Radius of curvature of the trajectory of points A and B

[Davidllerenav]

It seems to me that @TSny's acceleration/velocity/radius approach gets there and by a rather simpler route.
A's motion can be viewed as a linear velocity plus a rotation about C so its acceleration is rω²=v²/r. This is entirely centripetal.
Its velocity in the ground frame is 2v.
Viewed as a rotation about the centre of curvature, radius R, the centripetal acceleration is (2v)²/R.
Equating the accelerations, R=4r.

For point B one has to be a little careful, being sure to equate only the centripetal accelerations.

Why the velocity of A in the ground frame is 2v? I don't get that part

 

[haruspex]

Why the velocity of A in the ground frame is 2v? I don't get that part

The centre of the disc has velocity v, the bottom has velocity 0, so the top has velocity 2v.

 

[Davidllerenav]

The centre of the disc has velocity v, the bottom has velocity 0, so the top has velocity 2v.

Is it related to the radius then? How does it relate to post #3?

 

[haruspex]

Is it related to the radius then? How does it relate to post #3?

Not sure what part of post #3 you have in mind.
A useful way to think of rolling motion is as the sum of the linear motion of the disc as a whole and a rotation about its centre, with the relationship v=ωr (or -ωr, depending on your sign convention).
For a point on the disc at radius a, that gives the horizontal velocity v plus a tangential velocity aω at some angle.
For point O, the two counteract, leading to a net velocity of v-ωr=0. For point A, they add, leading to v+ωr=2v. For point B they are perpendicular.

 

[Davidllerenav]

Not sure what part of post #3 you have in mind.
A useful way to think of rolling motion is as the sum of the linear motion of the disc as a whole and a rotation about its centre, with the relationship v=ωr (or -ωr, depending on your sign convention).
For a point on the disc at radius a, that gives the horizontal velocity v plus a tangential velocity aω at some angle.
For point O, the two counteract, leading to a net velocity of v-ωr=0. For point A, they add, leading to v+ωr=2v. For point B they are perpendicular.

Oh, I see. It has horizontal velocity because the whole cilinder is moving foward, rihgt?
I was talking about $\vec v_A = \vec v_C + \vec v_{A/C}$, that was pointed out on post #3.

 

[haruspex]

Oh, I see. It has horizontal velocity because the whole cilinder is moving foward, rihgt?
I was talking about $\vec v_A = \vec v_C + \vec v_{A/C}$, that was pointed out on post #3.

Yes, each point can be thought of as having the v of the full cylinder plus a tangential component.
That vector equation is the general statement of relative velocities. I explained that in post #46. Not sure I can make it any clearer.

 

[Davidllerenav]

That vector equation is the general statement of relative velocities. I explained that in post #46. Not sure I can make it any clearer.

Is it the same that you just said? Because $\vec v_c$ would be the horizontal velocity or the velocity of C with respect to point O, a tangential velocity. $v_{A/C}$ would be the velocity of A with respec to C, the tangential velocity of A, so the velocity of A with respect to O $v_A$ woudl be the sum of both. Is that what this is saying?

 

[haruspex]

Oh, I see. It has horizontal velocity because the whole cilinder is moving foward, rihgt?
I was talking about $\vec v_A = \vec v_C + \vec v_{A/C}$, that was pointed out on post #3.

I think I misinterpreted your question here.
You are asking how that equation relates to the way I decomposed the velocity at A.
For any point S at vector $\vec a$ displacement from C, $\vec v_{S/C}=\vec a\times\vec \omega$, so $\vec v_S=\vec v+\vec a\times\vec \omega$.

 

[Davidllerenav]

I think I misinterpreted your question here.
You are asking how that equation relates to the way I decomposed the velocity at A.
For any point S at vector $\vec a$ displacement from C, $\vec v_{S/C}=\vec a\times\vec \omega$, so $\vec v_S=\vec v+\vec a\times\vec \omega$.

So, if I understand it correctly, it is the same I said on post #57?

 

[haruspex]

Is it the same that you just said? Because $\vec v_c$ would be the horizontal velocity or the velocity of C with respect to point O, a tangential velocity. $v_{A/C}$ would be the velocity of A with respec to C, the tangential velocity of A, so the velocity of A with respect to O $v_A$ woudl be the sum of both. Is that what this is saying?

