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Found in: Page 292

### Physics For Scientists & Engineers

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 900 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.

See the step by step solution

## Step 1: Write the given data from the question.

The length of the rod is $L$.

The subtends angle, $\theta =90°$

## Step 2: Determine the formulas to calculate the distance of the centre of the mass that is outside the rod.

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.

## Step 3: Calculate the distance of the centre of the mass that is outside the rod.

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 900, take the limits for the angle from 450 to 1350.

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

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.

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

Hence the centre of the mass is 0.0634L.