Transition from inertial to circular motion

1. Apr 8, 2014

analyst5

Suppose that we have a body that is moving at a straight line, inertially wrt to another frame. If it starts to move in a circular way after that, what can be said about the motions of its points. Do all points have to deccelrate to achieve the circular motion, but in a different manner, since during circular motion the points which are at a greater radius from the centre of the circle in which the body moves have a greater velocity? Can anybody explain this interesting effect to me?

Regards

2. Apr 8, 2014

abitslow

?
Which is larger (1,0) or (0,1) ? Which is larger 5 + 3i or 33 ? Velocity is a VECTOR, not a scalar!
A change in velocity is an acceleration. So given a velocity of (3,4,5) is the change to (5,4,3) an acceleration or a deceleration??? We need to only use the term "deceleration" in very simple cases. Most physicists prefer to use the unambiguous term 'acceleration' to mean any change in the velocity vector, whether of simple magnitude (1,1,1)→ (2,2,2), or of direction, or both. Its magnitude can be positive or negative.
-----------------
That said, you need to explain what "interesting effect" you mean. Why would a force acting differentially on the particles of a body be expected to do anything else BUT result in different accelerations? If the object is rigid, the force will be countered by the interparticle forces, so that there will be stresses induced.
-=-=-=-=-
Lets say the initial velocity is (1,1) [in 2-d space]. After attaining its circular orbit the velocity is (dr, dΘ)
which is a different coordinate system. dr varies from particle to particle of the object, dΘ does not.
transformation of the polar coordinates back to orthonormal euclidean ones is trivial x = rcos(Θ), y = rsin(Θ)
since we know both x and y are changing with time, but x²+y² is not (for a circular orbit), we can solve for velocity in terms of either dx/dt and dy/dt or dΘ/dt. As long as the object is not spinning, tumbling, r is constant for each particle. What else do you want to know?

3. Apr 8, 2014

analyst5

I'm afraid I didn't understand the second part of your post. Can you precisely explain how does the object come to the state of circular motion from the state of inertial motion, what are the differences between each of its points during that event?

4. Apr 8, 2014

Staff: Mentor

This isn't necessarily true. For simplicity I will assume that you were interested in rigid body motion from an initially inertial body.

If the object experiences a net force but no net torque then each point will have the same speed as every other point.

If the object experiences a net torque but no net force then different point will have different speeds.

Circular motion is caused by a centripetal net force, so it does not result in different points having different speeds. Only in the case that a net torque is also involved will the different points attain different speeds.

5. Apr 8, 2014

analyst5

But don't the more distant points have a greater radius in their circular motion so they have to travel with a greater velocity to remain at test wrt to closer ones? As in the example where the earth's equator travels with a greater tangential velocity than some point between it and the poles? Wouldn't the situation where all points have the same speed lead to different parts 'trailing' in space relative to other parts of the body? And how does the body come to this position, if it was moving with the some speed inertially can it continue to move with the same speed just in a circular way after the inertial part of the motion?

6. Apr 8, 2014

Staff: Mentor

No, they travel in a circle with a different center but the same size. Unless there is a torque in addition to the centripetal force they will not travel in different sized circles.

The earth is not an initially inertial object. It is rotating to begin with so some points are already travelling at different speeds in an inertial frame.

I thought your interest was in a rigid object which was initially moving inertially.

No.

Yes, assuming that the net force is centripetal and there is no net torque.

7. Apr 9, 2014

analyst5

My basic interest was the situation like that in the twin paradox where the moving twin undergoes the u-turn and a sort of circular motion, I really wanted to know if all of his points have the same speed while doing the turn. Also, if Earth was not rotating, would its motion around the sun look like the first scenario you mentioned, without torque and with all of its points circulating at the same speed or otherwise, like all of its points having different speeds?

8. Apr 9, 2014

Staff: Mentor

I don't see how this is relevant to the twins scenario. The twins are generally considered point particles undergoing 1 dimensional movement.

Yes.

9. Apr 9, 2014

Staff: Mentor

There doesn't have to be any rotation or circular movement in this case: imagine a car braking to a stop, shifting into reverse, and backing up. If that simplifies the twin paradox for you, go for it, but you could still ask what happens during the linear acceleration.

Each individual molecule in his body moves according to Newton's $F=ma$, where the $F$ is the sum of the inter-molecular forces and whatever part of the external force is acting on that molecule. Remember, the external force doesn't act on the body as a whole; when I lift an object by a handle I'm applying an external force to the handle and the handle is applying force to rest of the object.

If the inter-molecular forces are strong enough to hold the body together, then all of the molecules will always have pretty much the same velocity relative to one another. Because they are traveling along slightly different trajectories, they will be experiencing different accelerations which just tells us that they're experiencing slightly different forces - the stress in different parts of the body is different.

If the inter-molecular forces are not strong enough to hold the body together, which is to say strong enough to accelerate every molecule enough to keep it reasonably close to its neighbor as the neighbor accelerates, the body tears apart.