Not quite.
$\vec v_C$ is the horizontal velocity of C in the lab frame. Its velocity relative to O would be written $\vec v_{C/O}$, but since O is stationary (instantaneously) that equals $\vec v_C$.
The velocity of A wrt O would be $\vec v_{A/O}=\vec v_{A/C}+\vec v_{C/O}=\vec v_{A/C}-\vec v_{O/C}$

 

[Davidllerenav]

Not quite.
$\vec v_C$ is the horizontal velocity of C in the lab frame. Its velocity relative to O would be written $\vec v_{C/O}$, but since O is stationary (instantaneously) that equals $\vec v_C$.
The velocity of A wrt O would be $\vec v_{A/O}=\vec v_{A/C}+\vec v_{C/O}=\vec v_{A/C}-\vec v_{O/C}$

Wouldn't $\vec v_{A/O}=\vec v_{A/C}+\vec v_{C/O}=\vec v_{A/C}-\vec v_{O/C}=0$? Since \$vec v_{A/C}=\vec v_{O/C}$. Can't we write $\vec v_{A/O}=\vec v_{A/C}+\vec v_{C/O}=\vec v_{A/C}+v_C$?

 

[TSny]

$\vec v_A = \vec v_C + \vec v_{A/C}$ where $\vec v_A$ and $\vec v_C$ are the velocities of $A$ and $C$ with respect to the "lab frame" (i.e., with respect to the fixed surface on which the cylinder is rolling). The two velocities on the right-hand side of the equation are illustrated below

The left figure shows the velocity of $C$ relative to the lab.

The figure on the right shows the velocity of $A$ relative to $C$. In this figure, $C$ is at rest while the cylinder rotates around $C$ with angular speed $\omega$.

You should be able to express both $v_C$ and $v_{A/C}$ in terms of $\omega$ and the radius of the cylinder, $r$.

 

[Davidllerenav]

$\vec v_A = \vec v_C + \vec v_{A/C}$ where $\vec v_A$ and $\vec v_C$ are the velocities of $A$ and $C$ with respect to the "lab frame" (i.e., with respect to the fixed surface on which the cylinder is rolling). The two velocities on the right-hand side of the equation are illustrated below

View attachment 239530

The left figure shows the velocity of $C$ relative to the lab.

The figure on the right shows the velocity of $A$ relative to $C$. In this figure, $C$ is at rest while the cylinder rotates around $C$ with angular speed $\omega$.

You should be able to express both $v_C$ and $v_{A/C}$ in terms of $\omega$ and the radius of the cylinder, $r$.

So I would end up with $\vec v_A=2v_C=2\omega r$? Is it correct the way I understand why $\vec v_A = \vec v_C + \vec v_{A/C}$?

 

[haruspex]

Wouldn't $\vec v_{A/O}=\vec v_{A/C}+\vec v_{C/O}=\vec v_{A/C}-\vec v_{O/C}=0$?

No. $\vec v_{A/C}=-\vec v_{O/C}$, so $\vec v_{A/C}-\vec v_{O/C}=2\vec v_{A/C}=2\vec v_{C/O}$

 

[Davidllerenav]

No. $\vec v_{A/C}=-\vec v_{O/C}$, so $\vec v_{A/C}-\vec v_{O/C}=2\vec v_{A/C}=2\vec v_{C/O}$

And since $2\vec v_{C/O}=2 \vec c = 2\omega r$, am I right?

 

[haruspex]

And since $2\vec v_{C/O}=2 \vec c = 2\omega r$, am I right?

Yes.

 

[Davidllerenav]

Yes.

Ok, thanks. To sum up, The point A will have two velocitie, $v_{A/C}$ which is the tangential velocity and $v{_A/O}$ which is the velocity with respect to the ground (correct me if I'm wrong, but I think that I can say that $v_{A/O}$ is the linear velocity, right? So it will have to be the same as the velocity $v_C$ that is the velocity of the center of mass). Then, I need to sum both to find the total velocity of A when the cylinder is rolling, right?

 

[haruspex]

Ok, thanks. To sum up, The point A will have two velocitie, $v_{A/C}$ which is the tangential velocity and $v{_A/O}$ which is the velocity with respect to the ground (correct me if I'm wrong, but I think that I can say that $v_{A/O}$ is the linear velocity, right? So it will have to be the same as the velocity $v_C$ that is the velocity of the center of mass). Then, I need to sum both to find the total velocity of A when the cylinder is rolling, right?

