50

There's a very interesting way to find the angular velocity of a wheel that's spinning so fast that you can't measure using a stopclock. We'll be using a strobe light (a light that flashes on and off repeatedly) and a very interesting concept known as the Wagon Wheel effect under stroboscopic conditions. A little bit about the concept: The Wagon Wheel ...


28

From Wikipedia, the sidereal rotation period is 3.91 hrs, so the angular frequency of its rotation is $2\pi/(3.91 \, \mathrm{hrs}) = 4.48 \times 10^{-4} \, \mathrm{rad/s}$, or $7.12 \times 10^{-5} \, \mathrm{Hz}$. Its "semimajor axis" is about 1000 km, so this corresponds to a surface speed of $2\pi (1000 \, \mathrm{km})/(3.9 \, \mathrm{hrs}) \approx 1600 \,...


28

Note that this answer, like the tags of the question, is only focusing on Newtonian mechanics. The definition of linear momentum $\mathbf p$ is expressed in terms of the linear velocity $\mathbf v$ using $$\mathbf p=m\mathbf v$$ Since $m$ is just a scalar quantity, the vectors $\mathbf p$ and $\mathbf v$ are obviously parallel. i.e. $$p_x=mv_x$$ $$p_y=mv_y$...


27

Consider a thin rectangular block with width $w$, height $h$ resting along the xy plane as shown below. The mass of the block is $m$. The mass moment of inertia (tensor) of the block about point A is $$ {\bf I}_A = m \begin{vmatrix} \frac{h^2}{3} & -\frac{w h}{4} & 0 \\ -\frac{w h}{4} & \frac{w^2}{3} & 0 \\ 0 & 0 & \frac{w^2+h^2}{3}...


26

A laser tachometer: https://www.amazon.com/Neiko-20713A-Digital-Tachometer-Non-contact/dp/B000I5LDVC Make a mark at one point on the motor, then set it spinning. Point the tachometer at it. By shining a laser on the surface and measuring the changes in returned light as the mark passes, it can determine the RPM.


25

Model the tree as a point mass $m$ located some height $h$ above the ground --- that is, forget the mass of the trunk and assume all the mass of the tree is in the branches and leaves above the ground. Then the moments of inertia of the tree before and after felling are \begin{align} I_\text{tree,up} &= m \left( (R+h)\cos\theta \right)^2 \\ I_\text{...


24

You have to remember that the entire wheel is also moving. Think of this. Where the wheel meets the ground, the velocity of the contact point must be 0, otherwise the wheel would be skidding. Another way of looking at it is that at the contact point the forward velocity of the wheel is cancelled by the backward velocity of the point. On the other hand, at ...


24

One idea for an approach may be to record the sound the motor produces and then Fourier transform that signal. The assumption is that the frequency you look for will be prominently visible in the spectrum of that signal. Of course, it is not clear whether this frequency is as easily identifiable as it sounds.


22

In the no slipping condition the translational speed $v$ of the centre of mass of the wheel and the angular speed of rotation $\omega$ of the wheel are related. $v= r \omega$ where $r$ is the radius of the wheel. So one can find the vector sum (blue) of the translational velocity of the wheel at any point (red) and the tangential velocity of the wheel at any ...


21

Print out a disk that looks like this: Attach it to the spinning object and then observe it with a 60 Hz strobe light. By determining which rings look like they are stopped by a 60 Hz strobe you can deduce the RPM of the object. As you can see, the inner ring has 8 segments (4 white, 4 black), the next ring has 10 segments etc. If you rotate the disk by ...


21

They have the same units of $\mathrm s^{-2}$ only if you don't use $\mathrm{rad}$ unit for bookkeeping (which you can indeed avoid because radians are technically dimensionless, similarly to turns and other auxiliary units). But if you do try to distinguish angles from dimensionless numbers, then they are not the same: the unit of angular acceleration is $\...


20

Angular momentum is most naturally a skew two-index tensor $L^{ab} = x^ap^b-x^bp^a$. Indeed, in higher dimensions, the two-index language is essential. In 3-d the $SO(3)$ invariance of the Levi-Civita symbol $\epsilon^{abc}$ gives us the option of converting $L^{ab}$ to index object to a one-index (cartesian vector) object $L^a= \epsilon^{abc} L^{bc}$ -...


19

The other answers that say the difference arises because there is an inertia tensor in the rotation case are all perfectly correct, but i think we can go deeper and more intuitive that this and say: Translations commute. Rotations don’t. That’s the fundamental reason for the difference. And it’s that way that these two different behaviors bear upon Noether’...


15

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 ...


15

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 ...


13

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 ...


13

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 rotation ...


12

There's two possible views here. One of these is that you can, indeed, consider angular displacement or position as a vector in that it can be encoded with one: if you have $$\mathbf{\Theta} := \left<\theta_x, \theta_y, \theta_z\right>$$ then you can declare that this encodes an angular displacement with axis $\mathbf{\hat{\Theta}}$ and with ...


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 has ...


11

Both twist and wrenches are screws. "Screw" is the general term, and "Twist" is the specific application to motion whereas "Wrench" is the specific application to forces and momentum. All of them combine the linear and angular aspects of the thing they describe in one 6×1 object. The definitions pertain to rigid body mechanics in general and are not specific ...


11

The intended meaning of the hertz unit is that one hertz represents one complete occurrence of a cyclic phenomenon in one second. Angular frequency is not the number of complete rotations occurring per unit time, but instead the angle covered over that unit of time. Technically, angular frequency is not measured in $$\frac{\mathrm{1}}{\mathrm{s}}$$ but ...


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 ...


10

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 ...


10

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 ...


10

Let's take this in two parts: first an exhibition of how this works for a trained physicist who has access to the tools of multivariate calculus, and second an examination of how you might explain this to students in an introductory class based on algebra and trigonometry (no calculus). Sophisticated view Just as the proper formulation of the Newtonian ...


9

I happen to have done this exact experiment over the weekend - detailed write-up below. Basically it involved a laser pointer, a photo diode, two resistors, a transistor, and an Arduino. Set the Arduino timer to run at 10 kHz, and the output of the photodiode (into the interrupt input of the Arduino) triggers a "read" of the timer. When at least a second has ...


9

I am assuming that by $I$ you meant the moment of inertia relative to a given axis. Then you are right, the relation $\vec L=I\vec ω$ does not hold in general. So for example a particle in circular motion about the $z$-axis at the plane $z=z_0$ has angular momentum relative to the origem which does not point in the $z$ direction. Moreover, it precesses about ...


9

Not quite an answer to the question asked, but a clarification of one of the ideas you used to build the question. As I have noted elsewhere right- (or left-)hand rules tend to appear in pairs when computing physical behaviors, so the resulting physics is reflection invariant. Yes, you use a right hand rule to compute the angular momentum of a spinning ...


9

Angular momentum has a magnitude, and is about some axis (thus has some sense of "direction). So vectors are used to represent the quantity, and much of the machinery of vectors applies to it. However, there are ways in which "vector" is a misleading term when applied to it, and some people characterize them as being in other mathematical structures, such as ...


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