# Tag Info

## New answers tagged rotation

1

The ISS travels from West to East (with an inclined orbit). That means it crosses the international date line 15 times per 24 hour period, with time "going backwards" as they do so. This also means they must cross "midnight of the New Year" that many times. Although - since the orbit is 15.5 revolutions per 24 hours it's possible that they sneak in a 16th ...

3

There should be no preferred reference frame and from an observer on the sun, the sun will rotate around the Earth. Aside: You have the sense backwards. From the perspective of an observer on the Earth, it's the Sun that appears to rotate around the Earth, and at three frequencies, once a day, once a year, and once every 26,000 years. An observer on the ...

10

I prefer to think of it that the Earth and Sun actually orbit around their combined center of mass, which just so happens to be very deep inside the sun. The same can be said for the Earth-Moon system.

7

People say the Earth rotates around the sun and not vice versa because a reference frame attached to the center of the Sun more closely approximates an inertial reference frame than a reference frame attached to the center of the Earth. Yes, there is the issue of the path of the other planets, but even if the Earth were the only planet, we would still say ...

8

The reason that you get slip at even the smallest forces results not from the fact that the tire is slipping against the ground, but that the tire is elastic. There is no way to completely eliminate slip with an elastic tire. Let's see why this is. To measure the slip, lets put twenty little green splotches of die evenly spaced on the circumference of the ...

0

Both graphs have 0% force at 0% slip. That implies that no force is being applied to the tire. As applied force is increased until a certain point the increase in slip remains nearly linear because the coefficient of friction remains nearly constant. Thus effective force also increases nearly linearly. After that point the coefficient of friction gets ...

0

My analysis is that the object would not rotate at any velocity. My reasoning is that if the object is completely flat on the floor then there would be no net torque. Say r is the distance from the bottom left corner (as in your diagram) to the point at which the net frictional force acts (directly beneath center of mass, at a point where the object and ...

0

1- yes for a body to experience uniform circular motion the net force must be directed towards center of the the circle. In a gravitron ride the walls are slanted. Therefore normal acts at an angle with the horizontal. The vertical component of normal balances gravitational force while the horizontal component provides the necessary centripetal force to the ...

1

If the motion is uniform (no angular acceleration) you do not need any tangential force, at least ideally. It's during the acceleration that you feel the force throwing you out of the chair. Also related: When does centripetal force cause constant circular motion?. I'm not sure I understand your second question. The centripetal force is the net force on the ...

0

What this refers to is the Rotation Reversal Theorem - rotating first about axis z with angle az , and then about the rotated y axis by angle ay , followed by rotation by the now twice rotated z axis by angle bz is the same as rotating first about the original z axis by bz, followed by rotation about the ORIGINAL y axis by ay and then finally about the ...

0

The way I see it this problem is similar the problem of column of water and the pressure rising due to gravity pulling down on the liquid - in which case pressure, $P$, is given by $P = h \rho g$ where $h$ is the height of liquid, $\rho$ is the density and $g$ the acceleration due to gravity - but you probably knew this already. The point I want to make ...

2

Angular speed is a vector (a pseudovector actually), and as such it changes relative to you under a rotation of coordinates. Angular speed is defined as the vector $\boldsymbol \omega =\boldsymbol r$ x $\boldsymbol v/|r|^2$, and it will change direction in an angle of $\pi$ as, expected, if your system of coordinates rotates by $\pi$ too (as it was in your ...

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