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Draw a picture of the geometry, including the cloud, the earth, and the galactic center. The earth and the cloud are rotating around the center at the same speed (that is your flat curve). You have two sides and one angle of the triangle. The radial velocity gives you the projection of the difference of the velocity vectors onto the line of sight.

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"Equation that is all over the internet"... You started at http://thatsmaths.com/2014/06/26/balancing-a-pencil/ and from there, you linked to http://arxiv.org/pdf/1406.1125v1.pdf which was the source for the former. In the third paragraph of that paper, it states We model the pencil as an inverted simple pendulum with a bob of mass m at one end of ...

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What I cannot understand is, why acceleration, a=lθ¨ and not lθ¨/2? The equation you wrote doesn't mention anything about the linear acceleration. Is the center of mass located at its top and not the center? Or is there something else I am missing? The center of mass of the pencil is in the middle, not the top. There is likely something else ...

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Moving bodies have inertia which means that they will continue to move at a constant velocity unless acted upon by an external force (this is Newton's first law of motion). Similarly, rotating bodies have a moment of inertia, meaning that they will continue to rotate unless acted upon by an external force (torque). Therefore, torque is only required to ...

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It seems helpful to consider an extremely simple scenario. Suppose an astronaut is floating near two balls of lead; in this case the closed system consists of the astronaut together with the balls. She can pull the balls together without changing the momentum or angular momentum of the system. She can then rotate them in the center with almost no change, and ...

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John correctly stated that this is possible because re-configuring our bodies allows us to change our moment of inertia, but not our mass. As the question was about an intuitive explanation, consider adding a series of floating weights to get an analogous situation for translational motion: The astronaut stretches their arms above the head, grabs a weight, ...

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Here is a simpler answer: if something can change shape, then it doesn't really have an orientation.

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It is because the moment of inertia is not a conserved quantity. The statement that an isolated body can't change its position is more precisely the statement that an isolated body cannot change the position of its centre of mass. The position of the centre of mass, ${\bf R}$, is given by: $${\bf R} = \frac{1}{M}\sum m_i {\bf r}_i$$ where $M$ is the ...

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Let us denote $\ P$ the point corresponding to the end of the rod with no force applied, then we apply a force $\ \overrightarrow{F}$ on the other end of the rod, such that is parallel to the plane and perpendicular to the rod. Now, with respect to the center of mass, the only force whose torque is different from zero is $\ \overrightarrow{F}$: ...

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I'm afraid your separation of these two types of friction is not entirely useful in this situation. Rolling and sliding are, when you look very closely and over small time scales, just about the same. (The friction applied to a wheel, though, causes some rotation, which can lead to the wheel traveling over the surface). If I had to use your model of sliding ...

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Even if I ignore wind and the drag forces and only consider the rotation of the earth the bullet will not hit the ground at the same place from where it was projected. There will be Coriolis effect. Coriolis effect: The Coriolis effect is a deflection of moving objects when the motion is described relative to a rotating reference frame. I suggest this ...

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This builds partially off of TZDZ's answer. In order to fully understand the moment of inertia, you need to fully understand the idea of torque. The only sensible way to fully understand torque that I've seen is to first understand it as the derivative of rotational work $W_R$ being done with respect to the angle $\theta$ through which the work is done. So: ...

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Since there are no extra rules for black holes they should follow the same laws. In general relativity this effects are called the de Sitter- and the Lense Thirring-effect which have been verified with big trumpets call by the famous Gravity Probe B.

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Assuming earth mass, procession would be reduced practically to zero because tidal effects would be practically zero. Procession, at least according to this site, has to do with tidal effects and a planet being mailable. A black hole - 2/3rds of an inch in diameter would experience essentially zero tidal effects and it would be far less prone to bulging ...

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The dynamics of a ball rolling down an incline is interesting. Let's start by figuring out the forces that come into play for the non-slipping case (mass m, radius R, angle of ramp $\theta$): If we consider the motion of the ball as a rotation about point $P$, then the torque is given by $$\Gamma = mgR\sin\theta$$ and the moment of inertia about $P$ is ...

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To understand rolling without slipping, first consider the case of rolling only case about the center of mass.In this case, a point on the top rim will have a speed $v=\omega\cdot R$ and a velocity $v=-\omega\cdot r$, at bottom of the rim, as observed by you. However in case of rolling without slipping, We observe the velocity of center of mass is $v$ (only ...

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Here is the approach: Rolling and slipping simultaneously is not much different from rolling without slipping. All the equations are the same except you can't use $a=R\alpha$. Here is the method for solving the problem: Friction: $f=\mu mg\cos(\theta)$ Torque to find angular acceleration: $\tau=I\alpha=fR$ Linear acceleration: $ma=mg\sin(\theta)-f$ Final ...