10. Apr 9, 2014

Andrew Mason

The centripetal acceleration of a point on the body is ac = -ω2r where r is the radial vector (from the centre of curvature to the point on the rotating body). r is slightly different for each point in the body.

The mass as a whole experiences a force:

(1) $F_c = -m\omega^2\vec{r_{com}}$ where $\vec{r_{com}}$ is the radial vector to the centre of mass and m is the mass of the body.

An element of mass, mi, experiences an accleration aci = -ω2ri where ri is the radial vector to the element mi. ri = rcom + Ri where Ri is the vector from the centre of mass of the body to mi. So the force on mi is:

(2) $F_{ci} = -m_i\omega^2\vec{r_i} = -m_i\omega^2(\vec{r_{com}}+\vec{R_i})$

The total force acting on the body is:

$F_c = \sum F_{ci} = -\sum m_i\omega^2(\vec{r_{com}}+\vec{R_i}) = -(\omega^2(\vec{r_{com}}\sum m_i +\sum m_i\omega^2\vec{R_i}) = -m\omega^2\vec{r_{com}} - \sum m_i\omega^2\vec{R_i}$

It follows from (1) that:

(3)$\sum m_i\omega^2\vec{R_i}= 0$

This latter term consists of the forces within the body arising because each part of the body undergoes a slightly different centripetal acceleration.

This latter term (3) describes the sum of all the forces within the body due to the rotation of all the mi s about the centre of mass of the body. This makes sense because a rotating rigid body can be thought of as centripetal acceleration of each part of the body about its centre of mass which itself is undergoing centripetal acceleration toward the centre of curvature, both rotations having the same angular speed.

AM

11. May 7, 2014

analyst5

So if there was no rotation, would the Earth stay rigid during circular movement, since all of its points are moving with the same velocity around the Sun? I ask this from a relativistic perspective, since it seems to me that each point on the object undergoing circular motion will have a different simultaneity perspective.

12. May 7, 2014

A.T.

In the Classical Physics forum?

Relative simultaneity is very tricky when gravity is involved, because of gravitational time dilation. You cannot always synchronize distant clocks, even if they are at relative rest, because they might run at different rates.

13. May 7, 2014

analyst5

I know this is the classical physics forum, but nonetheless, I'm trying to connect the concepts of circular motion in classical physics and in relativity, that's why I'm asking the question about the velocities of the points of a body undergoing circular motion. And only considering Special relativity, for now.

14. May 7, 2014

Andrew Mason

Not all points in the earth would move with the same velocity.

If the earth did not rotate (ie. relative to an inertial frame such as the stars) only the centre of mass (we are assuming there is no moon or other third bodies) would prescribe a circle. The other parts would wobble off from a circular path. So their speeds would vary as well. This means that there would be varying tidal forces between parts of the earth as the earth moved around the sun.

If the earth's angular speed of rotation about its axis was the same as its angular speed of rotation about the sun (like the moon always facing earth as it goes around the earth), all points in/on the earth would maintain the same distance from the centre of rotation at all times. The speed of a given part would be constant ωr, where r is the distance from that part to the centre of rotation about the sun (assuming no effects from third bodies including the moon).

That is true, but given that the speeds are not relativistic the effect would be extremely small.

AM

15. May 11, 2014

analyst5

@Andrew Mason: I think Dale said that all points would move with the same velocity, and what does actually happen then, does Earth stay rigid while revolving around the Sun?

16. May 11, 2014

Staff: Mentor

For a rigid object which is not rotating, by definition all of the points are moving with the same velocity.

Andrew Mason must have been talking about a rotating or a non-rigid earth.

17. May 11, 2014

Andrew Mason

How is that? We are talking about circular motion of a rigid body about a centre of rotation at a constant angular speed. The rigid body has a finite size so points on it will be at different distances from the centre of rotation.

AM

18. May 11, 2014

A.T.

There is no center of rotation, if the body is not rotating. There is just circular motion, and each point of the body orbits a different center of circular motion.

All points have the same distance to their own center of circular motion.

Last edited: May 11, 2014
19. May 11, 2014

Staff: Mentor

In addition to A.T.'s correct explanation, consider the fact that rigid body motion has 6 degrees of freedom. If you constrain 3 of them by setting rotation = 0 on all three axes, then there are only 3 degrees of freedom remaining. This only allows you to specify a single velocity for all points on the body.

Regardless of the motion of the center of mass, be it linear, circular, helical, or whatever, if there is no rotation then all points on a rigid object will have the same velocity at all times.

20. May 11, 2014

analyst5

So does the body stay rigid during circular motion? I mean if all of its points have the same velocity it seems that there are no velocities between those points, and that the proper length remains the same.

edit: I ask this from a relativistic viewpoint. It's clear that in linear acceleration the only way the body will stay rigid is if it is undergoing Born rigid acceleration. But what about circular motion, since while undergoing it each point on the body will have a different simultaneity perspective.