Reference frames, center of rotation, etc

[anuttarasammyak]

@gen x
Let us see the case of moving-spinning disk as sketched below.

Orbit of O is a line. The disk is rotating around O with $\Omega$ along the orbit.
Orbit of A is a helix. The disk is rotating around A with the same $\Omega$ along the orbit.

 

[Dale]

If we choose center as B, B is now fixed and everything rotate around B, so how CoM still move in stright line in this anylize?

No, you specified

Frame fixed to the table:

So there is no change to the motion. We are only changing how we decompose the motion into translation and rotation.

The motion is a vector field, specifically the field of the velocity of each tiny piece of the object.

You can decompose a vector into x and y components. Similarly you can decompose a vector field for rigid motion into translation and rotation components.

Decomposing a vector into x and y components does not change the vector, it is just a different representation. Decomposing a rigid motion vector field into rotation and translation does not change the vector field, it is just a different representation.

There is not just one pair of x and y axes, you can choose your axes in a different direction, each direction producing a different pair of x and y components. There is not just one center of rotation, you can choose your center in a different location, each center location producing a different pair of rotation and translation fields.

 

[weirdoguy]

This whole thread rotates around the point that @gen x can't see a mathematical fact about kinematics without using math. I think that to adopt a frame in which this point moves 'forward', @gen x needs to learn the math

 

[Dale]

I agree. I think we have explained it every way possible. @gen x just needs to actually sit down and do the math.

 

[gen x]

I agree. I think we have explained it every way possible. @gen x just needs to actually sit down and do the math.

So answer from John Alexiou is not correct?

https://physics.stackexchange.com/q...he-bar-rotate-about-the-center-of-mass-or-not

From ground frame velocity at bottom part of car wheel is zero, that mean centripetal force is zero..
Here must exist centripetal force, why ground frame here show wrong result?

For top part of wheel I use peripheral veliocity or peripheral veliocity+ translation velocity to calculate centripetal force?
I think translation velocity dont have any effect on centripetal force so I use only peripheral veliocity

 

[jbriggs444]

So answer from John Alexiou is not correct?

https://physics.stackexchange.com/q...he-bar-rotate-about-the-center-of-mass-or-not

From ground frame velocity at bottom part of car wheel is zero, that mean centripetal force is zero..

You are invoking the formula for centripetal acceleration, $a = \frac{v^2}{r}$, I suppose. That formula does not apply to rotation about a continuously changing point such as the succession of instantaneous centers of rotation.

[I am guessing that the "succession of instantaneous centers of rotation" is the straw man that you are attacking. That is not the only alternative notion of rotation that someone might consider. But it is the one that I will consider in the next paragraph].

If you know the rotation rate and the [motionless by definition] instantaneous center of rotation then you can compute the instantaneous velocities of all the pieces comprising the rigid object in question. But simply knowing those two pieces of information (instantaneous center of rotation and rotation rate) is not sufficient to determine the accelerations of all of those pieces. The trajectory of the sequence of instantaneous centers of rotation would enter into such a calculation.

 

[gen x]

You are invoking the formula for centripetal acceleration, $a = \frac{v^2}{r}$, I suppose. That formula does not apply to rotation about a continuously changing point such as the succession of instantaneous centers of rotation.

[I am guessing that the "succession of instantaneous centers of rotation" is the straw man that you are attacking. That is not the only alternative notion of rotation that someone might consider. But it is the one that I will consider in the next paragraph].

If you know the rotation rate and the [motionless by definition] instantaneous center of rotation then you can compute the instantaneous velocities of all the pieces comprising the rigid object in question. But simply knowing those two pieces of information (instantaneous center of rotation and rotation rate) is not sufficient to determine the accelerations of all of those pieces. The trajectory of the sequence of instantaneous centers of rotation would enter into such a calculation.

In answer are some parts that you(members in my thread) claim are wrong.

I will qoute them:

"To make the point of rotation the center of mass then there must be zero net force. Any rigid body with a pure torque applied (force couple) is going to rotate, and since the center of mass is not going to translate (zero net force), the center of rotation is the COM."

"Note: By definition the center of rotation is the point on the body (or the extended frame) that does not translate."

 

[jbriggs444]

In answer are some parts that you(members in my thread) claim are wrong.

That is not a substantive response. Since you are not interested in a genuine discussion, I am out.

I will qoute them:

"To make the point of rotation the center of mass then there must be zero net force. Any rigid body with a pure torque applied (force couple) is going to rotate, and since the center of mass is not going to translate (zero net force), the center of rotation is the COM."

"Note: By definition the center of rotation is the point on the body (or the extended frame) that does not translate."

Unattributed straw men devoid of context. Not substantive.

 

[A.T.]

From ground frame velocity at bottom part of car wheel is zero, that mean centripetal force is zero..
Here must exist centripetal force, why ground frame here show wrong result?

You haven't posted the math, that gave you the "wrong result" Are we supposed to guess what you got wrong this time around?

In the ground frame, a point on the wheel circumference is not moving on a circle, so how exactly do you define and compute the centripetal force in the ground frame?

