Book edition 9th Edition

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

Pages 1624 pages

ISBN 9781133947271

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Short Answer

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.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Write the given data from the question.

The length of the rod is $L$ .

The subtends angle, $θ = 90 °$

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.

$y_{C M} = \frac{1}{M} \int y d m$ …… (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,

$L = \frac{1}{4} (2 π r) L = \frac{1}{2} (π r) r = \frac{2 L}{π}$

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

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

$\frac{d m}{r d θ} = \frac{M}{L} d m = \frac{M r}{L} d θ$

The vertical since is given by,

$\sin θ = \frac{y}{r} y = r \sin θ$

Substitute $r S i n θ$ for and $\frac{M r}{L} d θ$ for into equation.

$y_{C M} = \frac{1}{M} \int r \sin θ \frac{M r}{L} d θ y_{C M} = \frac{M r^{2}}{M L} \int \sin θ d θ$

Since the given angle of subtends is 90⁰, take the limits for the angle from 45⁰ to 135⁰.

$y_{C M} = \frac{r^{2}}{L} \int_{45 °}^{135 °} \sin θ d θ y_{C M} = \frac{r^{2}}{L} {(- \cos θ)}_{45 °}^{135 °} y_{C M} = \frac{r^{2}}{L} (- \cos 135 ° - (- \cos 45 °)) y_{C M} = \frac{r^{2}}{L} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})$

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

$y_{C M} = \frac{{(\frac{2 L}{π})}^{2}}{L} \frac{2}{\sqrt{2}} y_{C M} = \frac{4 L^{2}}{π^{2} L} \frac{2}{\sqrt{2}} y_{C M} = \frac{4 \sqrt{2}}{π^{2}} L$

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

$y'_{C M} = r - y_{C M}$

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

${y'}_{C M} = \frac{2 L}{π} - \frac{4 \sqrt{2}}{π^{2}} L {y'}_{C M} = \frac{2 L}{π} (1 - \frac{2 \sqrt{2}}{π}) {y'}_{C M} = \frac{2 L}{π} \times 0.0996 {y'}_{C M} = 0.0634 L$

Hence the centre of the mass is 0.0634L.

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Short Answer

Step by Step Solution

Step 1: Write the given data from the question.

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

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

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Physics For Scientists & Engineers

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Short Answer

Step by Step Solution

Step 1: Write the given data from the question.

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

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

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