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Funcionamiento del Músculo
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Calculations
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Equations
(ID 3326)
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)
(ID 3702)
(ID 3705)
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)
(ID 4438)
Examples
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_i r ^2$ |
resulting in a moment of inertia as
| $I_t=\sum_kI_k$ |
.
(ID 4438)
$x_0=-l_s$
(ID 3705)
The relationship between
| $ c ^2= a ^2+ b ^2$ |
(ID 3326)
$L_z=\displaystyle\frac{L}{n_z}$
(ID 3695)
$l_s=\displaystyle\frac{L_z}{4}$
(ID 3697)
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)
$t_z=\displaystyle\frac{t}{2n_z}$
(ID 3696)
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)
$t_s=\displaystyle\frac{t_z}{2}$
(ID 3698)
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)
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)
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)
$V=\pi r^2h$
(ID 3702)
The angle
| $\theta=\arctan\displaystyle\frac{b}{a}$ |
The
To calculate the corresponding function can be used
(ID 3333)
$l_e=\displaystyle\frac{l_b}{2}$
(ID 3700)
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)
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)
ID:(318, 0)