New answers tagged

1

In general it changes although the reason is not exactly because its projections changes. For example. You start with a vector (let us say the electric field of a parallel plate capacitor) on the plane $xy$. Then you rotate the coordinate system by an angle. The components of the vector on the new coordinate system is changed. But the vector did not change ...


0

Direction of a vector is determined by the components themselves. Now if the components are changed the direction gets changed by the above definition. All this is with respect to one reference frame.


1

This sort of calculation, especially when the speed of the electrons from an observer's point of view is close to $c$, has to be done using special relativity, in that sense that the transformation between reference frames is determined by Lorentz rather than Galileo transformations. As you mentioned, you can put your reference frame origin at one of the ...


0

when you launched the ball upwards,as at that instant when the ball is in air and it falling possibilities should be on the palm but practically,I think its not possible it will return to your palm because no driver can maintain an exact uniform speed without changes.Only machines or robots can do so.The person also launching the ball cannot be so accurate ...


2

There isn't a "centripetal force" vector. As the car goes around the banked curve, the normal force on the car increases relative to what it would be on an un-banked straight road. The vertical component of the normal force supports the weight of the car, and the horizontal component of the normal force provides the centripetal force necessary to cause the ...


4

I'll give you a couple of ways to think about this. First, geometrically, the circle you are thinking about drawing should contain the entire circular path of the car. If we're assuming that the car is remaining at a constant "elevation" on the banked surface, then the center of that circle has to be at the same elevation: otherwise you'd be drawing a cone ...


0

P.S.2 - Done on Earth. One cannot watch the skies! If you insist on this, it cannot be done. Your condition is equivalent to us living on a permanently clouded world. In that case, we would experience alternating light and dark periods and would have no way of hypothesizing what was happening. Understanding the universe would have to wait for the ...


0

There is! It corresponds just to the tides due to the Sun. Let us suppose that the Earth is not rotating around the Sun, that is, we are not in free fall towards the Sun. In this case the liquid from oceans would accumulate nearest the Sun. The effect would be only one daily tide. On the other hand when we fall towards the Sun there are accumulation of ...


1

The most straightforward observation to show that the Earth moves is stellar parallax. If you take photographs of a groups of stars over a period of six months (half an orbit), some of the stars will seem to shift in position compared to the others. These stars are much closer to Earth and so seem to move more. This is similar to how, when you are riding in ...


1

Let two orthonormal systems $Oxyz$, $O'x'y'z'$ with a general motion (translational plus rotational) between each other and a point particle $\rm P$, see Figure. Symbol Conventions : 1.The vectors for position $\mathbf{R}$, velocity $\mathbf{U}$ and acceleration $\mathbf{A}$ of a particle with respect to $Oxyz$ expressed by coordinates of this same ...


1

Actually, it's NOT true that in SR the speed of light in vacuum is the same for all observers, regardless of the motion of the light source. This is true only for inertial observers. The same applies for GR, in which the generalization is a "freely falling frame" (a local inertial frame without effects of gravity). A good reference: Speed of Light


1

Correct; in general the speed of light is constant only as measured by local inertial observers. As an extreme example, consider a photon emitted from a galaxy far, far away, in our direction. Although it moves away from the galaxy in the direction of the Milky Way, the expansion of space makes it increase its distance from us. Eventually, however, it will ...


3

This is actually a famous theorem known as the Einstein Equivalence Postulate (sometimes Equivalence Principle). It's true that since Earth is spinning, acceleration in a spacecraft isn't quite the same situation we experience daily, but in general, yes, gravity is indistinguishable from uniform acceleration. Specifically, if you are in a box with no windows ...


0

Rotation is absolute. And in any non-rotating reference frame, the earth and the sun (ignoring all the other bodies in the solar system, and the rest of the universe) both revolve around their common center of mass. Since the mass of the sun is so much greater than the mass of the earth, it is pretty close to saying that the sun is stationary and the earth ...


0

If your two clocks $R$ and $R'$ are fixed to the same reference frame $S$ and synchronized between them, then they're stationary with respect to each other, and will read the same coordinate time for any event: $t_{R'}=t_{R}$ (the same applies for $t'_{R'}$ and $t'_{R}$) I suspect your confusion is due to your understanding that different time coordinates ...


2

The first figure shows the parallel and normal components of a vector $\mathbf{r}$ relatively to a direction $\mathbf{n}$. Based of this, the second figure shows the centripetal acceleration. In case of plane circular motion $\omega R = v$.


0

The 3×3 mass moment of inertia matrix only conveys information about the orientation of the principal axes and not their location. Their location is by definition on the center of mass. It is true that if rotating about an axis parallel to one of the principal axes, but not about the center of mass the MMOI matrix is world coordinates can still be diagonal ...