 

[gen x]

You haven't posted the math, that gave you the "wrong result" Are we supposed to guess what you got wrong this time around?

math: Cf=m x 0^2 /r = 0

In the ground frame, a point on the wheel circumference is not moving on a circle, so how exactly do you define and compute the centripetal force in the ground frame?

Yes that is good question. point t1 has centripetal force to the right toward center for frame fixed to center of wheel, but for ground frame centripetal force has different direction.

 

[A.T.]

math: Cf=m x 0^2 /r = 0

What is "r" supposed to be in the ground frame, where you don't have a circular path?

for ground frame centripetal force has different direction.

Again, how do you "define centripetal" force in the ground frame? And why do you expect it to have the same value as in the wheel center frame?

Only the total acceleration of the point must be the same in both inertial frames. But in the ground frame you also have tangential acceleration, because the point is not moving at constant speed. At the lowest point of the wheel, there is only tangential acceleration in the ground frame, vertically upwards, exactly equal to the centripetal acceleration in the wheel center frame.

 

[Dale]

I think that you need to work through the math for yourself. Why don’t you do it for the example you posted.

My assumption is that the desk is frictionless.

Frame fixed to the table:

After the strike, the center of mass (CoM) of the ruler will move in a straight line because the net force is zero.

Let’s simplify that the ruler is a rigid line of length $L$ and mass $m$. Use the variable $-L/2 \le l \le L/2$ to identify points on the ruler, with $l=0$ being the CoM.

Now, can you write a function $v_l(t)$ that describes the velocity of the point $l$ in the case that the CoM is stationary and the ruler is rotating with angular velocity $\omega$?

Can you write a function $v_l(t)$ that describes the velocity of the point $l$ in the case that the CoM is moving with velocity $v$ and the ruler is not rotating?

 

[gen x]

What is "r" supposed to be in the ground frame, where you don't have a circular path?

I thought maybe " r" of local curvature in specific point.

And why do you expect it to have the same value as in the wheel center frame?

Doesn’t the centripetal force have to be the same in both reference frames?

 

[gen x]

Now, can you write a function $v_l(t)$ that describes the velocity of the point $l$ in the case that the CoM is stationary and the ruler is rotating with angular velocity $\omega$?

vl(t) =ω⋅l

Can you write a function $v_l(t)$ that describes the velocity of the point $l$ in the case that the CoM is moving with velocity $v$ and the ruler is not rotating?

vl(t)=v

 

[A.T.]

I thought maybe " r" of local curvature in specific point.

OK, that's just the normal force. But as already said:

... in the ground frame you also have tangential acceleration, because the point is not moving at constant speed.

Doesn’t the centripetal force have to be the same in both reference frames?

No, as already said:

Only the total acceleration of the point must be the same in both inertial frames.

 

[gen x]

No, as already said:

Why is force frame dependent?
If acceleration is the same, and m is the same then why F is not equal? ..F=ma

 

[A.T.]

F=ma

That F is the total force, not just the normal component (perpendicular to velocity).

In the wheel center frame you only have a normal (centripetal) force for a wheel part on the circumference, because it's speed is constant and only direction changes, so tangential force is zero

In the ground frame both are changing: direction and speed. So you have both: normal and tangential force, which you have to add vectorialy to get the total force, which is the same in both frames.

 

[Dale]

vl(t) =ω⋅l


vl(t)=v

Excellent, except that you will want those as vectors, not just the magnitudes. So for the rotating $$v_l(t)=(-\omega l \sin(\omega t),\omega l \cos(\omega t))$$and for the translating $$v_l(t)=(0,v)$$

Then, the next two steps are as follows.

First, write the velocity $v_l(t)$ for a ruler that is both rotating and translating.

Second is the difficult step. You need to calculate the position $s_l(t)=(x_l(t),y_l(t))$

 

[gen x]

First, write the velocity $v_l(t)$ for a ruler that is both rotating and translating.

vl(t)=vtranslation+vrotation
vtranslation=(v,0)
vrotation=(−ωlsin(ωt),ωlcos(ωt))
total velocity of the point at position l:
vl(t)=(v−ωlsin(ωt),ωlcos(ωt))

Second is the difficult step. You need to calculate the position $s_l(t)=(x_l(t),y_l(t))$

stranslation=(vt,0)
srotation=(lcos(ωt),lsin(ωt))
total position of the point l at time t:
sl(t)=(vt+lcos(ωt),lsin(ωt)

What is physical meaning if we say Earth rotate around poin x, instead around CoM, isn't only point CoM physically correct and mathematically every point is correct?

From witch frame Earth rotate/spins around CoM?

 

[A.T.]

isn't only point CoM physically correct

No, as already stated many times:

- For mere kinematics, any "center of revolution" is as good as another.
- For dynamics, there are choices more convenient than others, but we still have the choice.

How many times will you ask the same question?

 

[gen x]

No, as already stated many times:

How can rotation about points A, B and C be the same, acelometer placed on Earth will read different values? Rotation for points B and C must have net force= non zero.

 

[A.T.]

