Chapter 8, Problem 15P

Many of the elements in horizontal-bar exercises can be modeled by representing the gymnast by four segments consisting of arms, torso (including the head), thighs, and lower legs, as shown in Figure P8.15a. Inertial parameters for a particular gymnast are as follows:

Note that in Figure P8.l5a r_cg is the distance to the center of gravity measured from the joint closest to the bar and the masses for the arms, thighs, and legs include both appendages. I is the moment of inertia of each segment about its center of gravity. Determine the distance from the bar to the center of gravity of the gymnast for the two positions shown in Figures P8.15b and P8.15c.

Figure P8.15

To determine

The distance from the bar to the centre of gravity of the gymnast for position b.

Answer to Problem 15P

Solution: The distance of the centre of mass from the bar is $1.01\text{m}$.

Explanation of Solution

Given info: The mass of the segments in kg (Mass), the lengths of the segments (Length), the distance of the centre of gravity to the nearest joint ($r_{eg}$) and the moment of inertia of each segment (I) are provided in the table.

The distance from the bar to the centre of mass of the body is given as,

$x=\frac{m_{\text{arms}}{x}_{\text{arms}}+m_{\text{torso}}{x}_{\text{torso}}+m_{\text{thighs}}{x}_{\text{thighs}}+m_{\text{legs}}{x}_{\text{legs}}}{m_{\text{arms}}+m_{\text{torso}}+m_{\text{thighs}}+m_{\text{legs}}}$

• $x$ is the x-co-ordinate of the centre of gravity
• $m_{\text{arms}}$ is the mass of the arm
• $m_{\text{torso}}$ is the mass of the torso
• $m_{\text{thighs}}$ is the mass of the thighs
• $m_{\text{legs}}$ is the mass of the legs
• $x_{\text{arms}}$ is the distance of the centre of gravity of arm from the bar
• $x_{\text{torso}}$ is the distance of the centre of gravity of torso from the bar
• $x_{\text{thighs}}$ is the distance of the centre of gravity of thighs from the bar
• $x_{\text{legs}}$ is the distance of the centre of gravity of legs from the bar

The different segments and positions of the gymnast is given in the picture below.

Consider the positive direction of the lengths to direct from the bar to the direction of the head.

The representative diagram of position b is given below.

From the above diagram,

$x_{\text{arms}}={\left(r_{\text{eg}}\right)}_{\text{arms}}$

$x_{\text{torso}}=L_{\text{arms}}+{\left(r_{\text{eg}}\right)}_{\text{torso}}$

$x_{\text{thighs}}=L_{\text{arms}}+L_{\text{torso}}+{\left(r_{\text{eg}}\right)}_{\text{thighs}}$

$x_{\text{legs}}=L_{\text{arms}}+L_{\text{torso}}+L_{\text{thighs}}+{\left(r_{\text{eg}}\right)}_{\text{legs}}$

Thus the formula for the distance of centre of mass from bar becomes,

$x=\frac{\left[\begin{array}{l}{m}_{\text{arms}}\left({\left(r_{\text{eg}}\right)}_{\text{arms}}\right)+m_{\text{torso}}\left(L_{\text{arms}}+{\left(r_{\text{eg}}\right)}_{\text{torso}}\right)+m_{\text{thighs}}\left(L_{\text{arms}}+L_{\text{torso}}+{\left(r_{\text{eg}}\right)}_{\text{thighs}}\right)\\ +m_{\text{legs}}\left(L_{\text{arms}}+L_{\text{torso}}+L_{\text{thighs}}+{\left(r_{\text{eg}}\right)}_{\text{legs}}\right)\end{array}\right]}{m_{\text{arms}}+m_{\text{torso}}+m_{\text{thighs}}+m_{\text{legs}}}$

Substitute $6.87\text{kg}$ for $m_{\text{arms}}$, $33.57\text{kg}$ for $m_{\text{torso}}$, $14.07\text{kg}$ for $m_{\text{thighs}}$, $7.54\text{kg}$ for $m_{\text{legs}}$, $0.239\text{m}$ for ${\left(r_{\text{eg}}\right)}_{\text{arms}}$, $0.548\text{m}$ for $L_{\text{arms}}$, $0.337\text{m}$ for ${\left(r_{\text{eg}}\right)}_{\text{torso}}$, $0.601\text{m}$ for $L_{\text{torso}}$, $0.151\text{m}$ for ${\left(r_{\text{eg}}\right)}_{\text{thighs}}$, $0.374\text{m}$ for $L_{\text{thighs}}$, $0.227\text{m}$ for ${\left(r_{\text{eg}}\right)}_{\text{legs}}$ to determine the distance of the centre of mass from the bar.

$\begin{array}{c}x=\frac{\begin{array}{l}\left(6.87\text{kg}\right)\left(0.239\text{m}\right)+\left(33.57\text{kg}\right)\left(0.548\text{m}+0.337\text{m}\right)+\left(14.07\text{kg}\right)\left(0.548\text{m}+0.601\text{m}+0.151\text{m}\right)\\ +\left(7.54\text{kg}\right)\left(0.548\text{m}+0.601\text{m}+0.374\text{m}+0.227\text{m}\right)\end{array}}{6.87\text{kg}+33.57\text{kg}+14.07\text{kg}+7.54\text{kg}}\\ =\frac{62.8\text{kg}\cdot \text{m}}{62.05\text{kg}}\\ =1.01\text{m}\end{array}$

Conclusion:

The distance of the centre of mass from the bar at position b is $1.01\text{m}$.

To determine

The distance from the bar to the centre of gravity of the gymnast for position c.

