\theta=1\cdot\theta=\theta, </math>  the wheel will roll forward the length of the arc, which is just <math style="vertical-align: 0px">\theta. </math>

Moreover, the axle's <math style="vertical-align: 0px">x </math> position will change in the same manner,
while the height of the axle will always be fixed at <math style="vertical-align: -4px">y=1 </math>. This
means we can describe the position of the axle as a function of <math style="vertical-align: -4px">\theta, </math>
or

:: <math>a(\theta)\,=\,(\theta,1). </math>

|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since the wheel is rotating, we also know that our point <math style="vertical-align: 0px">P </math> will rotate around the axle. As described in the problem, it is halfway
between the rim and the center/axle, so it is <math style="vertical-align: -4px">1/2 </math> unit away from the axle, and will rotate clockwise. Using our trig relations (while looking at the image), we find that the position of <math style="vertical-align: 0px">P </math> relative to the axle can be described as

:: <math>p(\theta)\,=\,\left(-\frac{1}{2}\sin\theta,-\frac{1}{2}\cos\theta\right). </math>

Notice that when <math style="vertical-align: -4px">\theta=0, </math> the point would be at the position

:: <math>p(0)\,=\,\left(-\frac{1}{2}\sin0,-\frac{1}{2}\cos0\right)\,=\,\left(0,-\frac{1}{2}\right), </math>

which is half a unit directly below the axle. This is shown as a gray "ghost" dot in the image, while the black triangle and circle represent the situation at <math style="vertical-align: -5px">\theta=\pi/4. </math>

|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|We therefore have a frame (the axle) that is moving, and a point <math style="vertical-align: 0px">P </math> that is moving relative to the frame. To get the movement relative to the stationary "world", we simply add them up to find

:: <math>P(\theta)\,=\,a(\theta)+p(\theta)\,=\,\left(\theta-\frac{1}{2}\sin\theta,1-\frac{1}{2}\cos\theta\right). </math>

|}
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
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:: <math>P(\theta)\,=\,\left(\theta-\frac{1}{2}\sin\theta,1-\frac{1}{2}\cos\theta\right). </math>
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|}
[[009C_Sample_Final_3|'''Return to Sample Exam''']]
 
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	<span class="exam">A wheel of radius 1 rolls along a straight line, say the  <math>x</math>-axis. A point  <math style="vertical-align: 0px">P</math>  is located halfway between the center of the wheel and the rim. As the wheel rolls,  <math style="vertical-align: 0px">P</math>  traces a curve. Find parametric equations for the curve.		<span class="exam">A wheel of radius 1 rolls along a straight line, say the  <math>x</math>-axis. A point  <math style="vertical-align: 0px">P</math>  is located halfway between the center of the wheel and the rim. As the wheel rolls,  <math style="vertical-align: 0px">P</math>  traces a curve. Find parametric equations for the curve.