1

Gravitational fields aren't homogeneous. Here on the Earth, a clock on the floor runs more slowly than a clock on the table, and we have clocks precise enough to measure such small differences due to the gravitational gradient. But doesn't a clock in an accelerating spaceship run at the same rate no matter where in the ship you put it? See page ...


2

Let the body rotate about the $z$-axis, then by the definition of angular momentum $$\vec{L}=\vec{\omega} I_z.$$ where $\omega$ is the angular velocity about the $z$-axis. So we could take the parallel axis theorem and multiply it by $\omega$: $$\vec{\omega}I_{z}=\vec{\omega}I_{cm}+\vec{\omega}ma^2$$ Now ponder the terms in it. If I understand the ...


0

Not sure what you are asking but here goes. With respect to observer on ground car would seem to rotate with $\vec{\omega_{car}}+\vec{\omega_{disc}}$, where angular velocity is the vector summation, whereas it remains for the disc with respect to ground observer $\vec{\omega_{disc}}$


2

You don't have to, but it makes the equations easier to deal with because you don't have to account for the moment of acceleration terms. See the 2nd part this this answer about deriving Newton's laws on an abitrary point not the center of mass. So finally the equations of motion of a rigid body, as described by a frame A not on the center of gravity C ...


0

The acceleration of an object by fictitious forces is by the definition of what a fictitious force is due to the acceleration of the non inertial reference frame only and does not depend on its mass. In order to make Newtonian mechanics work in those non inertial frames these fictitious forces are introduced and defined to be equal to the mass times the ...


0

Black holes don't form. And you should keep your hand on your wallet when the word clearly is involved. If you have a spherically symmetric star and a spherically symmetric shell around it and you are inside the shell but outside the star then you notice the mass of the star but not the shell. When the shell contracts and gets to where you are it changes ...


0

To move velocities between two points A and B do the following: $$ \vec{v}_B = \vec{v}_A + \vec{\omega} \times (\vec{r}_B- \vec{r}_A) $$ The derivative of the above moves the material accelerations from A to B $$ \vec{a}_B= \vec{a}_A + \vec{\alpha} \times (\vec{r}_B- \vec{r}_A) + \vec{\omega} \times (\vec{v}_B - \vec{v}_A) $$ There is also the spatial ...


1

Take a free particle moving on a plane in polar coordinates $$ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} r \cos \theta \\ r \sin \theta \end{pmatrix}$$ The velocity is found from the chain rule, with clear separation for radial and tangential components: $$\begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \begin{vmatrix} \cos \theta & ...


3

Here is one way of looking at it via a velocity-dependent potential.$^1$ The Coriolis potential is $$\tag{1} U_{\rm cor} ~=~ -m({\bf v} \times {\bf \Omega})\cdot{\bf r} ~=~-{\bf v}\cdot ({\bf \Omega}\times{\bf r} ),$$ cf. Ref. 1. The factor $2$ comes from two different terms in the corresponding force formula $$\tag{2} {\bf F}~=~\frac{\mathrm d}{\mathrm ...


1

But why must $d\hat{u}$ be orthogonal to $\vec{\Omega}$ too (i.e. be tangential to a circle orthogonal to $\vec{\Omega})$? To get such a precession there must be a clockwise torque in the plane of the screen acting on the system which means that the torque vector must be pointing into the screen. That torque produces a change in the angular momentum ...


1

I think sections 4.1.2 and 4.1.3 of this lecture on dynamics explains it by looking at each component separately. Since $\frac{{\rm d}}{{\rm d}t} \sin \theta = \dot{\theta} \cos\theta$ and $\frac{{\rm d}}{{\rm d}t} \cos \theta = -\dot{\theta} \sin\theta$ the components of ${\rm d}u$ are perpendicular to $u$. Our first step is to choose cartesian axes, ...


0

It is acceleration, change in direction in this case, which makes the difference as does the fact that graviton all attraction is a non-contact force. In space away from any large mass going at 1000m/hr in a straight line does not require a force to be acting on you. When you go around a corner it is a localised contact force that provides your centripetal ...


-6

"Does a black hole really slow down time?" No. Gravitational time dilation is an absurd concept. General relativity predicts that gravitational time dilation occurs even in a HOMOGENEOUS gravitational field ("the homogeneous gravitational field is the gravitational field which, in every point, has the same gradient of the potential. Such a field is produced ...


4

Fictitious forces do not exist in inertial frames. Fictitious forces result from force-fitting Newtonian mechanics to non-inertial frames.