No, still not right.
You can think of the solidus, /, as a minus sign here. $\vec v_{A/C}=\vec v_{A}-\vec v_{C}$.
So you can see that adding $\vec v_{A/C}$ and $\vec v_{A/O}$ would not make much sense.

$\vec v_{A}=\vec v_{A/O}+\vec v_{O}$, but $\vec v_{O}=0$.

 

[Davidllerenav]

No, still not right.
You can think of the solidus, /, as a minus sign here. $\vec v_{A/C}=\vec v_{A}-\vec v_{C}$.
So you can see that adding $\vec v_{A/C}$ and $\vec v_{A/O}$ would not make much sense.

$\vec v_{A}=\vec v_{A/O}+\vec v_{O}$, but $\vec v_{O}=0$.

So, as I understand it is the mix between circular motion and linear motion, right? So with linear motion every point in the cyinder would have the same velocity, and with the circular motion, every point would have its tangential velocity and angular velocity. So both must sum.

 

[haruspex]

So, as I understand it is the mix between circular motion and linear motion, right? So with linear motion every point in the cyinder would have the same velocity, and with the circular motion, every point would have its tangential velocity and angular velocity. So both must sum.

That’s a bit vague/unclear. E.g. I don't know what "both must sum" means.
There are many ways of decomposing motion into sums. For a rotating body it is often convenient to decompose the motion of a point P within it as the sum of the linear velocity of the mass centre plus a tangential velocity, tangential in the sense of a rotation about the mass centre. In the present context that's $\vec v_P=\vec v_C+\vec v_{P/C}$.

 

[Davidllerenav]

That’s a bit vague/unclear. E.g. I don't know what "both must sum" means.
There are many ways of decomposing motion into sums. For a rotating body it is often convenient to decompose the motion of a point P within it as the sum of the linear velocity of the mass centre plus a tangential velocity, tangential in the sense of a rotation about the mass centre. In the present context that's $\vec v_P=\vec v_C+\vec v_{P/C}$.

And that's exactly how we defined $v_A$ right?

 

[haruspex]

And that's exactly how we defined $v_A$ right?

We did not define $\vec v_A$ that way, we decomposed it that way.

 

[Davidllerenav]

We did not define $\vec v_A$ that way, we decomposed it that way.

Oh, ok. And why both $v_C$ and $v_{A/C}$ are equal, i.e. they ar both ##\omega r?

 

[Vivek098]

Please see the below image for solution...

 

[Davidllerenav]

Please see the below image for solution...

I understand the solution. What I want to know is why $\vec v_A = \vec v_C + \vec v_{A/C}=2\vec v_C=2\omega R$.

 

[Vivek098]

Point A is rotating about centre of mass with angular velocity w and also translating with velocity v. So net velocity of top point is v + wR and since v=wR. So Vnet = 2v = 2wR

 

[Davidllerenav]

Point A is rotating about centre of mass with angular velocity w and also translating with velocity v. So net velocity of top point is v + wR and since v=wR. So Vnet = 2v = 2wR

Ok. Thanks, one last question. Why is the translatong velocity $v$ also $\omega R$?

 

[Vivek098]

Lowest point on body is not slipping with respect to the point in contact on ground means there relative velocity is zero. Velocity of lowest point on body is v-wR and that on ground is zero. So, v-wR=0

 

[haruspex]

Ok. Thanks, one last question. Why is the translatong velocity $v$ also $\omega R$?

What about the radius of curvature for point B? Can you solve that now?

 

[Davidllerenav]

What about the radius of curvature for point B? Can you solve that now?

Yes, I think I can. I just need to use proyections, right? It would be almost the same process of A.

 

[haruspex]

Yes, I think I can. I just need to use proyections, right? It would be almost the same process of A.

Nearly, but as I wrote, you must be careful to use only the centripetal component of the acceleration.

 

[Henryk]

I just saw something like this on calculus. I think tha my physics teacher would like a more physical approach, using the definitions and formulas of velocity, angular velocity, etc

Well, it isn't physics problem, it's mathematical one. You are asking for a curvature and that's a curvature of a trajectory (of the two points). A trajectory is a curve in space, it has nothing to do with velocity or acceleration.
Second point: in science we don't do what we like, we do what we can prove and what works.

How did you manage to define the coordinate poins of A and B like that?

Simple, It is a combination of linear motion and rotation. Since there is no slippage, the angle of rotation is (in radians) distance traveled / radius. The rest is an elementary geometry.