Equivalent Mass and Inertia

Tip

This is basically applied to terms which have an acceleration or angular acceleration attached to it

Equivalent Mass

Basically the idea here, is to use an equation like KE (which includes linear and rotational kinetic energy), and combine terms (you will likely need to put all terms in terms of a single acceleration) to form something that looks like this
$KE=\frac{1}{2}\left(m_1+m_2+\cdots \right)x^{{}^{\prime \prime }}$
Where the masses in the brackets are your equivalent mass which can be used in other equations like N2L.

Equivalent Inertia

From Gears

Kinetic Energy of a Shaft and Spur Gear System

KE = $\frac{1}{2}{I}_1{\omega }_1^2+\frac{1}{2}{I}_2{\omega }_2^2$
and
${\omega }_2={\omega }_1\frac{n_1}{n_2}$
KE = $\frac{1}{2}\left[I_1+I_2{\left(\frac{n_1}{n_2}\right)}^2\right]{\omega }_1^2$

Which is equivalent inertia felt on the input shaft similar to equivalent mass.
$KE=\frac{1}{2}{I}_e{\omega }_1^2$
$I_e$ can be modelled as:
$I_e=I_1+I_2{\left(\frac{n1}{n_2}\right)}_{}^2$ and transferring the inertia to the first shaft, leaving the second shaft with no inertia attached which makes shafts $I_1$ and $I_2$ a |Lumped System as $I_e$.

Inertia of Rods

The Inertia of a rod pivoting at its end is:
$I=m{L}^2$

Mechanical Drivers

• Gears
• Rack and Pinions
• Sliding vs Rolling Motion

Equivalent Mass and Kinetic Energy for Spring-Mass Systems

See Week 5 notes for visuals Say we have a mass on a lever at point A
$KE=\frac{1}{2}{m}_A{y}_A^{\prime 2}+\frac{1}{2}{I}_o{\theta }^{\prime 2}$ Where
$y_A=\theta l_A$ and I can be found based on the geometry of the problem (for a rod)

You can substitute y and I into the kinetic energy expression yielding…
$\frac{1}{2}\left(m_A{l}_A^2+\frac{1}{3}m{l}_B^2\right){\theta }^{{}^{\prime }2}$
Which is the equivalent mass where $I_B$ is the length of the lever.

We can also find an equivalent system with one mass at the end of the lever and no inertia for the lever

The kinetic energy of this system would be
$KE=\frac{1}{2}{m}_B{y}_B^{{}^{\prime }2}=\frac{1}{2}{m}_B{l}_B^2{\theta }^{{}^{\prime }2}$
Where by equating the equivalent masses we can solve for $m_b$ which is the the equivalent mass needed to oppose $m_A$?

$m_B=\frac{l_A^2}{l_B^2}{m}_A+\frac{1}{3}m$
This effectively behaves the exact same as the previous system but is much simpler in terms of the equivalent mass because we can dodge having to deal with inertias of the rod