Question about circular movement.

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A user asks whether a car moving at constant velocity on Earth experiences acceleration due to the Earth's rotation. The discussion clarifies that while the car maintains a constant speed, it does have centripetal acceleration because it is moving in a circular path relative to the rotating Earth. The conversation also touches on the roles of gravity, normal force, and fictitious forces like centrifugal and Coriolis forces in this context. Ultimately, centripetal acceleration is necessary to change the direction of the car's velocity, even if its speed remains constant. The user expresses gratitude for the insights gained during the discussion.
cosmic_tears
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Hey everyone, I'm new here.
I'll try to be as clear as I can, but I'm warning you I'm not sure my science-English is accurate enough. I'll try though.

So, It's a pretty basic question, but I still find it confusing:
Say you have a body moving in a constant velocity v in relation to a rotating axis S. (i.e. a car driving with no acceleration on earth).
Does the moving body have an acceleration in relation to earth?
I get the impression (from exercises) that the body actually has a centripetal acceleration ((u^2)/R), but I do not understand it -
Why doesn't the body move in a constant velocity in relation to S?

Thank you very much for reading, hope I was clear.
Tomer.

(these forums are exciting! :biggrin:)
 
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Why doesn't the body move in a constant velocity in relation to S?
Gravity is pulling (accekeration) the body toward the center. However, the surface the Earth is a barrier, so the car just stays on the surface.
 
thanks for replying.

However... that's not what I meant.
I didn't ask what's preventing it from falling towars Earth... I asked if it has a centripetal acc. due to the rotation of the system (Which enables it to move in a circular orbit), in relation to the system...

Thanks again though :)
 
cosmic_tears said:
Say you have a body moving in a constant velocity v in relation to a rotating axis S. (i.e. a car driving with no acceleration on earth).
Does the moving body have an acceleration in relation to earth?

If you say you have a body moving in a constant velocity v with respect to some reference point, then the acceleration of that body respect to that reference frame must be zero.

However, I have feeling that's not what you really wanted to ask.

Give us more detail. Define your system and reference point at which you want to describe the motion of your system.
 
Ok, fine:
There is a car moving with constant velocity (in relation to Earth) on Earth.
(imagian the picture - a small car drawn somewhere on a circle with it's velocity vector tangental to the circle.)
So, the car is moving in a constant speed.
Does it have centripetal acc.?
If it's that important - assume the X axis is the "tangent" one, the Y axis is the "radial" one, and the Z axis is "coming from inside the page" (excuse my lame description).

I hope my question is more clear now :-\
And thanks again.
 
It has acceleration in two different directions. The first is the result of the Earth's gravity. It pulls the car down toward the center of the Earth, so it keeps the car on the surface of the Earth. This one is centripetal The second one moves the car along with the Earth's surface because of the spin of the Earth. This is caused by the force of friction between the ground and the car. This one is linear.

I hope that answered your question!
 
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I'll just try and be as specific as I can...

There's an excercise describing just what I did - a car with a const. velocity, it's mass M, moving on Earth. (say it's standing on top of the circle, with it's velocity vector U pointing rightwards). w, U, M, and g are given. (w - angular velcoity)
Here is what the lecturer did:
Solving it in the rotating Earth Axis, he mentioned that you have to add two "imagianary forces", which are the the centrifugal force and Coriolis force.
He marked a "-mwXu" vector towards the center of the Earth (which is coriolis vorce), and a "-mwX(wXu)" vector pointing outside the Earth (that's the centrifugal force). There's also a Normal force pointing up (the car's weight), and Gravity as Ateowa mentioned. However, there is no friction mentioned in the excerise...
He eventually wrote:

mg + 2mwu - N - m(w^2)R = m(u^2)/R

what bothers me the most is the right side of the equation - I don't understand why is has centripetal acc.

Thanks everyone, sorry for nagging :)
 
I agree, I find that confusing as well. The Coriolis effect (Which was the frictional force i was talking about, by the way) would not contribute to a centripetal acceleration. It would always be perpendicular to the centripetal acceleration caused by the combination of gravity and the normal force. At the very least, they would not be able to be put in the same equation like that.

I don't know what to tell you. Hopefully someone with more insight than I can help you.

Sorry!
 
Ateowa, thanks anyway!
Usually, the Coriolis force does indeed usually involves a 3D problem, but in this case, since the velocity vector is pointed rightwards, wXu (or -2mwXu) is indeed in the normal-gravity axis. That's why it can enter the equation...
 
  • #10
cosmic_tears said:
He eventually wrote:

mg + 2mwu - N - m(w^2)R = m(u^2)/R

what bothers me the most is the right side of the equation - I don't understand why is has centripetal acc.
It has centripetal acceleration because, with respect to the rotating earth, it's going in a circle. (Anything moving in a circle is centripetally accelerated.)
 
  • #11
It's a centripetal acceleration, whose job is to change the direction of the velocity and the magnitude, so car keeps on traveling with same speed but having varying direction.

One more thing, centripetal force must be perpendicular to the instantaneous velocity, to ensure that it doesn't change it's speed.
 
  • #12
I think I'm beginning to understand it. A friend of mine helped me also.

Thanks everyone!
 
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