New answers tagged

0

The above answers do not take into account the vector nature of $\omega$. I venture to provide an answer that hopefully should satisfy the questioner. We Consider a point at colatitude $\lambda$ on the Earth's surface in the Northern hemisphere. We draw a local coordinate system at this point such that the x-axis points south, the y-axis points east, and the ...


0

Let's just say, for the sake of argument, that the equator has a linear eastern velocity of 1,000 mph (actually 1,039). If you release a ball (presumably fired from a cannon) southward from a latitude of 70° north, your ball has an eastern velocity of about 342 mph (cosine of 70° = 0.342). As your ball travels over 45° north, the ground beneath it will have ...


1

The direction of deflection of the ball should be independent where the observer is. Left or right deflection is with respect to an observer facing in the direction the ball is moving. Are you asking about a ball being thrown from the northern to the southern hemisphere? In that case the deflection would change from deflecting right to deflecting left as ...


1

This is indeed confusing. The confusion comes from this very peculiar hypothesis: What if the person doesn’t apply a tangential friction force at his feet? It implies there is a radial contact force at the person's feet (I prefer "contact" to "friction", which refers to movement). And, indeed, for the person to move radially inwards, or even to stay ...


1

The vector $\mathbf r$ does change, even though its magnitude is nearly constant. Most importantly, the component of $\mathbf r$ which is perpendicular to the rotation axis is decreasing in this example. This fact leads to the explanation for which you are searching. Another way of seeing this would be to use this definition of angular momentum, which ...


0

In my view the motion of the person, seen in lab frame, would be linear motion, because, at the beginning, the person has a tangential velocity $ωr$ and radial velocity $v$, and he will keep these two forever. But then does it makes sense to talk about conservation of angular momentum? I mean it will surely be conserved in the lab frame but the ...


0

After reading your question again I gave a try proving why and how angular momentum is conserved. In the example given by the question one needs to understand how the angular momentum of an object moving in a rotating, non-inertial frame is conserved. I will present in brief, my effort of proving how and why the angular momentum is an integral of motion. The ...


1

The error is just to consider an average speed $h\omega$. When the particle is at height $z$, its horizontal (relative to the Earth) speed is $v=2z\omega$. The time of of flight is $$t=\sqrt{\frac{2z}{g}}.$$ Differentiating this expression we get the time taken by the particle to move a distance $dz$, $$dt=\frac{dz}{\sqrt{2gz}}.$$ The horizontal distance ...


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 ...


4

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


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 ...



Top 50 recent answers are included