# Tag Info

## New answers tagged rotation

0

Note that "c" means linear velocity, not angular. Then, you would refer to tangential velocity. If an object rotated at relativistic (tangential) velocity, then each "shell" will have a different space and time compression. By contraction of the tangent lengthes, the perceived length of large circles would be smaller than $2\pi r$, leaning to 0 lenght at ...

1

Consider the reaction impulse $P$ from the rough ground (infinite friction) ruling the impulse $F$ as shown. The equation of motion are: Sum of impulses equal change in momentum (motion of the center of mass) $$P- F = m \Delta v$$ Sum of moments about center of mass equal change in angular momentum $$F (r \sin \theta) + P r = I \Delta \omega$$ Under ...

1

there is no question in your question, an increase in the earths spin would cause the distance between the pole and the equator to increase - if only by a fraction - as the size of the 'bulge' at the equator (the earth is an oblate spheroid) is proportional to the rate of spin Personally, I don't see why the Ancient Egyptians are idolised so much in ...

-1

By convention, North and South rank higher in the English language. Thus we have Northeast and not Eastnorth. You have presumed a flat earth. If you start 1.5 km from the North Pole, then after skiing 3 km east and 1.5 km north, you will be at the North Pole, 1.5 km North from your starting point.

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On a vortex mixer the liquid is swirling inside a test tube. In a centrifuge, the test tubes (containing the liquid) are held in a fixed position inside a rotor, and the rotor is spinning inside the centrifuge.

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It is straightforward matrix multiplication and the operators enter through from either side. $exp(iS_zϕ/ℏ)$ would act from the left on the kets while $exp(−iS_zϕ/ℏ)$ would act from the right on the bras. You can see it more clearly if you do it explicitly. Use the basis $|+⟩$ and $|−⟩$ to represent the operators as matrices $|+⟩$ = ...

2

A vortex is a local disturbance caused by turbulent flow. It carries fluid and particles inward toward a central axis of rotation. The central axis is not fixed. It moves within a larger body of fluid. The speed of the vortex is greater near the axis of rotation and decreases toward the edge of the vortex. Unlike a vortex, a centrifuge creates ...

1

Can an angle be defined as a vector? It depends on what you mean by "vector". If by "vector" you only mean something that has a magnitude and a direction, then yes, the axis-angle representation qualifies as a "vector". To a mathematician, a vector is something that is a member of a vector space. In this context, the axis-angle representation fails to ...

1

Describing a rotation as a vector, with the direction of the vector along the axis of rotation, and the magnitude of the vector as the angle, is known as the axis–angle representation.

1

Let $\alpha$ be the angle between the rod and the floor at point $B$ (see my drawing). Then we can write the points $A$ and $B$ with the length $L$ of the rod as $$\vec A = L \sin(\alpha) \begin{pmatrix} 0 \\ 1 \end{pmatrix} \,,\qquad \vec B = L \cos(\alpha) \begin{pmatrix} 1 \\ 0 \end{pmatrix} \,.$$ The center of gravity $M$ of the rod is just the ...

0

One way to see that $\vec{A}(t)$ rotates around $\vec{\alpha}$ is that its tangent (or derivative) is always perpendicular to $\vec{\alpha}$ and $\vec{A}$. In order to give more definitive proof it might be easier to write the differential equation is matrix form  \begin{bmatrix} \dot{A}_x \\ \dot{A}_y \\ \dot{A}_z \end{bmatrix} = \begin{bmatrix} 0 & ...

0

The equation is quite relevant in physics. Think of the precession of a magnetic dipole in a magnetic field, NMR etc. \begin{eqnarray*} \frac{d\mathbf{A}}{dt} &=&\mathbf{\alpha }\times \mathbf{A} \\ \frac{d\mathbf{\alpha \cdot A}}{dt} &=&\mathbf{\alpha \cdot \alpha }\times \mathbf{A}=0\Rightarrow \mathbf{\alpha \cdot A}\;\mathrm{const} \\ ...

2

Your claim is false (in $\mathbb R^3$). If $\vec{a}$ and $\vec{b}$ are fixed vectors (I assume they are not co-linear and satisfy $|\vec{a}|=|\vec{b}| \neq 0$) there are infinitely many rotations $R \in O(3)$ such that $R\vec{a}=\vec{b}$. One is the rotation $R$ of the angle between $\vec{a}$ and $\vec{b}$ performed around an axis orthogonal to the plane ...

0

Yes, but it will be of order $\frac {m_{elevator}}{m_{earth}}$ (quite small compared to your skater's arms example) and also depend on the latitude of where it is tethered.

1

Here is another way to look at this problem: we should consider the total torque relative to the two points where wheels touch the ground. Clearly in your picture you rotate the whole bicycle backwards regarding the back as well as front wheel (points where wheels touch the ground). Imagine that the wheels are nailed. If you would pull the rope forward the ...

1

In the reference frame fixed to the bike, a backward movement of the pedal by $dx$ moves the ground backward by $\alpha dx$. So the pedal moved backward $(1-\alpha)dx$ with respect to the ground. If $\alpha>1$ then this means it either moved forward, which would deliver energy to the string puller and thus makes no sense didn't move at all ($dx=0$) ...

1

I think the way to look at the problem is this: The tricky thing about this problem is that the force $F$ is doing two things. First, it is pulling the entire bicycle backwards and, secondly, it is exerting a torque on the pedals. So let's try to separate those two effects. Consider, then, a person just sitting normally on the bicycle with his feet on the ...

-1

It will definitely move backwards. The angular velocity of the wheel, multiplied by the radius of the wheel is much bigger then the angular velocity of the pedals, multiplied by their length (the ratio of angular velocities is constrained) so the backward motion of the bicycle is much more significant then the rotation of the pedals. When the horizontal ...

0

The energy of the system is not only proportionate to the force, applied to it, but is actually the work done by that force ($F\Delta s$) on the path that your system has travelled ($\Delta s$). In the first case the molecule has only the translational motion. In the second case, in addition to translational motion there is also a rotation, thus the path is ...

1

The mistake in your reasoning is to assume that the same force does the same work. That is simply not true - as the force operates over a different distance. To analyze the problem, you have to think in terms of impulse ($F\cdot \Delta t$) or work done ($F\cdot \Delta x$). Let's assume that the same impulse is applied. Then indeed the linear momentum of ...

0

The work done by the force in the first case is just translational, i.e. integral of the force over some distance and this gives rise to the translational kinetic energy. In the second case, there are two types of work done on the system, first being the usual translational and second is rotational. This is due to the that the object in question now has an ...

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