Chapter 10, Problem 58SP

Determine the moment of inertia (a) of a vertical thin hoop of mass 2 kg and radius 9 cm about a horizontal, parallel axis at its rim; (b) of a solid sphere of mass 2 kg and radius 5 cm about an axis tangent to the sphere.

To determine

The moment of inertia of the vertical thin hoop of mass $2\text{ kg}$ and radius $9\text{ cm}$ about a horizontal, parallel axis at its rim.

Answer to Problem 58SP

Solution:

$0.03\text{ kg}\cdot {\text{m}}^2$

Explanation of Solution

Given data:

The mass of the vertical thin hoop is $2\text{ kg}$.

The radius of vertical thin hoop is $9\text{ cm}$.

Formula used:

Write the expression of moment of inertia of vertical hoop about an axis through the center of mass.

$I=M{r}^2$

Here, $I$ is the moment of inertia of vertical hoop about an axis through the center of mass, $M$ is the mass of vertical hoop, and $r$ is the radius of hoop.

Write the expression of moment of inertia of object about any axis parallel to the axis passing through the center of mass.

$I=I_{cm}+M{h}^2$

Here, $I$ is the moment of inertia of object about any axis parallel to the axis passing through the center of mass, $I_{cm}$ is the moment of inertia of object about an axis passing through the center of mass, $M$ is the mass of the object, and $h$ is the perpendicular distance between two parallel axis.

Explanation:

The expression of moment of inertia of vertical hoop about an axis through the center of mass is,

$I=M{r}^2$

Substitute $\text{2 kg}$ for $M$ and $9\text{cm}$ for $r$

$\begin{array}{c}I=\left(2\text{ kg}\right){\left(9\text{ cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)\right)}^2\\ =0.0162\text{kg}\cdot {\text{m}}^2\end{array}$

The horizonal, parallel axis at its rim is at a distance equal to the radius of vertical hoop, from the center of the hoop.

The expression of moment of inertia of vertical hoop about horizonal, parallel axis at its rim is,

$I=I_{cm}+M{r}^2$

Substitute $0.0162\text{ kg}\cdot {\text{m}}^2$ for $I_{cm}$, $2\text{ kg}$ for $M$, and $9\text{cm}$ for $r$

$\begin{array}{c}I=\left(0.0162\text{ kg}\cdot {\text{m}}^2\right)+\left(2\text{ kg}\right){\left(9\text{ cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)\right)}^2\\ =\left(0.0162\text{ kg}\cdot {\text{m}}^2\right)+\left(0.0162\text{kg}\cdot {\text{m}}^2\right)\\ =0.03\text{ kg}\cdot {\text{m}}^2\end{array}$

Conclusion:

The moment of inertia of the vertical thin hoop of mass $2\text{ kg}$ and radius $9\text{ cm}$ about a horizontal, parallel axis at its rim is $0.03\text{kg}\cdot {\text{m}}^2$.

To determine

The moment of inertia of the solid sphere of mass $2\text{ kg}$ and radius $5\text{ cm}$ about an axis tangent to the sphere.

Answer to Problem 58SP

Solution:

$7\times {10}^{-3}\text{ kg}\cdot {\text{m}}^2$

Explanation of Solution

Given data:

The mass of the solid sphere is $2\text{ kg}$.

The radius of solid sphere is $\text{5 cm}$.

Formula used:

Write the expression of moment of inertia of solid sphere about an axis through the center of mass.

$I=\frac{2}{5}M{r}^2$

Here, $I$ is the moment of inertia of solid sphere about an axis through the center of mass, $M$ is the mass of solid sphere, and $r$ is the radius of solid sphere.

Write the expression of moment of inertia of object about any axis parallel to the axis passing through the center of mass.

$I=I_{cm}+M{h}^2$

Here, $I$ is the moment of inertia of object about any axis parallel to the axis passing through the center of mass, $I_{cm}$ is the moment of inertia of object about an axis passing through the center of mass, $M$ is the mass of the object, and $h$ is the perpendicular distance between two parallel axis.

Explanation:

The expression of moment of inertia of solid sphere about an axis through the center of mass is,

$I=\frac{2}{5}M{r}^2$

Substitute $\text{2 kg}$ for $M$ and $5\text{cm}$ for $r$

$\begin{array}{c}I=\frac{2}{5}\left(\left(2\text{ kg}\right){\left(\text{5 cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)\right)}^2\right)\\ =2\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

The axis tangent to the sphere is at a distance equal to the radius of solid sphere, from the center of sphere.

The expression of moment of inertia of solid sphere about an axis tangent to the sphere,

$I=I_{cm}+M{r}^2$

Substitute $2\times {\text{10}}^{-3}\text{ kg}\cdot {\text{m}}^2$ for $I_{cm}$, $2\text{ kg}$ for $M$, and $5\text{cm}$ for $r$

$\begin{array}{c}I=\left(2\times {\text{10}}^{-3}\text{ kg}\cdot {\text{m}}^2\right)+\left(2\text{ kg}\right){\left(\text{5 cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)\right)}^2\\ =\left(2\times {\text{10}}^{-3}\text{ kg}\cdot {\text{m}}^2\right)+\left(\text{5}\times {\text{10}}^{-3}\text{kg}\cdot {\text{m}}^2\right)\\ =7\times {10}^{-3}\text{ kg}\cdot {\text{m}}^2\end{array}$

Conclusion:

The moment of inertia of solid sphere about an axis tangent to the sphere is $7\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$.