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10

The rectangular prism is a rigid body. The equations of motion of a rigid body around its center of mass are given by: (Please, see for example: Marsden and Ratiu , (page 6). $$I_1\dot\Omega_1=(I_2-I_3)\Omega_2\Omega_3$$ $$I_2\dot\Omega_2=(I_3-I_1)\Omega_3\Omega_1$$ $$I_3\dot\Omega_3=(I_1-I_2)\Omega_1\Omega_2$$ Where $\Omega_1,_2,_3$ are the angular ...


9

The instability inherent in the medium length axis or $\prod_2 $ as shown above is discussed in detail in Marsden and Ratiu, which is where the image is from. The unstable homoclinic orbit that connect the two unstable points have intersting features. Not only are they interesting because of the chaotic solutions via the Poincare-Melnikov method that ...


8

One may prove for an arbitrary rigid body (and wrt. to an arbitrary choice of pivotal point for the rigid body) that the three moments of inertia $I_x$, $I_y$, and $I_z$, around the three principal axes (which we will call $x$, $y$, and $z$) satisfy the triangle inequality, $$\tag{1} I_x +I_y ~\geq~ I_z, \qquad I_y +I_z ~\geq~ I_x, \qquad I_z +I_x ~\geq~ ...


7

I think a reasonable first approximation can be made like this: choose an arbitrary orientation for the die, and figure out, if the die were released in that orientation with its lowermost point resting on a surface, which side would it fall on? That can be easily calculated; you just draw a line going straight down from the center of mass, and whichever ...


7

You can always decompose a motion like this into two parts: (1) rolling without slipping and (2) slipping without rolling. What is slipping without rolling? It means the object moves uniformly in one direction along the surface, with no angular velocity about the object's own center of mass. For instance, a box that is pushed along the ground can easily ...


6

Here's a straightforward but somewhat computational way. There are two steps. (1) Show how to define the angular velocity vector in terms of rotation matrices. (2) Write a general rotation in terms of Euler angles. (3) Combine (1) and (2) to get an expression for the angular velocity vector in terms of Euler angles. Step 1. Recall that if $\mathbf x(t)$ is ...


6

The moment of inertia is a rank 2 tensor not a scalar. You'll commonly see it written as a scalar, but this is because by choosing your axes to line up with the principal axes of the object the matrix representing the moment of inertia can be diagonalised: $$ {\bf I} = \left( \begin{matrix} I_{00} & 0 & 0 \\ 0 & I_{11} & 0 \\ 0 & 0 ...


5

For longitudinal waves, the effect moves with the speed of sound inside the matter which is equal to $$c=\sqrt{\frac{E}{\rho}}$$ where $E$ is the modulus of elasticity and $\rho$ is the mass density. For steel this is like $5000\;{\rm m/s}$. For beam transverse waves, it depends on which harmonic is excited and how many wavelengths fit in the length of the ...


5

You're correct that there is no such thing as a rigid body in reality. Any time a force is applied to an object at one point (such as a boulder being placed on one end of the see-saw), it only immediately applies to the molecules that it is touching. The displacement of those molecules propagates to the rest of the object as a "deformation wave," which is ...


5

You've duplicated constraints because if any one particle is constrainined in all three dimensions with all the other particles this constrains all the particles. The number of constraints is 3(N - 1). To give an example, take three particles a, b and c. If a is fixed relative to b and is also fixed relative to c, then b and c are fixed relative to each ...


5

Building up from the ground is essentially impossible. One reason is stability, as John Rennie points out in his answer, but a far more fundamental reason* has to do with compressive strength. When a space elevator is completed, the cable is under a lot of tension. However, if you build up from the ground then while you're building it it's under compression ...


4

Every rigid body has 3 translational dof. In addition, there are 0, 2, or 3 rotational dof, depending on the geometry, giving a total of 3, 5, or 6 dof. A spherically symmetric rigid body has no other dof. A rigid body with rotational symmetry around an axis has 2 rotational dof, namely two angles for orienting the symmetry axis along a direction. All ...


4

This paper (http://www-stat.stanford.edu/~cgates/PERSI/papers/dyn_coin_07.pdf) shows that the probability distribution of getting a head, if I toss with the head side up is given by: $p(ψ, φ) =\frac{1}{2}+\frac{1}{\pi} \sin^{-1} (\cot(φ) \cot(ψ))$ if $(\cot φ)(\cot ψ) ≤ 1$, =1 if $\cot(φ) \cot(ψ) ≥ 1$ where $\phi, \psi$ are the Euler angles.


