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18

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 ...


9

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 ...


8

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 ...


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

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 ...


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 ...


6

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. ...


5

Yes there will be a drag torque opposite the direction of spin. The name for this seems to be viscous torque. See e.g. this paper. Now 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 ...


5

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 ...


4

For the person not to slip, there must be a centripetal force of $mv^2/r = m r \omega^2$ towards the centre. Since $v$ varies with $r$ while $\omega$ is fixed ($v=r\omega$), it is probably easier to take the second form, in which case this force has to increase as $r$ increases. This forces comes from friction since there are no other forces in the plane ...


4

$a_c = \frac{v^2}{r}$ 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.


4

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 ...


4

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 ...


3

There are lots of different examples of oscillatory systems that have essentially the same mathematical form. Let's start by just looking at one type of differential equation: $a = \frac{d^2 x}{dt^2} = -\omega^2 x$ This equation has a general solution (you can check this) $x(t) = A \sin (\omega t + \phi)$ which oscillates with a period of ...


3

The g force is a unit of acceleration. 1 g is equal to 9.80665 m s-2. So the correct formula is $$ \text{G force} = \frac{\text{Acceleration in m s}^{-2}}{9.8}. $$ However, when describing uniform circular motion (i.e. $\boldsymbol\omega$ is constant) in free space, the only acceleration felt by the person rotating (in their frame of reference) is the ...


3

Assuming your rotating object (e.g. the Earth) is rotating at a steady speed the only way to change it's apparent speed of rotation is if you're rotating around it. You give the example of a geostationary satellite. This rotates around the Earth at the same angular velocity as the Earth rotates, so the Earth appears to be stationary (hence the name ...


3

The condition for staying in a circular orbit is the requirement for the centripetal force to be equal in magnitude to the gravitational pull. To be precise: $$F_g=F_c,$$ $$mg=\frac{mv^2}{r},$$ where $F_g$ is the absolute value of gravitational force, $F_c$ the absolute value of centripetal force, $g$ the gravitational acceleration, $m$ the mass of the ...


3

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 $\dot{\psi}$ ...


3

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 ...


2

Consider two rotating masses (1) & (2) with a torque $\tau$ applied on (1) only. If you define some sort of coupling between the two, with resulting angular velocities $\omega_2 = \gamma \omega_1 $ then since the power is conserved in the coupling then the two torques through are $T_2 = \frac{1}{\gamma} T_1$ such that the product $T_1 \omega_1 = T_2 ...


2

Suppose your robot walks vertically on two legs, and you want to mount a gyroscope in the center of the robot with a vertical spin axis. As a seat-of-the-pants engineer, I would ask how much rolling moment is needed to keep the robot from falling very far before it places a foot so it will stop falling. This depends on the robot's mass, how high it is, how ...


2

Always start with the units. They'll tell you a lot about the equations, and allow you to fix consistency errors. Incidentally, this is why I prefer Leibniz's notation over Newton's for derivatives, the units are immediately determined by examining the derivative, e.g. $dx/dt$ has units of distance over time assuming the usual definition of $x$ and $t$. ...


2

In general, the angular momentum for a rigid body is $\vec{L}=I\vec{\omega}$. For the special case of a point particle $\vec{r}$ from the axis of rotation, we have $I=mr^2$ and $\vec{\omega}=\frac{\hat{r}\times\vec{v}}{r}$, or $\omega=\frac{v}{r}\sin\theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{r}$. In this case, the angular momentum ...


2

The definition of rotational kinetic energy is $$ E_\text{rot}{}_{(i)} = \frac{1}{2} J_i \omega_i^2 = \frac{\;\; L_i^2}{2J_i} $$ where $J_i$ is moment of inertia, $\omega_i$ is angular velocity and $L_i = J_i\omega_i$ is angular momentum of the particle. If you select $\vec{r}_\text{cm}$ as the center of rotation these values can be calculated for each ...


2

It is simple. $\vec{\omega}_{[e]}$ are not the components of angular velocity seen in the reference frame attached to the rigid body itself. As you point out, that angular velocity is zero. It is the result of mathematical manipulation. You have a set of relations between the basis vectors of the inertial frame and the rotating frame, and you use that to ...


2

Always start with a nice clear diagram/sketch of the problem. It all follows from there. Here is a Free Body Diagram I made for you. Then you have (the long detailed way): Sum of the forces on body equals mass times acceleration at the center of gravity. $\sum_i \vec{F}_i = m \vec{a}_C $ $$ A_x = m a_x \\ A_y - m g = m a_y $$ Sum of torques about ...


2

The rate of change of the rotation matrix $\boldsymbol R$ is $$ \dot{\boldsymbol{R}} = \vec{\omega} \times \boldsymbol{R} $$ where $\vec\omega\times = \begin{pmatrix} 0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0 \end{pmatrix}$ is the cross product operator in 3x3 form. So if the position of a ...


2

I don't think it is difficult to derive analytically the shape of the Earth. Simply look for the shape of the surfaces of equal potential. The geometrical symmetry reduces the calculation to a 2-dimensional problem. Assume the rotation axis is vertical. The potential is the sum of the gravitational plus centrifugal: ...


2

Your angular velocity vector is $$ \vec{\omega} = \Omega \frac{ \vec{r}_D - \vec{r}_A }{|\vec{r}_D - \vec{r}_A|} $$ where $\vec{r}_A = (0,0.2,0.12)$, $\vec{r}_D = (0.3,0,0)$, $\vec{r}_B = (0.3,0.2,0.12) $ in meters and $\Omega = 90\;{\rm rad/s}$. Your velocity kinematics is $$ \vec{v}_B = \vec{\omega} \times ( \vec{r}_B - \vec{r}_A ) $$ And acceleration ...



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