How can rotation about points A, B and C be the same,

The same in what sense? As already said:

- For dynamics, there are choices more convenient than others, but we still have the choice.

acelometer placed on Earth will read different values? Rotation for points B and C must have net force= non zero.

Post a clear description and your math, that you think leads to disagreement with what an accelometer measures?

 

[Dale]

vl(t)=vtranslation+vrotation
vtranslation=(v,0)
vrotation=(−ωlsin(ωt),ωlcos(ωt))
total velocity of the point at position l:
vl(t)=(v−ωlsin(ωt),ωlcos(ωt))



stranslation=(vt,0)
srotation=(lcos(ωt),lsin(ωt))
total position of the point l at time t:
sl(t)=(vt+lcos(ωt),lsin(ωt)

Good! For simplicity and convenience, let’s temporarily choose $t=0$. At $t=0$ we can use the total $s$ to see that $y=0$. So now we can eliminate $l$ and express the velocity as a function of $x$ only. Do that for the translational, rotational, and total velocity.

 

[gen x]

Good! For simplicity and convenience, let’s temporarily choose $t=0$. At $t=0$ we can use the total $s$ to see that $y=0$. So now we can eliminate $l$ and express the velocity as a function of $x$ only. Do that for the translational, rotational, and total velocity.

sl(t)=(vt+lcos(ωt),lsin(ωt)) for t=0, cos0=1,sin0=0

sl(0)=(l,0) .... x=l

Translational velocity as a function of x:
vtrans=(v,0)
vtrans(x)=(v,0)

Rotational velocity as a function of x:
vlrot(t)=(−ωlsin(ωt),ωlcos(ωt))
t=0........vlrot(0)=(0,ωl)
l=x.... vrot(x)=(0,ωx)



Total velocity as a function of x:
v(x)=(v,0)+(0,ωx)

v(x)=(v,ωx)

Post a clear description and your math, that you think leads to disagreement with what an accelometer measures?

Centripetal force is Fc=mω2r
I see right away that r is different for cases, A(r), B(x+r), C(y)
Isnt it?

 

[A.T.]

Centripetal force is Fc=mω2r

Don't post a random formula. Post the full math, that shows disagreement with what an accelometer measures.

 

[gen x]

Don't post a random formula. Post the full math, that shows disagreement with what an accelometer measures.

ω=7.292×10−5 rad/s
R=6.378×106 m

earth rotate around point A(CoM):
ameasured=g−ac

ac=ω^2R............. ac=(7.292×10−5)2×6.378×106≈0.034 m/s^2

ameasured=9.81−0.034≈9.78 m/s2

Earth rotate around point B, where x=50 000km:
r=5.0×10^7+6.378×10^6=5.6378×10^7 m

ac=ω^2r.............ac=300m/s^2
ameasured=9.81−300≈−290 m/s^2

centrifugal acc, higher than gravity...

etc.....

 

[A.T.]

Earth rotate around point B, where x=50 000km:
r=5.0×10^7+6.378×10^6=5.6378×10^7 m

ac=ω^2r.............ac=300m/s^2

This acceleration is relative to an non-inertial point B, as opposed to the inertial point A.

ameasured=9.81−300≈−290 m/s^2

No. When you use a non-inertial center of rotation, then you cannot equate relative (coordiante) acceleration with what an accelerometer measures (proper acceleration). You have to account for the proper acceleration of the reference point, which makes the math more complicated. As already said:

- For dynamics, there are choices more convenient than others, but we still have the choice.

 

[Dale]

sl(t)=(vt+lcos(ωt),lsin(ωt)) for t=0, cos0=1,sin0=0

sl(0)=(l,0) .... x=l

Translational velocity as a function of x:
vtrans=(v,0)
vtrans(x)=(v,0)

Rotational velocity as a function of x:
vlrot(t)=(−ωlsin(ωt),ωlcos(ωt))
t=0........vlrot(0)=(0,ωl)
l=x.... vrot(x)=(0,ωx)



Total velocity as a function of x:
v(x)=(v,0)+(0,ωx)

v(x)=(v,ωx)

Excellent. Now, notice that $v(x)=v_{rot}(x)+v_{trans}(x)$. So this decomposition into rotational and translational components does not change $v$, it just expresses it differently.

Now, looking at the above, how do we determine the center of rotation? Is it the $x$ where $v(x)=0$ or where $v_{rot}(x)=0$?

 

[wrobel]

The composition of a translation and a rotation is written as follows
$$\boldsymbol r'=\boldsymbol c+A\boldsymbol r,$$
where $A$ is an operator (matrix) of rotation; $A\ne I.$
Pure rotation about a fixed point with position vector $\boldsymbol u$ is
$$\boldsymbol r'= B(\boldsymbol r-\boldsymbol u)+\boldsymbol u.$$
We must find a rotation operator $B$ and a center of rotation $\boldsymbol u$ such that
$$B(\boldsymbol r-\boldsymbol u)+\boldsymbol u=\boldsymbol c+A\boldsymbol r.$$
It is easy to see that $B=A$ and
$$\boldsymbol u=(I-B)^{-1}\boldsymbol c$$