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45

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 ...


32

It's a classical mechanics effect for sure although a really interesting one. Following links on "Dzhanibekov effect" one gets at Marsden and Ratiu's "Introduction to Mechanics and Symmetry" Chapter 15 Section 15.9 "Rigid Body Stability" treating this with use of the Casimir functions. From remark 1: A rigid body tossed about its middle axis will undergo an ...


17

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 ...


15

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 ...


13

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

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 ...


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 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 ...


9

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~ ...


9

You are making a major flaw here. The friction between the 2 blocks is not going to be $10N$. It is going to be something, but we will have to calculate it. Assume the friction force to be $f$ such that the acceleration for both the blocks is same. Now, equations for $m_2$ and $m_1$ separately are: $$F-f = m_2*a = 3a$$ $$f = m_1*a = 5a$$ Solving both ...


8

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 ...


8

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 ...


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

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 ...


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 ...


5

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 $\dot{\psi}$ ...


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 ...


5

This is an interesting paradox! I suggest resolving it in the following way: First, note that the coefficient of friction between the two blocks allows friction "up to 30N" to occur - but if the two blocks are not moving relative to each other, the friction will be "whatever it is". Second, look at the force balance on the top block. We have the force of ...


4

If you have the same mass, then the fluted beam will be more rigid because the second moment of area is larger - in the same way that an $\mathrm{I}$ beam is more rigid than a circular rod of the same mass. In the case of a gun barrel, lighter weight helps in a number of ways - portability, ability to hold the gun still as you aim... but making it lighter ...


4

You can look at tensor as a data file. The most important thing 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 two steps: Visualizing data - vector fields, scalar fields, oriented volume fields and sooo on. ...


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.


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 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

I worked on a physics engine written in C# that does just this. Here are my notes on this topic. Objects have both translational and rotational momentum. When two objects collide, the overall algorithm goes like this: 1> Find the total momentum of both objects. Calculate the translational and rotational momentum, the vector sum of this is the total ...


3

I) A Lagrangian variational principle for Euler's equations for a rigid body $$ \tag{1} (DL)_i ~=~M_i, \qquad i\in\{1,2,3\}, $$ is e.g. explained in Ref. 1. Here the angular momentum $L_i$, $i\in\{1,2,3\}$, along the three principal axes of inertia is tied to the angular velocity $\omega_i$, $i\in\{1,2,3\}$, by the formula $$\tag{2} L_i~:=~I_i ...



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