Hot answers tagged rotational-dynamics
22
It's a classical mechanics effect for sure although a really interesting one. Following links on "Dzhanibekov effect" one gets at Marsden and Ratiu's "Introduction to Mechanics and Symmetry" Chapter 15 Section 15.9 "Rigid Body Stability" treating this with use of the Casimir functions.
From remark 1: A rigid body tossed about its middle axis will undergo an ...
21
The reason is that you have a boundary layer on the surface of the blade of the fan. On the frame of the blade (the blade moves with some velocity, but at the frame of the blade the air moves) the boundary layer starts from the surface of the blade where the fluids velocity is zero and as you move away from the blade, the velocity increases up to the value ...
19
Your explanation is right: an earthquake can't change the axis of rotation, relative to a given inertial reference frame -- that is, the axis of rotation doesn't change relative to the "fixed stars" as a result of the earthquake. What the earthquake does is to move material around within the Earth, so that the position of the rotation axis relative to any ...
15
Applying the brakes makes the wheel stop turning in relation to the bicycle's frame but not in relation to the road. The bike's center of mass (especially with a rider pressing against the handle bars) is higher than the hub of the front wheel.
When the brakes are applied that mass has momentum toward the front of the bike that exerts a force on the front ...
13
What about this hypothesis:
Dust sticks everywhere, but since the propeller cuts through a lot of air, it meets more dust particles. Thus, more dust sticks to the propeller than elsewhere.
Evidence
I (Mark) took photos my the fan my room to support Damien's hypothesis. The first photo is of the leading edge of the fan blade, which impacts a lot of air, ...
13
Well, the angular momentum conservation is still the essence although it may be formulated in a different language.
The top is spinning around a vertical axis and the spinning around this axis can't disappear. if the top decided to fall, the spinning would either disappear or would be replaced by a totally different spinning around a horizontal axis, and ...
10
There was some doubt about Lubos' answer (which I've accepted), so this is just a verification.
I copied the method Lubos described and found the potential difference for an ellipsoid with different eccentricities. Sure enough, for an oblate spheroid, if you make the center-equator distance a fraction $e$ larger than the center-pole distance, the ...
10
Infinitesimal rotations don't commute exactly if you're accurate enough. An infinitesimal rotation may be written as
$$ \exp( i a A ) $$
where $a$ is an infinitesimal "angle" and $A$ is a combination of generators. Such an object doesn't commute with the analogous object $\exp(ibB)$ in general. Instead,
$$ \exp(iaA) \exp(ibB) = \exp(ibB) \exp(iaA) \exp(-ab ...
8
No, it's caused by conservation of angular momentum. Reducing air resistance won't cause her (or anything else) to speed up without an external force.
Like linear momentum ($m v$), angular momentum ($r \times mv$) is a conserved quantity, where $r$ is the vector from the center of rotation. For a skater holding a static pose, for each particle making up ...
8
Wind doesn't actually touch the surface. You can see the same effect on a car: even if you move at speeds beyond 70mph, the dust doesn't get blown away.
If you look closely, there is a boundary layer between the matter of the fan and the air around the fan. When you get closer to the fan blades, the air starts to move with the fan (the blade pulls it ...
7
When it is spinning its angular momentum is quite high. By conservation of angular momentum the spinning top is then more stable against small torques like the action of gravity on the top.
The angular momentum of the top is $J = I \omega$ where $I$ is the inertia tensor and $\omega$ is the Darboux vector, whose magnitude is proportional to the rotational ...
7
Joe's answer is of course right and I gave it +1. However, let me say some slightly complementary things.
Whenever the laws of physics don't depend on the orientation in space, a number known as the angular momentum is conserved. For a rotating body - including the body of a lady - the angular momentum $J$ may be written as the product of the moment of ...
7
Quite clearly this man’s theory is balderdash. A massive collision could change the tilt of the Earth’s axis, but that would be one hell of a torque. The collision would not be with an asteroid either, but with another planet that might be as big as Mars. Clearly nothing like this happened in recent geological history, though the Earth did suffer a ...
7
There is some confusion about the term axis. The axis about which the Earth rotates of course did not change. It requires some external torque or transfer of angular momentum for that to happen. However, the Earthquake changed the shape of the Earth slightly, which changed the figure axis of the Earth. Calculations also show the Japan quake should have ...
