Chapter 19, Problem 19.159RP

To determine

The centroidal mass moment of inertia.

The centroidal radius of gyration.

Answer to Problem 19.159RP

The centroidal mass moment of inertia is $0.76\text{lb}\cdot \text{ft}\cdot {\text{s}}^2/\text{rad}$.

The centroidal radius of gyration is $8.66\text{in}$.

Explanation of Solution

Given information:

The weight of the wheel is $47\text{lb}$, the torsional spring constant of the wire is $0.40\text{lb}\cdot \text{in}/\text{rad}$ and the period of the oscillation is $30\text{s}$.

Write the expression for the centroidal moment of inertia.

$I=K_{\text{t}}{\left(\frac{T}{2\pi }\right)}^2$ ...... (I)

Here, the torsional spring constant is $K_{\text{t}}$ and the time period is $T$.

Write the expression for the centroidal radius of the gyration.

$k=\sqrt{\frac{Ig}{W}}$

Here, the acceleration due to gravity is $g$ and the weight of the wheel is $W$.

Calculation:

Substitute $0.40\text{lb}\cdot \text{in}/\text{rad}$ for $K_{\text{t}}$, and $30\text{s}$ for $T$ in Equation (I):

$\begin{array}{c}I=0.40\text{lb}\cdot \text{in}/\text{rad}{\left(\frac{30\text{s}}{2\pi }\right)}^2\\ =0.40\text{lb}\cdot \text{in}/\text{rad}\times \frac{\frac{1}{12}\text{lb}\cdot \text{ft}/\text{rad}}{1\text{lb}\cdot \text{in}/\text{rad}}\times 22.8{\text{s}}^2\\ =0.76\text{lb}\cdot \text{ft}\cdot {\text{s}}^2/\text{rad}\end{array}$

Substitute $0.76\text{lb}\cdot \text{ft}\cdot {\text{s}}^2/\text{rad}$ for $I$, $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$, and $47\text{lb}$ for $W$ in Equation (II):

$\begin{array}{c}k=\sqrt{\frac{0.76\text{lb}\cdot \text{ft}\cdot {\text{s}}^2/\text{rad}\times 32.2\text{ft}/{\text{s}}^{\text{2}}}{47\text{lb}}}\\ =\sqrt{\frac{24.472\text{lb}\cdot {\text{ft}}^2/\text{rad}}{47\text{lb}}}\\ =0.721\text{ft}\times \frac{12\text{in}}{1\text{ft}}\\ =8.66\text{in}\end{array}$

Conclusion:

The centroidal mass moment of inertia is $0.76\text{lb}\cdot \text{ft}\cdot {\text{s}}^2/\text{rad}$.

The centroidal radius of gyration is $8.66\text{in}$.