Chapter 14, Problem 14D.4P

(a)

Interpretation Introduction

Interpretation:

The expression for moment of inertia and the radii of gyration of a uniform thin disk has to be derived.

Concept introduction:

The moment of inertia of a object about an axis passing through the centre of mass is directly proportional to the mass of each particle present in the object and the distance of each particle from center of mass.  The expression for moment of inertia is shown below.

  $I={\displaystyle \sum _i{m}_i{r}_i^2}$

Answer to Problem 14D.4P

The expression for moment of inertia of a uniform thin disk is shown below.

  $I=M\left(\frac{R^2}{2}\right)$

The expression for the radii of gyration of a uniform thin disk is shown below.

    $R_{\text{g}}=\frac{R}{{\left(2\right)}^{1/2}}$

Explanation of Solution

The structure of disk is shown below.

Figure 1

The total mass of the disk is given by the expression as shown below.

    $M=\rho \text{π}{R}^2$

Where,

• $\rho$ is the density of the disk.
• $R$ is the radius of the disk.

The mass of a particle in disk is given by the expression as shown below.

    $m=\rho \text{π}{r}^2$

Where,

• $\rho$ is the density of the disk.
• $r$ is the distance of the particle from center of mass.

The component of moment of inertia of disk in $x$ axis and $y$ axis is equal and component of moment of inertia in $z$ axis is zero.

The component of moment of inertia in $x$ is shown below.

    $I_x={\displaystyle {\int }_0^{R}mr}\text{d}r$

Substitute the value of $m$ in the above equation.

    $\begin{array}{c}{I}_x={\displaystyle {\int }_0^R\rho \text{π}{r}^{2}r}\text{d}r\\ =\rho \text{π}{\displaystyle {\int }_0^R{r}^3}\text{d}r\\ =\rho \text{π}{\left[\frac{r^4}{4}\right]}_0^R\\ =\rho \text{π}\left[\frac{R^4}{4}-0\right]\end{array}$

The above equation is further simplified as shown below.

    $\begin{array}{c}{I}_x=\rho \text{π}\left(\frac{R^4}{4}\right)\\ =\rho \text{π}{R}^2\left(\frac{R^2}{4}\right)\left[M=\rho \text{π}{R}^2\right]\\ =M\left(\frac{R^2}{4}\right)\end{array}$

The moment of inertia of uniform disk is given by the expression as shown below.

    $I=I_x+I_y+I_z$        (1)

Where,

• $I_x$ is the component of moment of inertia of disk in $x$ axis.
• $I_y$ is the component of moment of inertia of disk in $y$ axis.
• $I_z$ is the component of moment of inertia of disk in $z$ axis.

Substitute the values of $I_x$, $I_y$ and $I_z$ in the above equation.

    $\begin{array}{c}I=M\left(\frac{R^2}{4}\right)+M\left(\frac{R^2}{4}\right)+0\\ =M\left(2\right)\left(\frac{R^2}{4}\right)\\ =M\left(\frac{R^2}{2}\right)\end{array}$

The relation between moment of inertia and radius of gyration $\left(R_{\text{g}}\right)$ is shown below.

    $R_{\text{g}}=\sqrt{\frac{I}{M}}$

Substitute the value of $I$ and $M$ in the above expression.

    $\begin{array}{c}{R}_{\text{g}}=\sqrt{\frac{M\left(\frac{R^2}{2}\right)}{M}}\\ =\frac{R}{{\left(2\right)}^{1/2}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The expression for moment of inertia and the radii of gyration of a long uniform rod has to be derived.

Concept introduction:

As mentioned in the concept introduction in part (a).

Answer to Problem 14D.4P

The expression for moment of inertia of a uniform rod is shown below.

  $I=M\left(\frac{L^2}{3}\right)$

The expression for the radii of gyration of a uniform rod is shown below.

    $R_{\text{g}}=\frac{L}{{\left(3\right)}^{1/2}}$

Explanation of Solution

The structure of a long uniform rod is shown below.

Figure 2

The total mass of the rod is given by the expression as shown below.

    $M=\rho L$

Where,

• $\rho$ is the density of the rod.
• $L$ is the length of the rod.

The mass of a particle in rod is given by the expression as shown below.

    $m=\rho x$

Where,

• $\rho$ is the density of the rod.
• $x$ is the distance of the particle from center of mass.

The component of moment of inertia is shown below.