1

The 'swinging in a circle' creates a centripetal force that accelerates all water away from the center of the circle, with the force increasing with the distance from the center. The result will be that the water will orient so it is maximally distant from the center of the rotation. Hindered by the bucket from simply flying away, the resulting surface will ...


2

It is possible as soon as one is sure to be very distant from every body (gravitational source) in the universe. This is because all inertial forces behave as gravitational forces. If one is confined to stay in a closed room and observes the motion of bodies therein, he/she cannot decide whether the observed accelerated motion is due to a gravitational field ...


0

1) This is one of Einstein's relativity statement. An observer in an inertial frame cannot tell whether he is in motion or at uniform velocity as both corresponds to a non- accelerating situation. You can't "feel" velocity or rest, but acceleration, because an acceleration is accompanied by a force which changes your state of inertia (equilibrium). It is ...


0

In a non-inertial frame, since the frame is accelerating the person feels a force. A simple example is this. You just stand on a bathroom scale. While the frame moves, if the scale reading is not changing, then the frame is inertial. Otherwise it's non-inertial. This is how gravity is explained in general relativity. So if the person is aware of the ...


0

In a non-inertial frame, observers will see fictitious forces with no reaction pair. For example, in a frame accelerating linearly forward, there appears to be a force acting backwards, and one cannot find the reaction (or source) of this force.


0

If you are riding along the rigid body, the vector $\boldsymbol{A}$ is not fixed. It may change with time and you see this change as $\Delta \boldsymbol{A}_{S'}$. Now to find the change in $\boldsymbol{A}$ from an inertial (non-rotating) frame you have to add the effect of the rotation as the unit vectors of the S' coordinate frame change. This is the ...


2

A rotating reference frame is not an inertial reference frame: In the rotating frame, objects accelerate even though there are no forces acting on them. In your example, you can in fact determine easily whether you are rotating or the universe is rotating around you. In the first case there is artificial gravity on the ship, and in the second case there is ...


1

The $\Omega \times$ terms signify change due to the moving orientation of the item. So the velocity vector when attached to a moving body will cause a $\Omega \times v$ term in the acceleration vector.


0

If one takes a vector in an inertial frame say r and the same vector viewed from an observer in a rotating frame say r' (call it dashed frame) and observe the the rate of change with respect to time- Due to rotation of the dashed frame the tip of the vector starts rotating about the axis of rotation and any change in time dt leads to a change dr ...


0

Imagine two ficed chares on an axis that can rotate (like merry go round). On the one chair is your friend that holds a ball and on the other it's you. At some point the hole system starts rotating and your friend throws the ball towards you. After he throws the ball you have changes position so the ball doen't come to you. Instead according to you it moves ...


4

There are a few ways to justify it. First, you could look at the motion of the object as it rotates. In 2D, it turns out that all such motion can be decomposed into motion of the CM, and rotation about the CM. Therefore, rotating around the CM itself is the only way to guarantee the CM doesn't move, and hence is the least energetically costly. Another way ...


1

That the moment of inertia about an axis passing through the CM is minimized, with respect to any other parallel axes, is a consequence of the quadratic (squared) dependence of the moment of inertia on distance. In other words, the ${r^2}$ term in ${I=mr^2}$ makes it so that masses at farther distances are preferentially weighted in their contribution to the ...


0

This additional detail is in response to the comment thread on my earlier answer. It provides every step of the calculations, which are perfectly standard and the sort of thing you will find in any reputable textbook. They are also exactly equivalent to what's going on in the spacetime diagrams you're curiouosly reluctant to study. Start by labeling ...


0

Rapid acceleration does what to inertial frame clocks? Nothing much. Like I said in response to your previous question. I'm 10 light years from Earth and see Earth's clock/calendar reading 2000, including light travel time. In other words, I assume it's 2010 on Earth, but what I actually see is the year 2000 due to light travel time. OK, that's ...


0

Coordinate Systems There is a great paper by Fränz and Harper [2002] that describes all the coordinate systems used to describe planetary orbits. Most of the coordinates that would be relevant to your question are celestial and/or heliographic coordinate systems. The celestial systems use the First Point of Aries as a reference point and then the Earth's ...


2

Here is the question: what do we mean by "kinetic energy"? I assert that we mean "the amount of energy or 'work' I would need to make this object come to rest in my frame of reference". Consider a ball with mass $m$ in free space $${\huge\circ}$$ In it's inertial frame of reference it's velocity is zero so its kinetic energy is also zero. But all inertial ...


0

When you say the small mass will have a certain velocity, v, after converting the chemical energy - what frame of reference is that in? Not in this a frame of reference following the small mass because in that frame of reference the small mass's velocity must always be zero. For that step you're using what I'll call the page frame of reference - presumably ...



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