7
The source in the Einstein field equations is the stress-energy tensor, not the scalar mass-energy. Adding rotation will affect multiple elements of the stress-energy tensor. You can sometimes get rough estimates of effects in GR by using $E=mc^2$ and pseudo-Newtonian arguments, but sometimes these are way off. As an example where it's way off, two light ...
7
First I wanted to turn to textbooks to solve that, but felt this way is going to be boring (especially if we are talking Mass Effect). So I decided to derive equations of motion for our system from the first principles.
Further I need a lot of trigonometry, so I'll use short-hands for cosine and sine: $$c_x=\cos x,\;s_x=\sin x$$
Let us start by introducing ...
7
First break the problem up using two free body diagrams.
Then figure out the kinematics at point A
$$ \vec{r}_A = \begin{pmatrix} x \\ 0 \\ 0 \end{pmatrix} $$
$$ \vec{v}_A = \begin{pmatrix}\dot x \\ 0 \\ 0 \end{pmatrix} $$
$$ \vec{a}_A = \begin{pmatrix}\ddot x \\ 0 \\ 0 \end{pmatrix} $$
and point B
$$ \vec{r}_B = \vec{r}_A + \begin{bmatrix} ...
6
The answer to your question depends on precisely how it is interpreted. In my opinion, the clearest way of understanding a car driving on the road does in fact have the engine doing work on the car, but it is possible to define the system involved such that this is is not so. However, under this interpretation, the engine does no work on the car regardless ...
6
As you say, there's a perfectly sensible operational definition of an inertial frame: it's one in which free particles move with constant velocity. Even in general relativity, it makes sense to talk about inertial frames, but only locally. To be precise, an inertial frame is well-defined only in an infinitesimal neighborhood of a spacetime point, although in ...
6
The short answer is that there's no wind near the blade. This is called no-slip condition in hydrodynamics of viscous fluids.
[Concession] It is actually more than that. There's minor van der waals sticking which contributes to this otherwise purely hydrodynamic phenomenon.
6
First taking only speed of the fan into the account. If the fan rotates slowly then the situation is obviously not very much different from if it weren't rotating at all. The centrifugal force on the dust particles is not big enough to throw them away off the fan.
Second, there's static electricity that has to be taken into the account. It's perfectly ...
6
Try looking at this problem microscopically:
You can imagine the ball consisting of a number of smaller pieces of matter. The total kinetic energy of the entire ball is the sum of the kinetic energies of its pieces:
$$T=\frac{1}{2}\sum_im_iv_i^2$$
Now, if it were to just fall straight down, all these little pieces would have velocity vector $v_i=v$ and ...
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.
...
6
I like your description of this cool bit of unintuitive physics. I find the best balance of $a$ to $b$ to $c$ to cost of the object involved is best for a (boxed) pack of playing cards.
The mathematical explanation for this (see also Wikipedia) is that when considered in the principal axis frame (i.e. the frame of reference that rotates with the body and ...
6
The center of mass of the bike+rider wants to keep moving forward (Newton's first law)
Since the center of mass is higher on the bike - when it goes forward the bike pivots around the point where the front wheel sticks to the road.
Imagine attaching a string to the middle of the wooden block and pulling it forward, while having the front wheel stuck to ...
6
There is simplified "cheat" method for thinking about this.
The hard way is to draw a free-body diagram of the bicycle, in which all the forces acting on it are made plain: the force of gravity acting through its center of mass, the force of friction from braking, and so on.
The cheat method is to visualize the bicycle in an accelerating frame of ...
5
The basic setup is correct, conservation of energy might be the quickest way to go.
$$(m_1 -m_2) g h = \frac 1 2 I \omega^2 + \frac 1 2 (m_1+m_2) v^2, I=\frac{MR^2}{2}, \omega = v/R$$ gives me one of your options as the result.
The two $m$ in your formula seem to refer to different quantities.
5
Let's begin by choosing coordinates. Let the orbit of the earth define the x-y plane. Assume that just before the collision the earth's orbital axis is pointing in the x-direction, that is $(1,0,0)$. I'll assume that the earth absorbs the asteroid, and that after the collision the axis has been changed to $(\sin(27),0,\cos(27))$ where 27 is the degree ...
5
The electric field is null: Because of the rotational symmetry assumed there, the magnetic induction $B$ is constant in time, so $\nabla\times\,E = 0$ by Faraday's law. On the other hand no electric charge is present, so $\nabla\cdot E = 0$. This is enough to make $E = 0$.
Moving permanent magnets do generate an electric field, "even in cases where ...
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