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Momento de Inercia
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Equations
The relationship between the angular Momentum ($L$) and the moment ($p$) is expressed as:
| $ L = r p $ |
Using the radius ($r$), this expression can be equated with the moment of Inertia ($I$) and the angular Speed ($\omega$) as follows:
| $ L = I \omega $ |
Then, substituting with the inertial Mass ($m_i$) and the speed ($v$):
| $ p = m_i v $ |
and
| $ v = r \omega $ |
it can be concluded that the moment of inertia of a particle rotating in an orbit is:
| $ I = m_i r ^2$ |
(ID 3602)
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
(ID 4432)
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)
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_c ^2$ |
.
(ID 4434)
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_c ^2)$ |
.
(ID 4435)
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_e ^2$ |
.
(ID 4436)
Examples
For a particle of mass the point Mass ($m$) orbiting around an axis at a distance the radius ($r$), the relationship can be established by comparing the angular Momentum ($L$), expressed in terms of the moment of Inertia ($I$) and the moment ($p$), which results in:
| $ I = m_i r ^2$ |
.
(ID 3602)
The moment of Inertia at the CM of a thin Bar, perpendicular Axis ($I_{CM}$) is obtained as a function of the body mass ($m$) and the length of the Bar ($l$):
| $ I_{CM} =\displaystyle\frac{1}{12} m l ^2$ |
(ID 4432)
The moment of Inertia at the CM of a Cylinder, Axis perpendicular to the Cylinder Axis ($I_{CM}$) is obtained as a function of the body mass ($m$), the cylinder Height ($h$) and the radius of a Cylinder ($r_c$):
| $ I_{CM} =\displaystyle\frac{1}{12} m ( h ^2+3 r_c ^2)$ |
(ID 4435)
The moment of Inertia at the CM of a Cylinder, Axis parallel to the Cylinder Axis ($I_{CM}$) is obtained as a function of the body mass ($m$) and the radius of a Cylinder ($r_c$):
| $ I_{CM} =\displaystyle\frac{1}{2} m r_c ^2$ |
(ID 4434)
The moment of Inertia at the CM of a thin Bar, perpendicular Axis ($I_{CM}$) is obtained as a function of the body mass ($m$), the length of the Edge of the Straight Parallelepiped ($a$) and the width of the Edge of the Straight Parallelepiped ($b$):
| $ I_{CM} =\displaystyle\frac{1}{12} m ( a ^2+ b ^2)$ |
(ID 4433)
The moment of Inertia at the CM of a Sphere ($I_{CM}$) is obtained as a function of the body mass ($m$) and the radio of the Sphere ($r_e$):
| $ I_{CM} =\displaystyle\frac{2}{5} m r_e ^2$ |
(ID 4436)
The moment of inertia for axis that does not pass through the CM ($I$) can be calculated using the moment of Inertia Mass Center ($I_{CM}$) and adding the moment of inertia of the body mass ($m$) as if it were a point mass at the distance Center of Mass and Axis ($d$):
| $ I = I_{CM} + m d ^2$ |
(ID 3688)
ID:(678, 0)