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Physics is the accumulation of observations, the devising and fitting mathematical models to these observations and study of the predictions of the mathematical models to validated them. In the progress of time, the mathematical models become theories, i.e. derived from a few postulates, also called laws, that define the framework for physical problems , ...

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Just like Anna V said, Conservation of angular momentum. It's sort of like how on Motorcycle races the racer's can do extremely sharp turns with their knee's almost touching the the ground. Once it's gained enough speed it begins to just well, support itself. (I understand that is a very bad word to use for this question but it's hard to explain)

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I studied this problem 20 years ago in a classical mechanics course, and my recollection is that when you solve the equations of motion, you get an imaginary exponential component that describes the motion for each axis. However, in the case of the intermediate axis, you get a multiplication of two imaginary numbers in the exponential, which then gives ...

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For a rotating ball you should use for energy $$E=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2+mgh$$ where $I$ is the moment of inertia which for a sphere is $$I=\frac{2}{5}mr^2$$ where $r$ is the radius of the sphere, and $\omega$ it's his angular velocity which is related to the velocity of the center of mass via the equation $$\vec{r}\times \vec{\omega}=\vec{v}$$ ...

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Of course you can specialise the derivation of the three dimensional Euler equations for your special case. However, unless you are expressly asked to do this, I doubt that your teacher means for you to take this path because it is very much an overkill. The yoyo is spinning in a plane about its axis of symmetry. The inertia tensor is now a simple scalar ...

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Integrating your equations (1) and (2) you get that $(\text i) \ \omega_1 = C_1$ a constant in time, while another constant in time. This equality you can re-write as $(\text {ii}) \ \omega_2^2 + \omega_3^2 = C_2^2$, As to the vector $J_{\Omega}\omega$ you know that its components are $\{J_1\omega_1, J_2\omega_2, J_2\omega_3\}$, therefore $(\text ... 0 This is what tidal power generators do - if earth did not rotate relative to the moon, there would be no lunar tidal motion (there would be a distortion of the earth's equipotential surface but it would not move). Since earth does rotate, you get motion of the oceans. At Fundy Bay, for example, significant power is extracted from this motion. See ... 0 Gravity pulls in the center of gravity. It will be in the middle of both a cube and a ball in question. There will be a normal force on the cube on the ends. There will be a normal force on the ball at the contact point. If the weight pulls in the center further to the left (for a left directed incline) than where the normal force pushes, then the object ... 0 Look at the forces -- and more specifically the torque. The sphere has the force of gravity pulling it downward, and the inclined plane pushing it out. The inclined plane exerts a force on the sphere at its contact point perpendicular to the plane. The force of gravity has some component along the plane, and some component perpendicular to the plane. The ... 0 You'd have to check to see if the sum of the moments around the toe causes a rolling rotation about the toe. In the case of a cylinder, the toe is positioned in such a way that the moment about the toe always causes rolling. In general, draw up a free body diagram that shows all the forces and moments, focusing on the moments about the toe. Include gravity, ... 1 I will answer the mass moment of inertia tension question. For an infinitesimal clump of mass${\rm d}m$located at$\vec{r}$its effect on the inertia tensor is $${\rm d}{\bf J} = -[\vec{r}\times][\vec{r}\times]{\rm d}m$$ where$[\vec{r}\times]$is the skew symmetric cross product operator $$\begin{pmatrix}x\\y\\z \end{pmatrix}\times = ... 5 John Rennie's answer is correct in the special case that angular momentum and angular velocity are parallel to one another. This is not always the case as moment of inertia is a second order tensor. Angular momentum is given by \boldsymbol L = \boldsymbol{\mathsf{I}} \boldsymbol \omega. Differentiating this gives the rotational analog of Newton's second ... 2 The equivalent to force is torque, \tau. So the second law would be:$$ \tau = I\dot{\omega} = I\ddot{\theta} $$1 Mathematicians usually are not worried about the conecpt of physical units. As such, a mathematician probably would argue that \mathbf M_O, \vec{OP} and \mathbf F belong to \mathbb{R}^3, as MyUserIsThis did in his comment. If this is not satisfactory to you, you could consider three distinct fields of numbers, say \mathbb{R}_F for forces, ... 1 Your intuition is correct: you can't find the centripetal force from an analysis of only forces. The centripetal force is found from an analysis of motion. The problem is that one of the forces varies in order to meet the requirements of the motion: it is a force of constraint. Consider a test particle on the rim of the disk. There are two forces on ... 0 Electron spin is not the result of a rotation of the electron around itself. In this case, of course (2) also doesn't hold. In fact, one can show that there is a double implication as follows: 1) if \vec v is defined as in (1) one gets$$ \frac {\vec r \times \vec v}{r^2} = \vec {\omega} - \vec r \frac {(\vec r \cdot \vec {\omega})}{r^2}. \tag{I}$\$ So, ...

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