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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 ...


15

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 ...


12

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 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 ...


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 ...


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~ ...


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

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 ...


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

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.


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 $\dot{\psi}$ ...


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 ...


3

The argument is that the air was flowing through the hole at around 700 mph, so the air inside the aircraft had a substantial velocity in the direction of the hole. The air velocity inside the plane would have been less than 700 mph because the flow was converging on the hole, but the speed of the air would still have been hundreds of mph. When the hole was ...


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 ...


3

What is done here and maybe not stated so clearly is that the rate of change of angular momentum vector is broken up in two parts. Change in angular momentum due to angular acceleration $I\dot{\vec{\omega}}$ Change in angular momentum due to inertia tensor rotation $\vec{\omega}\times I \vec{\omega}$ All this is explained in terms of a changing vector ...


3

$\vec\omega = I^{-1} \vec L$, and $\vec L$ is constant in the absence of external forces. The bit that I think you're missing is that $I$ rotates with the rigid body, so it is not constant in general and neither is $\vec\omega$. I played with your online example, and the angular velocity does seem to always remain constant when I'm not poking the block, ...


3

This can be answered in two different frames of reference. If we look at this problem from the perspective of an outside observer, then as the wheel moves forward, the bottom of the wheel doesn't move at all and the top of the wheel moves at twice the speed of the wheel itself. In this frame, as the wheel moves forward, the centripetal force provided by the ...


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 ...



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