@@ Line 20: / Line 20: @@
 !Step 1: &nbsp;
 |-
-|If a wheel of radius one is resting at the origin, its axis will be at the point <math style="vertical-align: -5px">(1,0). </math> For this solution, we will assume the point <math style="vertical-align: 0px">P </math> is below the axle, although the problem does not state the position of <math style="vertical-align: 0px">P </math>. When the wheel rotates clockwise, it will move to the right. Since the length of the arc defined by an angle <math style="vertical-align: 0px">\theta </math> on a circle of radius <math style="vertical-align: 0px">R </math> is <math style="vertical-align: -4px">L=R\cdot\theta=1\cdot\theta=\theta, </math> &thinsp;the wheel will roll forward the length of the arc, which is just <math style="vertical-align: 0px">\theta. </math>
+|If a wheel of radius one is resting at the origin, its axis will be at the point &nbsp;<math style="vertical-align: -5px">(1,0). </math>&nbsp; For this solution, we will assume the point &nbsp;<math style="vertical-align: 0px">P </math> &nbsp;is below the axle, although the problem does not state the position of &nbsp;<math style="vertical-align: 0px">P </math>. &nbsp;When the wheel rotates clockwise, it will move to the right. Since the length of the arc defined by an angle &nbsp;<math style="vertical-align: 0px">\theta </math> &nbsp;on a circle of radius &nbsp;<math style="vertical-align: 0px">R </math> &nbsp; is &nbsp; <math style="vertical-align: -4px">L=R\cdot\theta=1\cdot\theta=\theta, </math> &nbsp;the wheel will roll forward the length of the arc, which is just &nbsp;<math style="vertical-align: 0px">\theta. </math>
-Moreover, the axle's <math style="vertical-align: 0px">x </math> position will change in the same manner,
+Moreover, the axle's &nbsp;<math style="vertical-align: 0px">x </math> &nbsp;position will change in the same manner, while the height of the axle will always be fixed at &nbsp;<math style="vertical-align: -4px">y=1 </math>. &nbsp;This means we can describe the position of the axle as a function of &nbsp;<math style="vertical-align: -4px">\theta, </math> &nbsp;or
 :: <math>a(\theta)\,=\,(\theta,1). </math>
@@ Line 34: / Line 31: @@
 !Step 2: &nbsp;
 |-
-|Since the wheel is rotating, we also know that our point <math style="vertical-align: 0px">P </math> will rotate around the axle. As described in the problem, it is halfway
+|Since the wheel is rotating, we also know that our point &nbsp;<math style="vertical-align: 0px">P </math>&nbsp; will rotate around the axle. As described in the problem, it is halfway between the rim and the center/axle, so it is &nbsp;<math style="vertical-align: -5px">1/2 </math> &nbsp;unit away from the axle, and will rotate clockwise. Using our trig relations (while looking at the image), we find that the position of &nbsp;<math style="vertical-align: 0px">P </math> &nbsp;relative to the axle can be described as
 :: <math>p(\theta)\,=\,\left(-\frac{1}{2}\sin\theta,-\frac{1}{2}\cos\theta\right). </math>
-Notice that when <math style="vertical-align: -4px">\theta=0, </math> the point would be at the position
+Notice that when &nbsp;<math style="vertical-align: -4px">\theta=0, </math> &nbsp;the point would be at the position
 :: <math>p(0)\,=\,\left(-\frac{1}{2}\sin0,-\frac{1}{2}\cos0\right)\,=\,\left(0,-\frac{1}{2}\right), </math>
-which is half a unit directly below the axle. This is shown as a gray "ghost" dot in the image, while the black triangle and circle represent the situation at <math style="vertical-align: -5px">\theta=\pi/4. </math>
+which is half a unit directly below the axle. This is shown as a gray "ghost" dot in the image, while the black triangle and circle represent the situation at &nbsp;<math style="vertical-align: -5px">\theta=\pi/4. </math>
 |}
@@ Line 50: / Line 46: @@
 !Step 3: &nbsp;
 |-
-|We therefore have a frame (the axle) that is moving, and a point <math style="vertical-align: 0px">P </math> that is moving relative to the frame. To get the movement relative to the stationary "world", we simply add them up to find
+|We therefore have a frame (the axle) that is moving, and a point&nbsp; <math style="vertical-align: 0px">P </math> &nbsp;that is moving relative to the frame. To get the movement relative to the stationary "world", we simply add them up to find
 :: <math>P(\theta)\,=\,a(\theta)+p(\theta)\,=\,\left(\theta-\frac{1}{2}\sin\theta,1-\frac{1}{2}\cos\theta\right). </math>

@@ Line 12: / Line 12: @@
 Essentially, equations of motion can often be broken into parts, and
 then added up.
 |}
@@ Line 43: / Line 35: @@
 |-
 |Since the wheel is rotating, we also know that our point <math style="vertical-align: 0px">P </math> will rotate around the axle. As described in the problem, it is halfway
-between the rim and the center/axle, so it is <math style="vertical-align: -4px">1/2 </math> unit away from the axle, and will rotate clockwise. Using our trig relations (while looking at the image), we find that the position of <math style="vertical-align: 0px">P </math> relative to the axle can be described as
+between the rim and the center/axle, so it is <math style="vertical-align: -5px">1/2 </math> unit away from the axle, and will rotate clockwise. Using our trig relations (while looking at the image), we find that the position of <math style="vertical-align: 0px">P </math> relative to the axle can be described as
 :: <math>p(\theta)\,=\,\left(-\frac{1}{2}\sin\theta,-\frac{1}{2}\cos\theta\right). </math>