Formula: Thin Rod (Rotation Perpendicular to Symmetry Axis) Moment of inertia Mass Length
$$\class{brown}{I} ~=~ \frac{1}{12} \, \class{brown}{m} \, l^2$$
$$\class{brown}{I} ~=~ \frac{1}{12} \, \class{brown}{m} \, l^2$$
$$\class{brown}{m} ~=~ \frac{12\class{brown}{I}}{l^2}$$
$$l ~=~ \sqrt{ \frac{12\class{brown}{I}}{\class{brown}{m}} }$$
Moment of inertia
$$ \class{brown}{I} $$ Unit $$ \mathrm{kg} \, \mathrm{m}^2 $$
According to \( M ~=~ I \, \alpha \) (\(\alpha\): angular acceleration), the moment of inertia determines how hard it is to generate a torque \(M\) on the body. Moment of inertia \(I\) depends on the mass distribution and on the choice of the axis of rotation. Here we calculate the moment of inertia of a thin rod whose axis of rotation is perpendicular to the axis of symmetry and goes through the center.
Mass
$$ \class{brown}{m} $$ Unit $$ \mathrm{kg} $$
Total mass of a thin rod that is homogeneously distributed. The larger the mass, the larger the moment of inertia.
Length
$$ l $$ Unit $$ \mathrm{m} $$
Length of the rod. When the length is doubled, the moment of inertia is quadrupled.