Hot answers tagged angular-velocity
16
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 ...
7
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
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 ...
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
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 ...
5
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 ...
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
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
$$
The paper ...
3
There are lots of different examples of oscillatory systems that have essentially the same mathematical form. Let's start by just looking at one type of differential equation:
$a = \frac{d^2 x}{dt^2} = -\omega^2 x$
This equation has a general solution (you can check this)
$x(t) = A \sin (\omega t + \phi)$
which oscillates with a period of ...
3
The g force is a unit of acceleration. 1 g is equal to 9.80665 m s-2. So the correct formula is
$$ \text{G force} = \frac{\text{Acceleration in m s}^{-2}}{9.8}. $$
However, when describing uniform circular motion (i.e. $\boldsymbol\omega$ is constant) in free space, the only acceleration felt by the person rotating (in their frame of reference) is the ...
3
If want to describe the dynamics of the ball, you need to use the SO(3) matrix which describes the ball's orientation. This is a 3 by 3 matrix whose transpose is its inverse. These may be parametrized by Euler angles, and most of the literature on rigid rotating bodies uses this convention, but I think it is best just to use the matrix entries themselves ...
3
Assuming your rotating object (e.g. the Earth) is rotating at a steady speed the only way to change it's apparent speed of rotation is if you're rotating around it.
You give the example of a geostationary satellite. This rotates around the Earth at the same angular velocity as the Earth rotates, so the Earth appears to be stationary (hence the name ...
2
Consider two rotating masses (1) & (2) with a torque $\tau$ applied on (1) only.
If you define some sort of coupling between the two, with resulting angular velocities $\omega_2 = \gamma \omega_1 $ then since the power is conserved in the coupling then the two torques through are $T_2 = \frac{1}{\gamma} T_1$ such that the product $T_1 \omega_1 = T_2 ...
2
The definition of rotational kinetic energy is
$$
E_\text{rot}{}_{(i)} = \frac{1}{2} J_i \omega_i^2 = \frac{\;\; L_i^2}{2J_i}
$$
where $J_i$ is moment of inertia, $\omega_i$ is angular velocity and $L_i = J_i\omega_i$ is angular momentum of the particle.
If you select $\vec{r}_\text{cm}$ as the center of rotation these values can be calculated for each ...
2
Suppose your robot walks vertically on two legs, and you want to mount a gyroscope in the center of the robot with a vertical spin axis.
As a seat-of-the-pants engineer, I would ask how much rolling moment is needed to keep the robot from falling very far before it places a foot so it will stop falling.
This depends on the robot's mass, how high it is, how ...
2
Always start with the units. They'll tell you a lot about the equations, and allow you to fix consistency errors. Incidentally, this is why I prefer Leibniz's notation over Newton's for derivatives, the units are immediately determined by examining the derivative, e.g. $dx/dt$ has units of distance over time assuming the usual definition of $x$ and $t$.
...
2
It is simple. $\vec{\omega}_{[e]}$ are not the components of angular velocity seen in the reference frame attached to the rigid body itself. As you point out, that angular velocity is zero.
It is the result of mathematical manipulation. You have a set of relations between the basis vectors of the inertial frame and the rotating frame, and you use that to ...
2
I don't think it is difficult to derive analytically the shape of the Earth. Simply look for the shape of the surfaces of equal potential.
The geometrical symmetry reduces the calculation to a 2-dimensional problem. Assume the rotation axis is vertical. The potential is the sum of the gravitational plus centrifugal:
...
2
Your angular velocity vector is
$$ \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 ) $$
And acceleration ...
1
What we do when we have a rigid body in motion is to attach some coordinates to the body, so an origin $O'$ and three base vectors. Now, the position of a point relative to the origin of a fixed reference frame of origin $O$ is
$$ \vec R_i= \vec{OO'}+ \vec r_i $$
where $\vec r_i$ is the position vector respect to the origin of the body reference frame ...
1
The rate of change of the rotation matrix $\boldsymbol R$ is
$$ \dot{\boldsymbol{R}} = \vec{\omega} \times \boldsymbol{R} $$
where $\vec\omega\times = \begin{pmatrix}
0 & -\omega_z & \omega_y \\
\omega_z & 0 & -\omega_x \\
-\omega_y & \omega_x & 0 \end{pmatrix}$ is the cross product operator in 3x3 form.
So if the position of a ...
1
From Goldstein, chapter 4 eqn 4-92', for a finite rotation the change $\boldsymbol{\Delta r}$ caused by rotating a vector $\boldsymbol{r}$ through an angle $\Phi$ about a direction defined by a unit vector $\boldsymbol{n}$ ($\Phi$ positive for a counter-clockwise rotation), to a final position $\boldsymbol{r'}$ is given by:
$$ \boldsymbol{\Delta r} = ...
1
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 ...
1
Firstly, we are just talking about the magnitude of $v_{tip}$. This will not change with rotation of the velocity. In the question we do not care about the direction of the velocity. It just says it may never move faster than $270 m s^{-1}$. That is why the solution only uses $v_{tip}^2$.
Now to find $v_{tip}^2$ they decide to use a configuration (position) ...
1
I don't think your "drag on each point of the door is the same as on a free particle in the wind" is going to be a good simulation of what goes on when your door slams. Actually, with your model, the door would never start moving.
If you wanted a back of the envelope estimate, I would base it on the Venturi effect: the air is stalled on one side of the ...
1
There really isn't, any mechanical work involved. It is similar to what happens if you push really hard against a wall: you get tired, but without doing any actual work.
See, when you apply torque to your gyroscope, you do it applying a couple, i.e. two displaced, equal but opposite forces. You may e.g. push the top of the rotation axis to the right, and ...
1
I think what you have noticed is that oscillatory motion is a common behaviour amongst physical systems. The notation used is often customised to that appropriate to the case. Physicists like working in radians as they have a certain independence of the size of the system, and from there we get a factor of $2\pi$.
1
With regard to the 'Basic question', the $2\pi$ value is used as it is the radian representation of a complete cycle; 360deg in radians.
This is, of course, derived from the basic velocity equation $v=\frac{s}{t}$, which, when adapted for circular motion becomes $\omega=\frac{\phi}{T}$, where $\phi$ equals the distance travelled, which in the case of a ...
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