4

I assume you know about rotation matrices, and so for a sequence rotations about Z-X-Z with angles $\phi$, $\theta$ and $\psi$ repsectively you have $$ \vec{\omega} = \dot{\phi} \hat{z} + T_1 \left( \dot{\theta} \hat{x} + T_2 \left( \dot{\psi} \hat{z} \right) \right) $$ The logic here is apply a local spin of $\dot{\phi}$, $\dot{\theta}$ and $\dot{\psi}$ ...


3

As a starting point, Wikipedia says on this issue: The principal axes are often aligned with the object's symmetry axes. If a rigid body has an axis of symmetry of order $m$, meaning it is symmetrical under rotations of 360°/$m$ about the given axis, that axis is a principal axis. When , the rigid body is a symmetrical top. If a rigid body has at least ...


3

The laws you are looking for are conservation of momentum and conservation of angular momentum. If you stick the two bodies together both laws still must be fulfilled (inelastic effects neglected). In the end you will have a compound single object. With the parallel axis theorem (wikipedia) you can calculate the mass moment of inertia and together with the ...


3

Your statement is true. Proof: Let $\rho$ be the mass density of the rigid body. Remember that the tensor of inertia $I$ is given by: $$ \vec{v}^t I \vec{w} = \int d^3b\, \rho(\vec{b}) (\vec{v} \cdot \vec{w} - (\vec{v}\cdot \vec{b})(\vec{w}\cdot \vec{b})) $$ for all $\vec{v},\vec{w} \in \mathbb{R}^3$. Now take an orthogonal matrix $O$ which ...


3

It will not rotate around A, since the centrifugal forces will make the body rotate about its center of mass. You can move the torque, it is not fixed to a given point.


3

You have already answered your own question! There is a force between the hinge and the door. If the door weren't attached to the hinge, it would start flying away in addition to spinning. The only error you have is a mistake in your integral. The net force on the door is $\frac{aML}{2}$ in your notation. (Note: it is probably best not to use $a$ for both ...


3

The trouble with building columns is that they are fundamentally unstable i.e. if you perturb them away from the vertical the perturbation grows and the column falls over. This problem is worse the higher the column relative to it's base, so for very high columnar structures you need extra butressing to keep the column stable. By contrast perturb a suspended ...


3

A rigid body has 6 configuration degrees of freedom because its most general configuration can be obtained by translating (3 degrees of freedom) and rotating (3 degrees of freedom) its initial configuration. A mathy way of saying this is that its configuration manifold is $\mathbb R^3\times \mathrm{SO}(3)$. However, you are right that the phase space of a ...


3

Simply stated, in the area of contact dynamics, a unilateral (bilateral) constraint refers to a one-sided (two-sided) constraint, i.e. a constraint described$^1$ via an inequality (equality) of some constraint function $f(q,\dot{q},t)$, respectively. Here we are assuming that the variables $q^i$ are real (as opposed to complex). Technically, one demands ...


3

I'm guessing from your description that the left end of the rod is the end on contact with the ground, and the right end is the in the air at the point that the string is cut. I think that what will happen is this: When the string is cut, gravity will cause the rod to fall, such that its centre of gravity falls vertically. As the rod falls, it will rotate ...


2

The most important thing is conservation of momentum to describe the collisions. This part is actually quite straightforward, but before you get to collisions you should model the motion of single balls. Obviously, you will describe the balls classically and probably not at relativistic speeds (though that would be interesting...) so pretty much all you ...


2

I don't see anything wrong with what you've done. Conservation of momentum, and conservation of total angular momentum will hold exactly (as you assumed). But what you're doing is an "inelastic collision" so energy is not typically conserved. So the "unaccounted for" energy appears as heat.


2

You can look at tensor as a data file. Most important is not to visualize it but to understand what each element of this file means and HOW to TRANSFORM this file from one system (system of coordinates) to another. Visualization here can be done in 2 steps : 1) Visualizing data - vector fields, scalar fields, oriented volume fields and sooo on. 2) ...


2

I'm afraid your result is just right, although you are making things overly complicated by considering the full three dimensional rigid body dynamics... The center of mass will start accelerating in the direction of the force with $a=F/M$, plus the whole disk will start turning around a diameter, with an angular acceleration of $\alpha = F r/I$. You have ...


2

The problem is that you are double counting a lot of your constraints. If the (vector) displacements between particles A and B, and between B and C is fixed, then the displacement between A and C is fixed. Therefore the constraint on distance between A and C is redundant, and you can't count it separately.


2

You can spin a symmetric top around axes other than the symmetry axis: your text is considering the general case of an arbitrary rotation, and how the motion evolves in that case.



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