Answer to Problem 15P

Solution: The distance of the centre of mass from the bar at position c is $0.015\text{m}\text{towards the head}$.

Explanation of Solution

Given info: The mass of the segments in kg (Mass), the lengths of the segments (Length), the distance of the centre of gravity to the nearest joint ($r_{eg}$) and the moment of inertia of each segment (I) are provided in the table.

The distance from the bar to the centre of mass of the body is given as,

$x=\frac{m_{\text{arms}}{x}_{\text{arms}}+m_{\text{torso}}{x}_{\text{torso}}+m_{\text{thighs}}{x}_{\text{thighs}}+m_{\text{legs}}{x}_{\text{legs}}}{m_{\text{arms}}+m_{\text{torso}}+m_{\text{thighs}}+m_{\text{legs}}}$

• $x$ is the x-co-ordinate of the centre of gravity
• $m_{\text{arms}}$ is the mass of the arm
• $m_{\text{torso}}$ is the mass of the torso
• $m_{\text{thighs}}$ is the mass of the thighs
• $m_{\text{legs}}$ is the mass of the legs
• $x_{\text{arms}}$ is the distance of the centre of gravity of arm from the bar
• $x_{\text{torso}}$ is the distance of the centre of gravity of torso from the bar
• $x_{\text{thighs}}$ is the distance of the centre of gravity of thighs from the bar
• $x_{\text{legs}}$ is the distance of the centre of gravity of legs from the bar

Consider the positive direction of the lengths to direct from the bar to the direction of the head.

The representative diagram of position b is given below.

• ${\left(r_{\text{eg}}\right)}_{\text{thighs}}$ is the distance between the centre of mass of the thighs to the nearest hip joint
• ${\left(r_{\text{eg}}\right)}_{\text{torso}}$ is the distance between the centre of mass of the torso to the nearest shoulder joint
• ${\left(r_{\text{eg}}\right)}_{\text{arms}}$ is the distance between the centre of mass of the arms to the nearest shoulder joint
• ${\left(r_{\text{eg}}\right)}_{\text{legs}}$ is the distance between the centre of mass of the legs to the nearest knee joint
• $L_{\text{arms}}$ is the length of arm
• $L_{\text{torso}}$ is the length of torso
• $L_{\text{thighs}}$ is the length of thighs
• $L_{\text{arms}}$ is the length of the arm

From the above diagram,

$x_{\text{arms}}={\left(r_{\text{eg}}\right)}_{\text{arms}}$

$x_{\text{torso}}=L_{\text{arms}}-{\left(r_{\text{eg}}\right)}_{\text{torso}}$

$x_{\text{thighs}}=L_{\text{arms}}-L_{\text{torso}}-{\left(r_{\text{eg}}\right)}_{\text{thighs}}$

$x_{\text{legs}}=L_{\text{arms}}-L_{\text{torso}}-L_{\text{thighs}}-{\left(r_{\text{eg}}\right)}_{\text{legs}}$

Thus the formula for the distance of centre of mass from bar becomes,

$x=\frac{\left[\begin{array}{l}{m}_{\text{arms}}\left({\left(r_{\text{eg}}\right)}_{\text{arms}}\right)+m_{\text{torso}}\left(L_{\text{arms}}-{\left(r_{\text{eg}}\right)}_{\text{torso}}\right)+m_{\text{thighs}}\left(L_{\text{arms}}-L_{\text{torso}}-{\left(r_{\text{eg}}\right)}_{\text{thighs}}\right)\\ +m_{\text{legs}}\left(L_{\text{arms}}-L_{\text{torso}}-L_{\text{thighs}}-{\left(r_{\text{eg}}\right)}_{\text{legs}}\right)\end{array}\right]}{m_{\text{arms}}+m_{\text{torso}}+m_{\text{thighs}}+m_{\text{legs}}}$

Substitute $6.87\text{kg}$ for $m_{\text{arms}}$, $33.57\text{kg}$ for $m_{\text{torso}}$, $14.07\text{kg}$ for $m_{\text{thighs}}$, $7.54\text{kg}$ for $m_{\text{legs}}$, $0.239\text{m}$ for ${\left(r_{\text{eg}}\right)}_{\text{arms}}$, $0.548\text{m}$ for $L_{\text{arms}}$, $0.337\text{m}$ for ${\left(r_{\text{eg}}\right)}_{\text{torso}}$, $0.601\text{m}$ for $L_{\text{torso}}$, $0.151\text{m}$ for ${\left(r_{\text{eg}}\right)}_{\text{thighs}}$, $0.374\text{m}$ for $L_{\text{thighs}}$, $0.227\text{m}$ for ${\left(r_{\text{eg}}\right)}_{\text{legs}}$ to determine the distance of the centre of mass from the bar.

$\begin{array}{c}x=\frac{\left\{\begin{array}{l}\left(6.87\text{kg}\right)\left(0.239\text{m}\right)+\left(33.57\text{kg}\right)\left(0.548\text{m}-0.337\text{m}\right)+\left(14.07\text{kg}\right)\left(0.548\text{m}-0.601\text{m}-0.151\text{m}\right)\\ +\left(7.54\text{kg}\right)\left(0.548\text{m}-0.601\text{m}-0.374\text{m}-0.227\text{m}\right)\end{array}\right\}}{6.87\text{kg}+33.57\text{kg}+14.07\text{kg}+7.54\text{kg}}\\ =\frac{0.924\text{kg}\cdot \text{m}}{62.05\text{kg}}\\ =0.015\text{m}\end{array}$

Conclusion:

The distance of the centre of mass from the bar at position c is $0.015\text{m}$.