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92 P

Expert-verifiedFound in: Page 292

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**In the 1968 Olympic games, University of Oregon jumper Dick Fosbury introduced a new technique of high jumping called the “Fosbury flop.” It contributed to raising the world record by about 30 cm and is currently used by nearly every world-class jumper. In this technique, the jumper goes over the bar face-up while arching her back as much as possible as shown in Figure P9.92a. This action places her center of mass outside her body, below her back. As her body goes over the bar, her center of mass passes below the bar. Because a given energy input implies a certain elevation for her center of mass, the action of arching her back means that her body is higher than if her back were straight. As a model, consider the jumper as a thin uniform rod of length L. When the rod is straight, its center of mass is at its center. Now bend the rod in a circular arc so that it subtends an angle of 90 ^{0} at the center of the arc as shown in Figure P9.92b. In this configuration, how far outside the rod is the center of mass?**

The centre of the mass is 0.0634L.

The length of the rod is $L$.

The subtends angle, $\theta =90\xb0$

The expression to calculate the centre of the mass of an extended object in vertical direction is given as follows.

${y}_{CM}=\frac{1}{M}\int ydm$ …… (i)

**Here, $M$ is the total mass and is the mass of the element.**

Consider the following figure,

The length of the rod is given by,

$L=\frac{1}{4}\left(2\pi r\right)\phantom{\rule{0ex}{0ex}}L=\frac{1}{2}\left(\pi r\right)\phantom{\rule{0ex}{0ex}}r=\frac{2L}{\pi}$

Here, $r$ is the radius of the curve.

The incremental bit of rod at angle $\theta $ is given by,

$\frac{dm}{rd\theta}=\frac{M}{L}\phantom{\rule{0ex}{0ex}}dm=\frac{Mr}{L}d\theta $

The vertical since is given by,

$\mathrm{sin}\theta =\frac{y}{r}\phantom{\rule{0ex}{0ex}}y=r\mathrm{sin}\theta $

Substitute $rSin\theta $ for and $\frac{Mr}{L}d\theta $ for into equation.

${y}_{CM}=\frac{1}{M}\int r\mathrm{sin}\theta \frac{Mr}{L}d\theta \phantom{\rule{0ex}{0ex}}{y}_{CM}=\frac{M{r}^{2}}{ML}\int \mathrm{sin}\theta d\theta $

Since the given angle of subtends is 90^{0}, take the limits for the angle from 45^{0} to 135^{0}.

${y}_{CM}=\frac{{r}^{2}}{L}{\int}_{45\xb0}^{135\xb0}\mathrm{sin}\theta d\theta \phantom{\rule{0ex}{0ex}}{y}_{CM}=\frac{{r}^{2}}{L}{\left(-\mathrm{cos}\theta \right)}_{45\xb0}^{135\xb0}\phantom{\rule{0ex}{0ex}}{y}_{CM}=\frac{{r}^{2}}{L}\left(-\mathrm{cos}135\xb0-\left(-\mathrm{cos}45\xb0\right)\right)\phantom{\rule{0ex}{0ex}}{y}_{CM}=\frac{{r}^{2}}{L}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)$

Substitute $\frac{2L}{\pi}$ for r into above equation.

${y}_{CM}=\frac{{\left(\frac{2L}{\pi}\right)}^{2}}{L}\frac{2}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}{y}_{CM}=\frac{4{L}^{2}}{{\pi}^{2}L}\frac{2}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}{y}_{CM}=\frac{4\sqrt{2}}{{\pi}^{2}}L$

The centre of the mass is below the bar which is given by,

$y{\text{'}}_{CM}=r-{y}_{CM}$

Substitute $\frac{2L}{\pi}$ for and $\frac{4\sqrt{2}}{{\pi}^{2}}L$ for into above equation.

${y\text{'}}_{CM}=\frac{2L}{\pi}-\frac{4\sqrt{2}}{{\pi}^{2}}L\phantom{\rule{0ex}{0ex}}{y\text{'}}_{CM}=\frac{2L}{\pi}\left(1-\frac{2\sqrt{2}}{\pi}\right)\phantom{\rule{0ex}{0ex}}{y\text{'}}_{CM}=\frac{2L}{\pi}\times 0.0996\phantom{\rule{0ex}{0ex}}{y\text{'}}_{CM}=0.0634L$

Hence the centre of the mass is 0.0634L.

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