Trajectory in a non inertial frame

  • #1
LCSphysicist
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Homework Statement:
The hallmark of an inertial reference frame is that any object which is subject to zero net force will travel in a straight line at constant speed. To illustrate this, consider the following experiment: I am standing on the ground (which we shall take to be an inertial frame) beside a perfectly flat horizontal turntable, rotating with constant angular velocity ω. I lean over and shove a frictionless puck so that it slides across the turntable, straight through the center. The puck is subject to zero net force and, as seen from my inertial frame, travels in a straight line. Describe the puck's path as observed by someone sitting at rest on the turntable. This requires careful thought, but you should be able to get a qualitative picture. For a quantitative picture, it helps to use polar coordinates.
Relevant Equations:
r = r*ur
v = r*ur + wru(theta)
Well, make the center of the polar coordinates at the center of the turntable, so put r along the ro initial.
I am well known that to someone who is rest on the turntable, the equations will be the follow:

dr/dt = ro - vt

If the turntable route with angular velocity w,

dtheta/dt = -wt

We could go on with this and finish the question.

But i want to know about the other way that i think to resolve the question starting by the velocities:

We will have v = v = r*ur + wr*u(theta)
v = -v*ur - w(ro-vt)u(theta)

I don't know how should i proceed here..
 

Answers and Replies

  • #2
To make it simpler, you might consider the puck to be released from the centre of the table. If the (polar) coordinates of the puck in the lab frame are ##(r, \theta) = (vt, 0)##, then what are the polar coordinates of the puck in the rotating frame? It might help to draw a quick sketch at some time ##t##.

Once you have time parameterised ##r## and ##\theta##, you can then eliminate ##t##. What sort of shape do you get?
 
  • #3
LCSphysicist
636
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To make it simpler, you might consider the puck to be released from the centre of the table. If the (polar) coordinates of the puck in the lab frame are ##(r, \theta) = (vt, 0)##, then what are the polar coordinates of the puck in the rotating frame? It might help to draw a quick sketch at some time ##t##.

Once you have time parameterised ##r## and ##\theta##, you can then eliminate ##t##. What sort of shape do you get?
Just to simplify the shape

r = vt
theta = wt

r = v*theta/w

r = K*theta

A type of spiral.

Well, in my question and your example it's easier see the motion by take it directly, but i am confused if i tried to do it starting with the velocity.

Your example:
the minus sign because to someone in rest at the center, he doesn't route, the stone or whatever, that starts a moving this spiral.

v = v*ur ´- wr*utheta
v = v*ur - wvt*utheta

Make sense i try to integrate this? I think not, maybe is this a dead end and have to do the first approach?
 
  • #4
LCSphysicist
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1589276193387.png
 
  • #5
You can try a velocity approach as well. It's perhaps easiest in Cartesian coordinates. In the lab frame, the velocity might be ##\vec{v} = v \hat{x}##, or more succinctly ##v_x = v##.

You might then try and find what the components of velocity ##v_x'## and ##v_y'## are in a rotating frame (with rotating ##x'## and ##y'## axes.

But be careful: the ##\hat{x}'## and ##\hat{y}'## basis vectors are also functions of time...
 
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