Difference between revisions of "009C Sample Final 3, Problem 9"

Revision as of 18:54, 2 December 2017

A wheel of radius 1 rolls along a straight line, say the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} -axis. A point Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P} is located halfway between the center of the wheel and the rim. As the wheel rolls, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P} traces a curve. Find parametric equations for the curve.

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 <span class="exam">A wheel of radius 1 rolls along a straight line, say the &nbsp;<math>x</math>-axis. A point &nbsp;<math style="vertical-align: 0px">P</math>&nbsp; is located halfway between the center of the wheel and the rim. As the wheel rolls, &nbsp;<math style="vertical-align: 0px">P</math>&nbsp; traces a curve. Find parametric equations for the curve.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 3, Problem 9 Solution|'''<u>Solution</u>''']]
-|-
-|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> &thinsp;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
-&nbsp;&nbsp;&nbsp;&nbsp; <math>50-5\,\textrm{m/sec}=45\,\textrm{m/sec}. </math>
-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.
-|-
-|
-|-
-|
-|-
-|
-|-
-|
-|}
-'''Solution:'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!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>
-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: &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
-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: &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
-:: <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: &nbsp;
-|-
-|
-:: <math>P(\theta)\,=\,\left(\theta-\frac{1}{2}\sin\theta,1-\frac{1}{2}\cos\theta\right). </math>
-|&nbsp;&nbsp;
-|}
 [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009C Sample Final 3, Problem 9"

Revision as of 18:54, 2 December 2017

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