4. Moment of Inertia

Definitions

Moment of Inertia (I)

• A measure of an object's resistance to rotational acceleration about an axis, measured in kg⋅m²

$$I = \sum_i m_i r_i^2$$

where $r$ is the distance to the axis

Rotational Kinetic Energy

• The kinetic energy associated with rotational motion

$$K = \dfrac{1}{2} I \omega^2$$

 

Derivations

starting with the kinetic energy of a single particle

$$K = \dfrac{1}{2}mv^2 = \dfrac{1}{2}m\omega^2r^2$$

for a system of particles, sum over all particles (assuming $\omega$ is constant)

$$K_{\text{total}} = \sum_{i} \dfrac{1}{2}m_i r_{i}^2 \omega_{i}^2$$

factoring out $\dfrac{1}{2}\omega^2$

$$K_{\text{total}} = \dfrac{1}{2} \omega^2 \sum_{i} m_i r_{i}^2$$

by definition, $I = \sum_{i} m_i r_{i}^2$, therefore

$$\boxed{K_{\text{total}} = \dfrac{1}{2} I \omega^2}$$

 

Calculating Moment of Inertia

for a system of point masses

$$I = \sum_i m_i r_i^2$$

for a continuous body, divide it into small elements $(dm)$

$$I = \int r^2 \, dm$$

for 3D bodies, express $dm$ in terms of density $(\rho)$ and volume element $(dV)$

$$dm = \rho \, dV$$

then the moment of inertia becomes

$$I = \int r^2 \, dm = \int r^2 \rho \, dV$$

we choose $(dV)$ such that all points in the element are approximately the same distance $(r)$ from the axis

 

Useful Equations

$$I = \sum_i m_i r_i^2, \quad K = \dfrac{1}{2} I \omega^2$$ $$I = \int r^2 \, dm = \int r^2 \rho \, dV$$

 

Example 1

Two point masses of 2 kg and 3 kg are located at distances of 1 m and 2 m from an axis of rotation, respectively. What is the moment of inertia of the system?

Answer

Using $I = \sum_i m_i r_i^2$

$$I = m_1 r_1^2 + m_2 r_2^2$$ $$I = 2(1)^2 + 3(2)^2 = 2 + 12 = 14 \text{ kg⋅m}^2$$

 

Example 2

A rotating object has a moment of inertia of 5 kg⋅m² and angular velocity of 4 rad/s. What is its rotational kinetic energy?

Answer

Using $K = \dfrac{1}{2} I \omega^2$

$$K = \dfrac{1}{2}(5)(4)^2$$ $$K = \dfrac{1}{2}(5)(16) = 40 \text{ J}$$