    $I={\displaystyle {\int }_0^{L}m{x}^2}\text{d}x$

Substitute the value of $m$ in the above equation.

    $\begin{array}{c}I={\displaystyle {\int }_0^L\rho x^{2}x}\text{d}x\\ =\rho {\displaystyle {\int }_0^L{r}^3}\text{d}r\\ =\rho {\left[\frac{x^3}{3}\right]}_0^L\\ =\rho \left[\frac{L^3}{3}-0\right]\end{array}$

The above equation is further simplified as shown below.

    $\begin{array}{c}I=\rho \left(\frac{L^3}{3}\right)\\ =\rho L\left(\frac{L^2}{3}\right)\left[M=\rho L\right]\\ =M\left(\frac{L^2}{3}\right)\end{array}$

The relation between moment of inertia and radius of gyration $\left(R_{\text{g}}\right)$ is shown below.

    $R_{\text{g}}=\sqrt{\frac{I}{M}}$

Substitute the value of $I$ and $M$ in the above expression.

    $\begin{array}{c}{R}_{\text{g}}=\sqrt{\frac{M\left(\frac{L^2}{3}\right)}{M}}\\ =\frac{L}{{\left(3\right)}^{1/2}}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The expression for moment of inertia and the radii of gyration of a uniform sphere has to be derived.

Concept introduction:

As mentioned in the concept introduction in part (a).

Answer to Problem 14D.4P

The expression for moment of inertia of a uniform sphere is shown below.

  $I=\frac{2}{5}M{R}^2$

The expression for the radii of gyration of a uniform sphere is shown below.

    $R_{\text{g}}={\left(\frac{2}{5}\right)}^{1/2}R$

Explanation of Solution

The structure of sphere is shown below.

Figure 3

The total mass of the sphere is given by the expression as shown below.

    $M=\rho \left(\frac{4}{3}\text{π}{R}^3\right)$

Where,

• $\rho$ is the density of the sphere.
• $R$ is the radius of the sphere.

The moment of inertia of the small disk is given by the expression as shown below.

    $\text{d}I=\frac{\text{π}\rho y^4\text{d}z}{2}$

Where,

• $y$ is the distance of particle in $y$ axis from center of mass.
• $z$ is the distance of particle in $x$ axis from center of mass.

The relation between $y$ and $z$ is shown below.

    $y^2=R^2-z^2$

The moment of inertia is shown below.

    $I={\displaystyle \int \text{d}I}$

Substitute the value of $\text{d}I$ in the above equation.

    $I={\displaystyle {\int }_{-R}^R\frac{\text{π}\rho y^4\text{d}z}{2}}$

Substitute the value of $y$ in the above equation.

    $\begin{array}{c}I={\displaystyle {\int }_{-R}^R\frac{\text{π}\rho {\left(R^2-z^2\right)}^2\text{d}z}{2}}\\ ={\displaystyle {\int }_{-R}^R\frac{\text{π}\rho {\left(R^4+z^4-2R^2{z}^2\right)}^2\text{d}z}{2}}\\ =\frac{1}{2}\pi \rho {\left[R^{4}z-\frac{2}{3}{R}^2{z}^3+z^5\right]}_0^R\end{array}$

The above equation is further simplified as shown below.

    $\begin{array}{c}I={\displaystyle {\int }_{-R}^R\frac{\text{π}\rho {\left(R^2-z^2\right)}^2\text{d}z}{2}}\\ ={\displaystyle {\int }_{-R}^R\frac{\text{π}\rho {\left(R^4+z^4-2R^2{z}^2\right)}^2\text{d}z}{2}}\\ =\frac{1}{2}\pi \rho {\left[R^{4}z-\frac{2}{3}{R}^2{z}^3+z^5\right]}_0^R\\ =\frac{2\times 4}{5\times 3}\pi \rho R^3\times R^2\end{array}$

Substitute the value of the term $\rho \left(\frac{4}{3}\text{π}{R}^3\right)$ in the above equation.

    $I=\frac{2}{5}M{R}^2$

The relation between moment of inertia and radius of gyration $\left(R_{\text{g}}\right)$ is shown below.

    $R_{\text{g}}=\sqrt{\frac{I}{M}}$

Substitute the value of $I$ and $M$ in the above expression.

    $\begin{array}{c}{R}_{\text{g}}=\sqrt{\frac{\frac{2}{5}M{R}^2}{M}}\\ ={\left(\frac{2}{5}\right)}^{1/2}R\end{array}$