# Tag Info

21

Let me first list all of the possibilities I considered that I later rejected. This is far from exhaustive, and I'm looking forward to seeing other people's creativity. Bad Ideas Sit on a tire swing with the fan pointing to the side. Point the fan up, measure speed of rotation of the system on the tire swing. Get a laser or collimated flashlight. Point ...

12

This is a note on why angular velocities are vectors, to complement Matt and David's excellent explanations of why rotations are not. When we say something has a certain angular velocity $\vec{\omega_1}$, we mean that each part of the thing has a position-dependent velocity $\vec{v_1}(\vec{r}) = \vec{\omega_1} \times \vec{r}$. We might consider another ...

11

... each day is 1 second longer every about 1.5 years That figure is way off. According to this Scientific American article, the Earth's rotation rate just after the collision that formed the Moon was about once every 6 hours. At that time, the Moon would have been about 25,000 kilometers away. The tidal effect of the Moon is the major reason the day ...

10

There are no "other" examples. The condition that $\vec \omega$ and $$\vec L = I_{\rm tensor} \cdot \vec \omega$$ point to the same direction i.e. $$(\vec L=) I_{\rm tensor} \cdot \vec \omega = k \vec \omega$$ where $k$ is a real number (and no longer a tensor) is a definition of an eigenvector of $I_{\rm tensor}$: both $\vec \omega$ and $\vec L$ are ...

9

The proper derivation of the centripetal acceleration—without assuming any kinematic variables are constant—requires a solid understanding of both the stationary Cartesian unit vectors $\hat{i}$ and $\hat{j}$ as well as the rotating polar unit vectors $\hat{e}_r$ and $\hat{e}_\theta$. The Cartesian unit vectors $\hat{i}$ and $\hat{j}$ are stationary and ...

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 ...

8

Here is one simple way. A point is moving around a circle. It has a blue position vector and a red velocity vector, like this: The position vector stays the same length and rotates around and around in a circle. Because the position vector is changing, it has a derivative. That derivative is the velocity. Because we're always going the same speed, the ...

8

In the basic discussion of angular momentum where something is rotating around a fixed symmetrical axis $\vec{L}=\vec{r}\times\vec{p}$ reduces to $\vec{L}=I*\vec{\omega}$ Like in this animation where each vector is colored appropriately: However angular velocity and angular momentum can have different directions in two cases: If the axis of ...

8

There are actually several different ways to interpret that question, depending on what you mean by "vector" and "rotation". But here's a sense that I've often wondered about myself: in introductory physics, the velocity vector is defined as the time derivative of the position vector (relative to some fixed point). Why is the same not true of angular ...

7

The direction of angular velocity is different from that of regular velocity for (arguably) two reasons. First, it points out of the plane because of the nature of angular velocity. It signifies a rotation, as such, there is not any particular direction unit vector in every coordinate space that could represent it. In spherical or cylindrical coordinates, it ...

7

two concentric and counterrotating flywheels preclude all precession forces regardless of which plane the axis is rotated in. this is assuming the connection between the two flywheels is sufficiently strong--it make break from tension/compression due to each flywheel experiencing its own forces. refer to the diagram i just drew up. the black rectangles ...

7

You made a mistake in assuming that the angular acceleration ($\alpha$) is equal to $v^2/r$ which actually is the centripetal acceleration. In simple words, angular acceleration is the rate of change of angular velocity, which further is the rate of change of the angle $\theta$. This is very similar to how the linear acceleration is defined. ...

