Chapter 10, Problem 29SP

A homogeneous cylinder of radius R and mass m has a moment of inertia about its central axis given by. If a cylinder has a mass of 4000 g and a diameter of 20 cm, what is its moment of inertia about that central axis?

To determine

The moment of inertia about the central axis of a homogeneous cylinder having a mass of 4000 g and a diameter of 20 cm.

Answer to Problem 29SP

Solution:

$0.02\text{ kg}\cdot {\text{m}}^2$

Explanation of Solution

Given data:

The mass of the cylinder is $4000\text{ g}$.

The diameter of the cylinder is $20\text{ cm}$.

Formula used:

Write the expression for moment of inertia about the central axis of a homogeneous cylinder:

$I=\frac{1}{2}m{R}^2$

Here, $m$ is the mass of the homogeneous cylinder and $R$ is the radius of the cylinder.

Explanation:

Recall the expression for moment of inertia about the central axis of a homogeneous cylinder:

$I=\frac{1}{2}m{R}^2$

Substitute $4000\text{ g}$ for $m$ and $10\text{ cm}$ for $R$

$\begin{array}{c}I=\frac{1}{2}\left(4000\text{ g}\left(\frac{\text{1 kg}}{1000\text{ g}}\right)\right){\left(\left(10\text{ cm}\right)\left(\frac{1\text{ m}}{100\text{ cm}}\right)\right)}^2\\ =0.02\text{ kg}\cdot {\text{m}}^2\end{array}$

Conclusion:

The moment of inertia of the given homogeneous cylinder is $0.02\text{ kg}\cdot {\text{m}}^2$.