Chapter 10.8, Problem 84P

Determine the moment of inertia of the thin ring about the z axis. The ring has a mass m.

To determine

The moment of inertia of the thin ring about the z-axis.

Answer to Problem 84P

The moment of inertia of the thin ring about the z-axis is $\underset{\_}{m{R}^2}$.

Explanation of Solution

Given:

The radius of the ring is $R$.

The mass of the ring is $m$.

Explanation:

Show the intersection of the ring at the arbitrary point $\left(x,y\right)$ as in Figure (1).

Conclusion:

From Figure 1,

Calculate the mass of the ring.

$m=\rho V$ (I)

Here, the density of the material is $\rho$, and the volume of the ring is $V$.

Substitute $2\pi R$ for $V$ in Equation (I).

$m=\rho 2\pi R$ (II)

Here, the density of the material is $\rho$ and the radius is $R$.

Calculate the density of the material from the Equation (II).

$\begin{array}{l}m=\rho 2\pi R\\ \rho =\frac{m}{2\pi R}\end{array}$

Compute the moment of inertia about the $z$-axis.

$I_z={\displaystyle {\int }_0^{2\pi }\rho R^{2}Rd\theta }$ (III)

Substitute $\frac{m}{2\pi R}$ for $\rho$ in Equation (III).

$\begin{array}{c}{I}_z={\displaystyle {\int }_0^{2\pi }\left[\frac{m}{2\pi R}\right]R^{2}Rd\theta }\\ ={\displaystyle {\int }_0^{2\pi }\frac{m}{2\pi }{R}^{2}d\theta }\\ =\frac{m{R}^2}{2\pi }{\displaystyle {\int }_0^{2\pi }d\theta }\end{array}$

$\begin{array}{c}=\frac{m{R}^2}{2\pi }\left(2\pi \right)\\ =m{R}^2\end{array}$

Hence, the moment of inertia of the thin ring about the z-axis is $\underset{\_}{m{R}^2}$.