# Tag Info

1

The Coriolis acceleration is $2\vec{v}\times\vec{\omega}$, where the velocity vector $\vec{v}$ is measured in the rotating reference frame. In your example, if I'm reading you right, the particle is moving with the disc, so $\vec{v}=0$, so the Coriolis acceleration is zero. In general, if you have either a radial or tangential velocity component in the ...

-1

To make sense of this question you need to decide what an "object" means (is a rock an object or a conglomeration of a vast number of much smaller objects?). Once you've settled that, you need to decide whether you're averaging velocities or speeds. If there are three objects, and two of them are moving away from me at the same speed $v$ in opposite ...

0

No, there cannot be a most stationary object in the universe. One of the basic tenets of relativity is that every inertial reference frame is equivalent. This is what Einstein said on the matter, If a co-ordinate system $K$ be so chosen that when referred to it the physical laws hold in their simplest forms, these laws would be also valid when referred ...

1

Relativity just requires "constant speed of light in vacuum". It makes no claims about the speed of light in a medium. When you are moving relative to water, you will observe a different speed of light depending on your relative velocity. But you will still have all the other effects of relativity at work - such as time dilation.

0

Think of a non homogeneous rigid body. It is rotating about a axis, you have to find say Moment of inertiaabout the axis. So in that case you consider a axis for which it's easy to calculate, and which is parallel to the axis you have to find. For examples, for the axes having origin at the centre of mass, then you can find the M.I easily, then use that ...

-1

Actually, the assumption of a psuedo riemannian manifold doesn't require many tacit assumptions. Can you measure time and distances? Can you define a right angle? Ok, you now have a manifold equipped with a metric. Want to include time as a dimension? Now you have four dimensions. You can't turn around in time like you can in space, so you need the time ...

0

My interpretation is that you are raising the following objection to the black hole information paradox: According to observers distant from the hole, causal lines take infinite coordinate time to cross the event horizon. To these observers, infalling information is thus never lost, but only very strongly redshifted; in essence it remains "painted" on the ...

1

Imaging the balls on a string. You are launching N balls per second, at a velocity $u$. This means the distance between the balls is $u/N$. And $N$ balls per second will pass a certain point in space. Now if the car is moving at a velocity $v$ (same direction as $u$), fewer balls per second can hit it - because subsequent balls on the string have further to ...

0

The angular momentum $L$ of an object is always defined with regards to the axis about which rotation takes place: $$L=I\omega,$$ where $I$ is the inertial moment and $\omega$ the angular speed. The inertial moment is defined and can be calculated as shown here. Where it's necessary to calculate $I$ with respect to another axis parallel to the first one, ...

0

Angular momentum is a vector defined by $$\vec{L}_{cm} = I_{cm} \vec{\omega}$$ where $I_{cm}$ is a 3×3 mass moment of inertia about the center of mass. When the rotational velocity vector $\omega$ is about one of the principal inertial axes then the angular momentum vector is parallel to the motion axis and $\vec{L}_{cm} =m \rho^2 \vec{\omega}$ where $m$ is ...

2

Let you apply force $\bf F$ at point $P$ the coordinate of which is $\bf r$ measured from a specific point $O$ - the point about which you want to rotate. Let $\bf r$ and $\bf F$ be in the same plane. Now, if you were to rotate $P$ about $O$, it would rotate around some axis perpendicular to the plane in which the force and the point lies; if ...

1

A reference frame is equivalent to a choice of coordinates. So, choosing an accelerated frame in Minkowski space is equivalent to choosing a specific coordinate system on Minkowski space. Most importantly, this means that there is not genuine curvature in an accelerated frame, i.e. it is fundamentally different than gravity. The equivalence principle ...

3

Let's say we're sending a scouting mission to an Earth-like planet 100ly away to see if it's suitable for colonization. We could send the scouts out at near light speed, and due to time dilation they could easily survive the trip without dying of old age. If we send them out at .9c then the entire round trip will only take 20 years in their frame of ...

