Chapter 5, Problem 32P

(a)

To determine

The center of mass of the pole vaulter for the given mass distribution of the athlete at a particular instant.

Answer to Problem 32P

The x and y co-ordinates of center of mass of the pole vaulter are $x_{cm}=8.8125\text{cm}$ and $y_{cm}=-9.4375\text{cm}$ respectively.

Explanation of Solution

Given:

A pole vaulter’s distribution of mass with the coordinates. Seven mass points are given. Mass $m_1$ is 11 kg located at (25, -70) cm, $m_2$ is 18 kg located at (25, -20) cm, $m$ is 14 kg at (25, 12.25) cm, mass $m_4$ is 20 kg (0, 15) cm, mass $m_5$ is 6 kg locatedat (-25, 17.25) cm, $m_6$ is at 6 kg (- 20,0) cm, $m_7$ is 5 kg at (-20, -40) cm.

Formula used:

The x and y coordinates of the center of mass is given as below:

  $x_{cm}=\frac{m_1{x}_1+m_2{x}_2+\mathrm{....................}{m}_n{x}_n}{m_1+m_2+\mathrm{....................}{m}_n}$

  $y_{cm}=\frac{m_1{y}_1+m_2{y}_2+\mathrm{....................}{m}_n{y}_n}{m_1+m_2+\mathrm{....................}{m}_n}$

Calculation:

The x coordinate of the center of mass of the pole-vaulter is given as below:

  $x_{cm}=\frac{m_1{x}_1+m_2{x}_2+mx{}_3+mx{}_4+m_5{x}_5+mx{}_6+mx{}_7}{m_1+m_2+m_3+m_4+m_5+m_6+m_7}$

Putting the values of masses and their coordinates we can write as below:

  $x_{cm}=\frac{(11\times 25)+\left(18\times 25\right)+\left(14\times 25\right)+\left(20\times 0\right)+6\times \left(-25\right)+6\times \left(-20\right)+5\times \left(-20\right)}{11+18+14+20+6+6+5}$

  $x_{cm}=\frac{275+450+350+0-150-120-100}{80}=\frac{705}{80}$

  $x_{cm}=8.8125\text{cm}$

The y coordinate of the center of mass of the pole-vaulter is given as below:

  $y_{cm}=\frac{m_1{y}_1+m_2{y}_2+my{}_3+my{}_4+m_5{y}_5+my{}_6+my{}_7}{m_1+m_2+m_3+m_4+m_5+m_6+m_7}$

Mass $m_1$ is 11 kg located at (25, -70) cm, $m_2$ is 18 kg located at (25, -20) cm, $m$ is 14 kg at (25, 12.25) cm, mass $m_4$ is 20 kg (0, 15) cm, mass $m_5$ is 6 kg located at (-25, 17.25) cm, $m_6$ is at 6 kg (- 20, 0) cm, $m_7$ is 5 kg at (-20, -40) cm.

  $y_{cm}=\frac{\left(11\times \left(-70\right)\right)+\left(18\times \left(-20\right)\right)+\left(14\times 12.25\right)+\left(20\times 15\right)+\left(6\times 17.25\right)+\left(5\times \left(-40\right)\right)}{80}$

  $y_{cm}=-9.437\text{5}\text{cm}$

It is observed that the center of mass of the pole-vaulter lies below the bar at -9.4375 cm.

Conclusion:

The coordinates of the center of mass of the pole-vaulter are $x_{cm}=8.8125\text{cm}$ and. $y_{cm}=-9.4375\text{cm}$

(b)

To determine

The height of the bar for the given height of the center of mass.

Answer to Problem 32P

The height of the bar 6.1 m.

Explanation of Solution

Introduction:

Let say, the center of mass of the pole-vaulter while standing on the ground is 1 meter. As discussed in previous section that the height of the center of mass of the vaulter lies below the bar. Pole vaulter runs with a speed vand hence the Kinetic energy is $K.E=\frac{1}{2}m{v}^2$ where mis the mass and thevis the velocity of theVaulter. After reaching up in the air, kinetic energy transforms into potential energy.

  $\frac{1}{2}m{v}^2=mgh$

The height of the center of mass of the pole-vaulter is 5.1 m. The height of center of mass while standing on the ground is 1m. Hence, the height of the bar will be $(5.1+1)\text{ m}=6.1\text{m}$

Conclusion:

The height of the bar 6.1 m.