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Moment of inertia calculation method
Equation
The total moment of inertia $I_t$ of an object is calculated by summing the moments of inertia of its components that are comparable to the moment of inertia of an individual particle,
| $ I = m r ^2$ |
resulting in a moment of inertia as
 $I_t=\sum_kI_k$ |
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ID:(4438, 0)
Pythagoras
Equation
The relationship between
| $ c ^2= a ^2+ b ^2$ |
ID:(3326, 0)
Moment of inertia of bar length $l$ axis $\perp$
Equation
The moment of inertia of a rod rotating around a perpendicular ($\perp$) axis passing through the center is obtained by dividing the body into small volumes and summing them:
| $ I =\displaystyle\int_V r ^2 \rho dV $ |
resulting in
| $ I_{CM} =\displaystyle\frac{1}{12} m l ^2$ |
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ID:(4432, 0)
Cylinder moment of inertia, axis $\perp$
Equation
The moment of inertia of a cylinder rotating around a perpendicular ($\perp$) axis passing through the center is obtained by segmenting the body into small volumes and summing them:
| $ I =\displaystyle\int_V r ^2 \rho dV $ |
resulting in
| $ I_{CM} =\displaystyle\frac{1}{12} m ( h ^2+3 r ^2)$ |
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ID:(4435, 0)
Cylinder moment of inertia, axis $\parallel$
Equation
The moment of inertia of a cylinder rotating around an axis parallel ($\parallel$) to its central axis is obtained by segmenting the body into small volumes and summing them:
| $ I =\displaystyle\int_V r ^2 \rho dV $ |
resulting in
| $ I_{CM} =\displaystyle\frac{1}{2} m r ^2$ |
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ID:(4434, 0)
Moment of inertia of a right parallelepiped
Equation
The moment of inertia of a parallelepiped rotating around an axis passing through its center is obtained by partitioning the body into small volumes and summing them up:
| $ I =\displaystyle\int_V r ^2 \rho dV $ |
resulting in
| $ I_{CM} =\displaystyle\frac{1}{12} m ( a ^2+ b ^2)$ |
.
ID:(4433, 0)
Water density
Equation
When working with water, it's also crucial to consider the variable the water density ($\rho_w$), which is calculated using the mass of water in the soil ($M_w$) and the water Volume ($V_w$) with the following equation:
| $ \rho_w =\displaystyle\frac{ M_w }{ V_w }$ |
ID:(4730, 0)
Moment of inertia of a sphere
Equation
The moment of inertia of a sphere rotating around an axis passing through its center is obtained by segmenting the body into small volumes and summing:
| $ I =\displaystyle\int_V r ^2 \rho dV $ |
resulting in
| $ I_{CM} =\displaystyle\frac{2}{5} m r ^2$ |
.
ID:(4436, 0)
Arc Tangent
Equation
The angle
| $\theta=\arctan\displaystyle\frac{b}{a}$ |
The
To calculate the corresponding function can be used
ID:(3333, 0)
Steiner theorem
Equation
When the axis of rotation does not pass through the center of mass (CM), the moment of inertia $I$ can be calculated using the Steiner's Theorem. To do this, we start with the moment of inertia with respect to the center of mass, for example:
• For a bar with a perpendicular axis, it is calculated using length of the Bar $m$, mass of Object $kg$ and moment of Inertia at the CM of a thin Bar, perpendicular Axis $kg m^2$
| $ I_{CM} =\displaystyle\frac{1}{12} m l ^2$ |
• For a cylinder with a perpendicular axis, it is calculated using cylinder Height $m$, mass of Object $kg$, moment of Inertia at the CM of a Cylinder, Axis perpendicular to the Cylinder Axis $kg m^2$ and radius of a circle $m$
| $ I_{CM} =\displaystyle\frac{1}{12} m ( h ^2+3 r ^2)$ |
• For a cylinder with a parallel axis, it is calculated using mass of Object $kg$, moment of Inertia at the CM of a Cylinder, Axis parallel to the Cylinder Axis $kg m^2$ and radius of a circle $m$
| $ I_{CM} =\displaystyle\frac{1}{2} m r ^2$ |
• For a parallelepiped, it is calculated using length of the Edge of the Straight Parallelepiped $m$, mass of Object $kg$, moment of Inertia at the CM of Parallelepiped, Center to the Face $kg m^2$ and width of the Edge of the Straight Parallelepiped $m$
| $ I_{CM} =\displaystyle\frac{1}{12} m ( a ^2+ b ^2)$ |
• For a cube, it is calculated using
| $ I_{CM} =\displaystyle\frac{1}{6} m a ^2$ |
• For a sphere, it is calculated using mass of Object $kg$, moment of Inertia at the CM of a Sphere $kg m^2$ and radius of a circle $m$
| $ I_{CM} =\displaystyle\frac{2}{5} m r ^2$ |
Then, the product of mass and the square of the distance between the axis of rotation and the center of mass is added using mass of Object $kg$, moment of Inertia at the CM of a Sphere $kg m^2$ and radius of a circle $m$
| $ I = I_{CM} + m d ^2$ |
ID:(3688, 0)
Moment of inertia of a particle
Equation
For a particle with mass $m$ orbiting around an axis at a distance equivalent to a radius $r$, the relationship can be established by comparing the angular momentum expressed in terms of the moment of inertia and in terms of the moment, which is equal to:
| $ I = m r ^2$ |
La relación entre momento angular y momento es igual a
| $ L = r p $ |
se puede igualar a
| $ L = I \omega $ |
que tras remplazar
| $ p = m_i v $ |
y
| $ v = r \omega $ |
se puede concluir que la momento de inercia de una partícula girando en una órbita es
| $ I = m r ^2$ |
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ID:(3602, 0)