2

A space train leaves Mars at 14:00pm and arrives on Earth at 19:45. The train moves at 0.001C and has 40Km of length. How long will it take for the whole train to arrive on Earth? - Disregard re-entry and friction. Nobody on Earth will say the train is leaving mars now. Same thing with the light, just it moves faster and is smaller than the train above. ...

18

A physicist, me for example, identifies events by choosing a set of coordinates. For example I have a clock that I use to record time and a ruler that I can choose to measure distance. This allows me to set up some coordinates $(t, x, y, z)$ so I can assign every event to some point in my coordinate system. If I received a laser pulse from Mars at 16:05 ...

0

I think the fact that point in question is separated in a spacelike way for the two observers does not address the argument put forward by the paradox, as it explicitly states that the two observers are comparing their accounts in the distant future, after it would be possible for the invading fleet to arrive and affect them. Penrose states it as: "In fact ...

1

One thing you have to note is that speed is relative, Clock A would see clock B moving from A's point of reference, and B would see A moving in B's reference, so you shouldn't be using the word "stationary" in this context. Both the clocks would see the other clock tick slower, B would see A's future only if it returns back to A, this makes it obvious to A ...

0

Let's assume that they go with the speed $v$ wrt their earthy friends. So for them (now onwards 'they' refere to spaceship crew unless otherwise mentioned) the initial distance is $l_0 \sqrt{1-v^2/c^2}$ and the planet is coming towards them at the speed of $v$ . Where $l_0$ is 8 light years. Now in their frame the planet takes 8 years to reach them. ...

2

A pedagogical note: Because students regularly try to look for the centripetal force and get confused (as noted by other answers here), I try to emphasize that it is the acceleration which is centripetal rather than the force (even though the sum is center directed). Because the acceleration is always the result of a sum, I have found that fewer students in ...

0

You are a bit confused by the meaning of the centripetal and centrifugal force. First, in a reference frame at rest relative to the center of mass, the gravity IS the centripetal force (just because it points towards the center). My impression is that you assumed that there where two forces, gravity and the centrifugal force. Second, in a reference frame in ...

3

This is a very common "gotcha" for novices, and something that is not covered well at all in textbooks, in my opinion. The problem is one of semantics, not of physics. The term centripetal force does not name a physical force the way gravitational force and electrostatic force do. The words centripetal force stand for the net result of real forces in the ...

1

Centripetal Force isn't a real force. It is a construct for representing the sum of different forces that cause circular motion. In the case of an orbit, the gravitational force is the centripetal force, which is what you stated in the first equation. They are NOT opposites, as you asked in the question. They do not become arranged to have a net force of ...

4

After a comment it becomes clear that there is deep misunderstanding here and the question title has nothing to do with the actual problem. Let's get something basic clear: there is no "the centripetal force". That is no force out there that magically decides to come into being when an object moves in a circle. Rather "centripetal" is a label for those ...

1

Do you know Bernard Schutz's book: A First Course in General Relativity? Check out the first chapter of that book. There is a derivation of invariance of proper time using first principles in section 1.6. Basically, the idea is to start from expressing $\Delta \bar s ^2$ (interval in the barred frame) as a linear combination of $\Delta x_i$'s (vector ...

1

So I take particle A and place it in space, then I place particle B 1,000,000 light years away from particle A. Alright. But, just to be sure: In astronomy, and (prehaps somewhat more recently) in cosmology and in physics in general, we understand this measure of "having been apart" as chronometric mutual separation. In your example this means ...

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When OP said, ""Why does there seem to be zero inertia along the tangent line when that is the direction it is moving at any moment?"" The inertia is always away from you, so when you spin the sling and let it go when it is in front of you, that is the real trajectory of the shot, the circle path only exist while you are swinging it. The acceleration force ...

-3

"They call μrϕ˙2 the fictitious force or the centrifugal force. I'm quite hazy on my memory of non-inertial frames, but I was under the assumption that fictitious forces only appear in non-inertial frames." all "fictitious" forces still "transfer energy." Also anything that is an acceleration in math (distance/time^2) is indicative of energy transfer. ...

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