Chapter 19, Problem 19.159RP

An automobile wheel-and-tire assembly of total weight 47 lb is attached to a mounting plate of negligible weight that is suspended from a steel wire. The torsional spring constant of the wire is known to be K = 0.40 lb·in/rad. The wheel is rotated through 90° about the vertical and then released. Knowing that the period of oscillation is observed to be 30 s, determine the centroidal mass moment of inertia and the centroidal radius of gyration of the wheel-and-tire assembly.

Fig. P19.159

To determine

The centroidal mass moment of inertia $\left(\overline{I}\right)$ and the centroidal radius of gyration $\left(\overline{k}\right)$ of the wheel and tire assemble.

Answer to Problem 19.159RP

The centroidal mass moment of inertia $\left(\overline{I}\right)$ and the centroidal radius of gyration $\left(\overline{k}\right)$ of the wheel and tire assemble is $0.7599\text{lb}\cdot {\text{s}}^2\cdot \text{ft}$ and 8.66 in. respectively.

Explanation of Solution

Given information:

The total weight of the automobile (W) is 47 lb.

The total spring constant of wire (K) is $0.40\text{lb}\cdot \text{in}\text{.}/\text{rad}$.

The wheel rotates at an angle $\left(\theta \right)$ is $90°$.

The time period of oscillation $\left(\tau \right)$ is 30 sec.

The acceleration due to gravity (g) is $32.2\text{ft}/{\text{s}}^{\text{2}}$.

Calculation:

Calculate the frequency of oscillation (f) using the formula:

$f=\frac{1}{\tau }$

Substitute 30 s for $\tau$.

$\begin{array}{c}f=\frac{1}{30\text{s}}\\ =0.03333\text{Hz}\end{array}$

Calculate the natural circular frequency of oscillation $\left({\omega }_n\right)$ using the formula:

${\omega }_n=2\pi f$

Substitute 0.03333 Hz for f.

$\begin{array}{c}{\omega }_n=2\pi \left(0.03333\text{Hz}\right)\\ =0.20942\text{rad}/\text{s}\end{array}$

When an angular displacement of $\theta$ is given to the wheel-and-tire assembly, the steel wire from which the assembly is suspended is twisted. This twist causes a torque of $K\theta$ on the steel wire. The external forces in the system is torque due to the steel wire $-K\theta$ (negative sign indicates that it acts against the system).

Take moment for external forces as follows:

${\left(\sum M\right)}_{external}=-K\theta$

Here, ${\left(M\right)}_{external}$ is the moment for external forces.

Restoring couple acts on the system due to the angular displacement of $\theta$. The effective force in the system is restoring couple, $\overline{I}\ddot{\theta }$. Here, the moment of inertia of the assembly is $\overline{I}$ and the angular acceleration is $\ddot{\theta }$.

Take moment for effective forces as follows;

${\left(\sum M\right)}_{effective}=\overline{I}\ddot{\theta }$

Here, ${\left(M\right)}_{effective}$ is the moment for effective forces.

Equate the moment for external and effective forces in the system using the relation:

$\begin{array}{l}{\left(\sum M\right)}_{external}={\left(\sum M\right)}_{effective}\\ -K\theta =\overline{I}\ddot{\theta }\\ \ddot{\theta }=-\frac{K}{\overline{I}}\theta \\ \ddot{\theta }+\frac{K}{\overline{I}}\theta =0\end{array}$ (1)

The expression for the general differential equation of motion as follows:

$\ddot{x}+{\omega }_n^{2}x=0$ (2)

Find the expression for the natural circular frequency of vibration:

Compare the differential equations (1) and (2).

${\omega }_n^2=\frac{K}{\overline{I}}$

Substitute $0.20942\text{rad}/\text{s}$ for ${\omega }_n$, and $0.03333\text{lb}\cdot \text{ft}/\text{rad}$ for K.

$\begin{array}{l}{\left(0.20942\text{rad}/\text{s}\right)}^2=\frac{0.40\frac{\text{lb}\cdot \left(\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)}{\text{rad}}}{\overline{I}}\\ \overline{I}=\frac{0.03333}{{\left(0.20942\right)}^2}\\ \overline{I}=0.7599\text{lb}\cdot {\text{s}}^2\cdot \text{ft}\end{array}$

The expression for the centroidal mass moment of inertia as follows:

$\overline{I}=m{\overline{k}}^2$ (3)

Here, m is the mass of the wheel-and-tire assembly and $\overline{k}$ is the radius of gyration of the wheel-and-tire assembly.

Calculate the mass of the wheel-and-tire assembly (m) using the formula:

$m=\frac{W}{g}$

Substitute 47 lb for W and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$\begin{array}{c}m=\frac{47\text{lb}}{32.2\text{ft}/{\text{s}}^{\text{2}}}\\ =1.4596\text{lb}\cdot {\text{s}}^2/\text{ft}\end{array}$

Substitute $0.7599\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $\overline{I}$ and $1.4596\text{lb}\cdot {\text{s}}^2/\text{ft}$ for m in equation (3).

$\begin{array}{l}\left(0.7599\text{lb}\cdot {\text{s}}^2\cdot \text{ft}\right)=\left(1.4596\text{lb}\cdot {\text{s}}^2/\text{ft}\right){\overline{k}}^2\\ {\overline{k}}^2=\frac{0.7599}{1.4596}\\ \overline{k}=\sqrt{\frac{0.7599}{1.4596}}\\ \overline{k}=0.7215\text{ft}\times \frac{12\text{in}\text{.}}{1\text{ft}}\\ \overline{k}=8.66\text{in}\text{.}\end{array}$

Therefore, the centroidal mass moment of inertia $\left(\overline{I}\right)$ and the centroidal radius of gyration $\left(\overline{k}\right)$ of the wheel and tire assemble is $0.7599\text{lb}\cdot {\text{s}}^2\cdot \text{ft}$ and 8.66 in..