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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. |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
| + | [[009C Sample Final 3, Problem 9 Solution|'''<u>Solution</u>''']] |
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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 | |
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| − | <math>50-5\,\textrm{m/sec}=45\,\textrm{m/sec}. </math>
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| − | Essentially, equations of motion can often be broken into parts, and
| + | [[009C Sample Final 3, Problem 9 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | then added up.
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| − | '''Solution:'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |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>
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| − | Moreover, the axle's <math style="vertical-align: 0px">x </math> position will change in the same manner,
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| − | while the height of the axle will always be fixed at <math style="vertical-align: -4px">y=1 </math>. This
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| − | means we can describe the position of the axle as a function of <math style="vertical-align: -4px">\theta, </math>
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| − | :: <math>a(\theta)\,=\,(\theta,1). </math>
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |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
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| − | 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
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| − | :: <math>p(\theta)\,=\,\left(-\frac{1}{2}\sin\theta,-\frac{1}{2}\cos\theta\right). </math>
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| − | Notice that when <math style="vertical-align: -4px">\theta=0, </math> the point would be at the position
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| − | :: <math>p(0)\,=\,\left(-\frac{1}{2}\sin0,-\frac{1}{2}\cos0\right)\,=\,\left(0,-\frac{1}{2}\right), </math>
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| − | 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>
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |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
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| − | :: <math>P(\theta)\,=\,a(\theta)+p(\theta)\,=\,\left(\theta-\frac{1}{2}\sin\theta,1-\frac{1}{2}\cos\theta\right). </math>
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !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|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