7

There is indeed a term involving the time derivative of the changing coupling between the masses. First, let's derive the equation for a single mass. $$L = \frac{1}{2} I\, \dot{\theta}^2 - V(\theta)$$ $$\frac{\partial L}{\partial \dot{\theta}} = I\, \dot{\theta}$$ $$\frac{\partial L}{\partial \theta} = -\frac{dV}{d\theta} = \tau$$ $$\tau = \frac{d}{dt} ... 7 For a single particle, yes they're parallel. For a system of particles,$$\sum_i \frac{{\bf r}_i\times \dot{\bf r}_i}{\|{\bf r }_i\|^2}\neq \alpha\sum_i{m_i{\bf r}_i\times \dot{\bf r}_i}$$(you can come up with a specific counterexample but it should be obvious the two sides don't have to be proportional/collinear -- each vector in the sum is weighted ... 7 I will attempt to answer this question with some basic dynamics and some contact mechanics. There are two special cases here. a) There is sufficient friction to keep the base of the pin A fixed (imparting a reaction impulse J_A when hit by the ball, or b) The floor is smooth and the pin will translate and rotate at the same time with J_A=0. There is ... 6 I'm not an engineer, but this is how id do it. According to your rules, we can use a computer, and Audacity. You get a pair of headphones. You plug the headphones into the microphone jack on your computer. You open Audacity. You get a very small magnet. You glue the very small magnet to one of the fan blades. You turn the fan on. You hold one of the ... 6 a_c = \frac{v^2}{r} isn't angular acceleration. It's the magnitude of the linear acceleration towards the centre of an object following a circular path at constant angular velocity. Angular acceleration is the derivative of angular velocity, and the analogue of Newton's second law is that angular acceleration equals torque divided by moment of inertia. 6 Defining properties of vectors are that you can add them and multiply them by constants. These both make sense for angular velocities. On the other hand, adding rotations doesn't make sense. What you can do with two rotations is compose them: first rotate one way, then rotate another. This operation doesn't look like addition of any sort. For one thing, it ... 5 Yes there will be a drag torque opposite the direction of spin. The name for this seems to be viscous torque. According to this paper, the viscous torque on a spinning sphere of radius R in a fluid with viscosity \eta spinning with constant angular velocity \vec\Omega is$$ \vec\tau = -8\pi R^3\eta\vec\Omega $$The paper goes on to describe how to ... 5 I assume you know about rotation matrices, and so for a sequence rotations about Z-X-Z with angles \phi, \theta and \psi repsectively you have$$ \vec{\omega} = \dot{\phi} \hat{z} + T_1 \left( \dot{\theta} \hat{x} + T_2 \left( \dot{\psi} \hat{z} \right) \right) $$The logic here is apply a local spin of \dot{\phi}, \dot{\theta} and \dot{\psi} ... 5 Yes, B does rotate when seen from a static frame of coordinates outside the disk: As to velocities and accelerations, see the article in Wikipedia. It says,$$\vec {v_s} = \vec {v_r} + \vec {\Omega} \times \vec r,$$where v_s is the velocity in the static frame and v_r in the rotating. If you apply this formula for both points A and B, their ... 5 If I understand your question correctly you are saying that:$$ v = r\omega $$and therefore:$$\begin{align} v_A &= r\omega \\ v_B &= \tfrac{1}{2}r\omega \\ v_A &= 2v_B \end{align} $$but how can A and B have different velocities when they are both attached to the disk so the separation between is fixed? The answer is that A and ... 5 High angular momentum presents a barrier preventing collapse to a black hole (at least until this angular momentum is radiated away). The parameter on which the formation of black hole depends is the ratio q of angular momentum (J) to the square of mass (M). If q=J/M^2 < 1 (in relativistic units with G=1, c=1), then the black hole ... 5 A day is currently about 86400.002 seconds long. If we could just increase the Earth's rotation rate by a mere 2 milliseconds per day we would get rid of the need for those pesky leap seconds. No problem! We only need something that rotates with an angular momentum of 1.4×1026 joule-seconds about an axis pointing due south. One way to do this would be to ... 4 If you solve the problem for the two forces the vertical and the horizontal force (which is required for the circular motion) you obtain the relation$$\omega^2=\frac{g}{L\cos\theta}$$Hence the minimum required \omega is \sqrt{g/l}, the same as the angular frequency for motion of a bob in a plane. 4 I have reproduced your calculation. If \theta is the angle from the central vertical axis of the hemisphere to the ball, the tangential downward force is g \sin \theta = g\frac rR The tangential upward force due to rotation is \omega ^2 r \cos \theta=\omega ^2 r \sqrt {1-\frac {r^2}{R^2}} When \omega \lt \sqrt{\frac gR} the downward force is ... 4 The moment of inertia tensor is not constant in the external reference frame (http://en.wikipedia.org/wiki/Precession#Torque-free ) 4 Actually we only gain 1.3 milliseconds every 96-100 years, not 1 second every 1.5 years! :) the shortest known Earth day was 6 hours and the longest is 24 hours & 2.5 milliseconds (today's current day), in 1820 the day was exactly 24 hours, but since it's been nearly 200 years we've gained 2.5 milliseconds to our day. So the days get longer just very ... 4 Use the spatial inertia to relate linear/angular momentum to changes in linear/angular speed$$\begin{pmatrix} \vec{L}\\ \vec{H}_A \end{pmatrix} = \begin{bmatrix} m {\bf 1}_{3×3} & -m [\vec{c}\times] \\ m [\vec{c}\times] & I_{cm}-m [\vec{c}\times][\vec{c}\times] \end{bmatrix} \begin{pmatrix} \Delta\vec{v}_A\\ \Delta \vec{\omega} \end{pmatrix} ...

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