New answers tagged rigid-body-dynamics
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Because the definition of rigid body requires that the distance between its points doesn't change. If we follow that model, only translations and rotations are possible.
If we modify the model from a rigid body to an elastic body, mechanical vibrations are included.
For a more complete description, heat transfer must be also included, resulting from external ...
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we add the total translational kinetic energy and the total rotational kinetic energy of the constituent particles. Why is the total vibrational kinetic energy of the constituent particles left out?
We don’t actually leave out vibrational kinetic energy if the centre of mass of the system is vibrating.
This is a matter of scales at which you’re analysing ...
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A net force or torque on a rigid body will not affect its internal energy. As it remains constant before and after the application of the force/torque, it is not relevant to the equations of motion.
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The triangle inequality for the moment of inertia tensor ultimately stems from the positivity of inertial mass (as shown in Eli's answer.) A body whose inertia tensor did not satisfy the triangle inequalities would necessarily have some region where $\rho < 0$.
We could then consider this body as composed of two sub-bodies, one with strictly positive ...
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A rigid body's principal moments of inertia are obtained from these equations :
$$I_1=\int_V\,(x_2^2+x_3^2)\,\rho\,dV$$
$$I_2=\int_V\,(x_3^2+x_1^2)\,\rho\,dV$$
$$I_3=\int_V\,(x_1^2+x_2^2)\,\rho\,dV$$
where $x=x_1~,y=x_2~,z=x_3$ and the inertia tensor is:
$$I= \left[ \begin {array}{ccc} I_{{1}}&0&0\\ 0&I_{{2}}&0
\\ 0&0&I_{{3}}\end {...
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The rod has six degrees of freedom: X,Y,Z location of its center of mass, two degrees for the direction the axis points, and one degree for rotation around the axis. The speck can move anywhere on the surface of the rod (a two dimensional surface), giving another two degrees of freedom. Presumably the speck is a dimensionless point, so does not have ...
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Three coordinates $(x, y, z)$ are required to describe the pole's position. Two coordinates $(\theta, \phi)$ are required to describe the pole's orientation (think latitude and longitude). Finally, one coordinate is required to describe the position of the speck on the pole. That makes six total degrees of freedom.
Edit: To clarify, this answer assumes that ...
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The moment due to a force $\boldsymbol{F}$ acting through some location $\boldsymbol{r}_1$ in space is $$\boldsymbol{M}_0 = \boldsymbol{r}_1 \times \boldsymbol{F} \tag{1}$$
Now transfer the moment from the origin 0 to point 1 where the force applies
$$ \boldsymbol{M}_1 = \boldsymbol{M}_0 + (- \boldsymbol{r}_1) \times \boldsymbol{F} = \boldsymbol{r}_1 \times \...
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Two systems of forces are dynamically equivalent if they have the same sum $\bf R$ (net force)
and the same moment $\bf M$ about a given pole $O$. It's worth noting that
$\bf M\cdot R$ is an invariant independent of the choice of the pole $O$, because the transformation rule of the moment is $\bf M' = \bf M + (O-O')\times \bf R $.
If $\bf M\cdot R = 0$ and $...
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If net external moment of some forces is zero about a point, then the net external force passes through the point
It means that when n number of forces are acting on a body,and if about certain point the net torque is zero (say P), then you find the net force on the body and apply it at that point, P (in free body diagram) and you will get the same dynamics ...
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@JAlex Answered the question in the comments. $\eta$ is NOT a cartesian vector. The attitude matrix converts a cartesian angular rate of the rotating frame (say $\omega_{BN}^{N}$) to the rigid-body-orientation-parameter-representation rate of change $\dot{\eta}$! I call them parameters because their derivative ($\dot{\eta}$) is not to be confused with an ...
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