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In order for a body to move with uniform velocity in a circular path, there must exist some force towards the centre of curvature of the circular path. This is centripetal force. By Newton's Third Law, there must exist a reactive force that is equal in magnitude and opposite in direction. True, although the adjective "reactive" is meaningless. There is ...


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The term centripetal force is only relevant in an inertial frame. However it is always created by some physical force, gravitation or the friction from a seat or something. In the rotating system, the fictional centrifugal force has to be invoked. The physical force that makes up the centripetal force also exist in the rotating frame and here it is exactly ...


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Lots of places state that the Earth's gravity is stronger at the poles than the equator for two reasons: The centrifugal force cancels out the gravity minimally, more so at the equator than at the poles. The poles are closer to the center due to the equatorial bulge, and thus have a stronger gravitational field. TL;DR version: There are ...


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Here's a simple argument that doesn't require any knowledge of fancy stuff like equipotentials or rotating frames of reference. Imagine that we could gradually spin the earth faster and faster. Eventually it would fly apart. At the moment when it started to fly apart, what would be happening would be that the portions of the earth at the equator would at ...


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The point is that if we approximate Earth with an oblate ellipsoid, then the surface of Earth is an equipotential surface,$^1$ see e.g. this Phys.SE post. Now, because the polar radius is smaller than the equatorial radius, the density of equipotential surfaces at the poles must be bigger than at the equator. Or equivalently, the field strength$^2$ $g$ ...


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Yes, the gradient in spherical coordinates contains angular terms. It has to, because the force need not point to the center. The relevant equation is $$\Delta U=\frac {\partial U}{\partial r}e_r+\frac {\partial U}{r\partial \theta}e_\theta+\frac {\partial U}{ r \sin \theta \partial\phi}e_\phi$$ You have no $\phi$ variation, so can ignore the last term.


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The potential energy is $U\left(r\right) = k \left(r - r_0\right)^2 / 2$, where $r_0$ is the equilibrium position of the spring. Since this is a central potential (depends only on $r$), angular momentum $L$ is conserved, and we can use the concept of an effective potential energy (see equation 335 here) $$ \begin{eqnarray} U_{eff}\left(r\right) &=& ...


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At 1:02 in that video, you see the "people" at the end of the swing do something like a half revolution with a radius of 15 meter in 0.3 seconds (rough estimate of dimensions and time - feel free to come up with your own estimates by counting frames etc - then substitute those numbers in the equations below). The centrifugal acceleration is given by $$a = ...


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would the steering wheel reach the end of travel at a smaller angle? NO. the range of travel of the wheel is set by mechanical parts in the chain. They don't know how fast you are going. There has been talk of "steer by wire" where a computer reads the angle of the steering wheel and decides how much to move the front wheels. As opposed to the throttle, ...


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Regardless of whether you are driving at 65mph or 1mph, as long as your steering angle is the same, you will travel the same path. (Of course in real life understeer or oversteer has to be considered) However, due to the much greater centripetal acceleration experienced, where it is proportional to the square of the velocity, $$a_c = \frac{v^2}{r}$$ You ...



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