Chapter 8, Problem 16P

To determine

The position of the centre of gravity of the gymnast.

Answer to Problem 16P

Solution: The x component of the centre of gravity is $0.459\text{m}$. The y component of the centre of gravity is $0.103\text{m}$.

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 ( $r_{\text{eg}}$) and the moment of inertia of each segment (I) are provided in the table.

The x co-ordinate of the centre of gravity is given by,

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

• $x$ 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 x component of the distance of the centre of gravity of arm from the bar
• $x_{\text{torso}}$ is the x component of the distance of the centre of gravity of torso from the bar
• $x_{\text{thighs}}$ is the x component of the distance of the centre of gravity of thighs from the bar
• $x_{\text{legs}}$ is the x component of the distance of the centre of gravity of legs from the bar

The y co-ordinate of the centre of gravity is given by,

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

• $y$ is the y-co-ordinate of the centre of gravity
• $y_{\text{arms}}$ is the y component of the distance of the centre of gravity of arm from the bar
• $y_{\text{torso}}$ is the y component of the distance of the centre of gravity of torso from the bar
• $y_{\text{thighs}}$ is the y component of the distance of the centre of gravity of thighs from the bar
• $y_{\text{legs}}$ is the y component of 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 the gymnast 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,

$x_{\text{arms}}=0$

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

$x_{\text{thighs}}=L_{\text{torso}}+{\left(r_{\text{eg}}\right)}_{\text{thighs}}\cos 60.0°$

$x_{\text{legs}}=L_{\text{torso}}+L_{\text{thighs}}\cos 60.0°+{\left(r_{\text{eg}}\right)}_{\text{legs}}$

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

$y_{\text{torso}}=0$

$y_{\text{thighs}}={\left(r_{\text{eg}}\right)}_{\text{thighs}}\sin 60.0°$

$x_{\text{legs}}=L_{\text{thighs}}\sin 60.0°$

Thus the x co-ordinate of the centre of gravity is given by,

$x=\frac{\left[\begin{array}{l}{m}_{\text{arms}}\left(0\right)+m_{\text{torso}}\left({\left(r_{\text{eg}}\right)}_{\text{torso}}\right)+m_{\text{thighs}}\left(L_{\text{torso}}+{\left(r_{\text{eg}}\right)}_{\text{thighs}}\cos 60.0°\right)\\ +m_{\text{legs}}\left(L_{\text{torso}}+L_{\text{thighs}}\cos 60.0°+{\left(r_{\text{eg}}\right)}_{\text{legs}}\right)\end{array}\right]}{m_{\text{arms}}+m_{\text{torso}}+m_{\text{thighs}}+m_{\text{legs}}}$

Thus the y co-ordinate of the centre of gravity is given by,

$y=\frac{\left[\begin{array}{l}{m}_{\text{arms}}\left(L_{\text{arms}}-{\left(r_{\text{eg}}\right)}_{\text{arms}}\right)+m_{\text{torso}}\left(0\right)\\ +m_{\text{thighs}}\left({\left(r_{\text{eg}}\right)}_{\text{thighs}}\sin 60.0°\right)+m_{\text{legs}}\left(L_{\text{thighs}}\sin 60.0°\right)\end{array}\right]}{m_{\text{arms}}+m_{\text{torso}}+m_{\text{thighs}}+m_{\text{legs}}}$

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 centre of gravity along  x and y component.

The centre of gravity along x component is,

$\begin{array}{c}x=\frac{\left[\begin{array}{l}\left(6.87\text{kg}\right)\left(0\right)+\left(33.57\text{kg}\right)\left(0.337\text{m}\right)+\left(14.07\text{kg}\right)\left(0.601\text{m}+\left(0.151\text{m}\right)\cos 60.0°\right)\\ +7.54\text{kg}\left(0.601\text{m}+\left(0.374\text{m}\right)\cos 60.0°+\left(0.227\text{m}\right)\right)\end{array}\right]}{6.87\text{kg}+33.57\text{kg}+14.07\text{kg}+7.54\text{kg}}\\ =\frac{28.5\text{kg}\cdot \text{m}}{62.05\text{kg}}\\ =0.459\text{m}\end{array}$

The centre of gravity along y component is,

$\begin{array}{c}y=\frac{\left[\begin{array}{l}\left(6.87\text{kg}\right)\left(0.548\text{m}-0.239\text{m}\right)+\left(33.57\text{kg}\right)\left(0\right)\\ +\left(14.07\text{kg}\right)\left(\left(0.151\text{m}\right)\sin 60.0°\right)+\left(7.54\text{kg}\right)\left(\left(0.374\text{m}\right)\sin 60.0°\right)\end{array}\right]}{6.87\text{kg}+33.57\text{kg}+14.07\text{kg}+7.54\text{kg}}\\ =\frac{6.41\text{kg}\cdot \text{m}}{62.05\text{kg}}\\ =0.103\text{m}\end{array}$

Conclusion:

The x component of the centre of gravity is $0.459\text{m}$. The y component of the centre of gravity is $0.103\text{m}$.