Chapter 14, Problem 1AP

The radius of gyration of the thigh with respect to the transverse axis at the hip is 54% of the segment length. The mass of the thigh is 10.5% of total body mass, and the length of the thigh is 23.2% of total body height. What is the moment of inertia of the thigh with respect to the hip for males of the following body masses and heights?

	MASS (kg)	HEIGHT (m)
A	60	1.6
B	60	1.8
C	70	1.6
D	70	1.8

(Answers: A = 0.25 kg m²; B = 0.32 kg m²; C = 0.30 kg m²; D = 0.37 kg m²)

Summary Introduction

To determine: The moment of inertia of the thigh with respect to the hip for males A, B, C, and D.

Answer to Problem 1AP

The moment of inertia of the thigh with respect to the hip for males A, B, C, and D are $0.25\text{kg}\cdot {\text{m}}^2$, $0.32\text{kg}\cdot {\text{m}}^{\text{2}}$, $0.30\text{kg}\cdot {\text{m}}^{\text{2}}$, and $0.37\text{kg}\cdot {\text{m}}^{\text{2}}$.

Explanation of Solution

Calculation:

Expression for the moment of inertia is,

$I=m{k}^2$ (1)

Here, $I$ is moment of inertia of the thigh, $m$ is mass of the thigh, and $k$ is radius of gyration.

The radius gyration is certain percentage of length of the thigh and length of the thigh is certain percentage of total body height.

The moment of inertia of male A:

Rewrite equation (1).

$I=m{k}^2$

Substitute $\left(10.5\%\times 60\text{kg}\right)$ for $m$ and $\left(54\%\times 23.2\%\times 1.6\text{m}\right)$ for $k$ to find $I$.

$\begin{array}{c}I=\left(10.5\%\times 60\text{kg}\right){\left(54\%\times 23.2\%\times 1.6\text{m}\right)}^2\\ =0.253\text{kg}\cdot {\text{m}}^{\text{2}}\\ \approx 0.25\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

Thus, the moment of inertia of male A is $0.25\text{kg}\cdot {\text{m}}^{\text{2}}$.

The moment of inertia of male B:

Rewrite equation (1).

$I=m{k}^2$

Substitute $\left(10.5\%\times 60\text{kg}\right)$ for $m$ and $\left(54\%\times 23.2\%\times 1.8\text{m}\right)$ for $k$ to find $I$.

$\begin{array}{c}I=\left(10.5\%\times 60\text{kg}\right){\left(54\%\times 23.2\%\times 1.8\text{m}\right)}^2\\ =0.3204\text{kg}\cdot {\text{m}}^{\text{2}}\\ \approx 0.32\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

Thus, the moment of inertia of male B is $0.32\text{kg}\cdot {\text{m}}^{\text{2}}$.

The moment of inertia of male C:

Rewrite equation (1).

$I=m{k}^2$

Substitute $\left(10.5\%\times 70\text{kg}\right)$ for $m$ and $\left(54\%\times 23.2\%\times 1.6\text{m}\right)$ for $k$ to find $I$.

$\begin{array}{c}I=\left(10.5\%\times 70\text{kg}\right){\left(54\%\times 23.2\%\times 1.6\text{m}\right)}^2\\ =0.295\text{kg}\cdot {\text{m}}^{\text{2}}\\ \approx 0.30\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

Thus, the moment of inertia of male C is $0.30\text{kg}\cdot {\text{m}}^{\text{2}}$.

The moment of inertia of male D:

Rewrite equation (1).

$I=m{k}^2$

Substitute $\left(10.5\%\times 70\text{kg}\right)$ for $m$ and $\left(54\%\times 23.2\%\times 1.8\text{m}\right)$ for $k$ to find $I$.

$\begin{array}{c}I=\left(10.5\%\times 70\text{kg}\right){\left(54\%\times 23.2\%\times 1.8\text{m}\right)}^2\\ =0.374\text{kg}\cdot {\text{m}}^{\text{2}}\\ \approx 0.37\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

Thus, the moment of inertia of male B is $0.37\text{kg}\cdot {\text{m}}^{\text{2}}$.

Conclusion

Therefore, the moment of inertia of the thigh with respect to the hip for males A, B, C, and D are $0.25\text{kg}\cdot {\text{m}}^2$, $0.32\text{kg}\cdot {\text{m}}^{\text{2}}$, $0.30\text{kg}\cdot {\text{m}}^{\text{2}}$, and $0.37\text{kg}\cdot {\text{m}}^{\text{2}}$.