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44

What luck! Just yesterday I was thinking about this exact same phenomenon whilst watching the film 'The Imitation Game'; the title sequence contained a moving tank, more on that later. When I was little, I used to observe this all the time; not in wheels however, but in caterpillar tracks: Notice how, when a segment of the track touches the ground, it ...


14

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 ...


14

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 ...


11

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 ...


10

Consider a point $P$ on the surface of the wheel. If you look at the horizontal velocity of that point in the frame of reference of the wheel (axis stationary), then for a wheel of radius $r$ with angular velocity $\omega$ that point will have horizontal component of velocity $$v_h = r\omega\cos(\omega t)$$ The linear velocity of the wheel $v = \omega r$. ...


9

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 ...


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~ ...


8

you wrote: "How can [the velocity of the contact point] be zero when it's in continuous motion?". However, you should keep in mind that motion is relative and therefore your question should be actually read as: "How can the velocity of the contact point be zero relative to the contact surface, when it's in continuous motion relative to its axis of ...


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

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

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 ...


6

The internal forces come in equal-and-opposite pairs (Newton's 3rd), and therefore result in no net force on the object. If you did take the vector sum of all of them, they would just cancel out. Including them therefore doesn't change the final expression. There is no need for idealization - just the acceptance of classical mechanics. Note that this 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

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

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

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

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.


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

The contact point of the disk with the plane has null instantaneous velocity This implies that there is no slippage, and as such there are no non-conservative forces doing work on the disk. Assuming the disk is perfectly rigid and is not being subjected to any linear or angular accelerations, the disk will continue to roll forever, and will not come to ...


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

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

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

If the wheel is rolling without slipping, what's the velocity of the point at the base of the wheel?? It is... zero! Convince yourself that the velocity must be zero. Since if it wasn't zero, the wheel wouldn't be rolling without slipping. So far the explanation is correct. "No slipping" refers really to some non-zero interval of time, and to the state of ...


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 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

If want to describe the dynamics of the ball, you need to use the SO(3) matrix which describes the ball's orientation. This is a 3 by 3 matrix whose transpose is its inverse. These may be parametrized by Euler angles, and most of the literature on rigid rotating bodies uses this convention, but I think it is best just to use the matrix entries themselves ...


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 angular and linear momentum of the two masses A and B are not necessarily conserved individually; it is the momenta of the system $S_{AB}$ that is conserved. If you know the conditions of the system at any particular time $t$, draw a free body diagram and work out the momentums for the system. Knowing that these values are conserved, you can use them ...



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