Radius of curvature of the trajectory of points A and B

[haruspex]

Yes.

It would be $OB=\sqrt{r^2+r^2}=r\sqrt{2}$ right?

 

[Davidllerenav]

Yes.

So that would be the radius of curvature of point B?

 

[haruspex]

So that would be the radius of curvature of point B?

Yes.

 

[Davidllerenav]

Yes.

Ok, thanks. How do I find the radius of curvature for A?

 

[TSny]

The distance between O and B is not equal to the radius of curvature of the trajectory of point B when B is at the position shown.

 

[Davidllerenav]

The distance between O and B is not equal to the radius of curvature of the trajectory of point B when B is at the position shown.

Why?

 

[TSny]

Why?

You have to be careful. People like to say that point $B$ is instantaneously rotating about point $O$ with angular speed $\omega$. That’s OK if you’re determining the velocity of $B$ at that one instant. But the radius of curvature is a “higher order” concept that involves the acceleration as well as the velocity through $R= v^2 /a_n$.

You cannot determine the acceleration of point $B$ by treating $B$ as if it were in pure rotation about point $O$. A point of the cylinder that is instantaneously in contact with the table has zero velocity but it has nonzero acceleration.

This is the main reason I suggested finding the velocity and acceleration of points $A$ and $B$ by using the relative velocity equations involving point $C$. For the velocities, however, you can use the idea of rotating about point $O$ instead, if you want.

 

[Davidllerenav]

You have to be careful. People like to say that point $B$ is instantaneously rotating about point $O$ with angular speed $\omega$. That’s OK if you’re determining the velocity of $B$ at that one instant. But the radius of curvature is a “higher order” concept that involves the acceleration as well as the velocity through $R= v^2 /a_n$.

You cannot determine the acceleration of point $B$ by treating $B$ as if it were in pure rotation about point $O$. A point of the cylinder that is instantaneously in contact with the table has zero velocity but it has nonzero acceleration.

This is the main reason I suggested finding the velocity and acceleration of points $A$ and $B$ by using the relative velocity equations involving point $C$. For the velocities, however, you can use the idea of rotating about point $O$ instead, if you want.

So the radius of curvature depends on the acceleration? It does't relate with the radius of the circle?

 

[TSny]

So the radius of curvature depends on the acceleration?

Yes, you gave the correct formula $R = v^2/a_n$ in your first post.
[EDIT: The radius of curvature of the trajectory of $A$ or $B$ does not depend on whether or not the cylinder has angular acceleration. The trajectory will be the same cycloid independently of the angular speed or acceleration of the cylinder. Thus, it is simplest to consider the cylinder as rolling with constant angular speed. But, of course, even with constant $\omega$, points $A$ and $B$ will have nonzero accelerations which will be used in $R = v^2/a_n$.]

It does't relate with the radius of the circle?

The radius of curvature is the radius of the circle that "best fits" the trajectory curve. However, in the case of point $B$, this best-fit circle is not a circle that has radius $OB$. Likewise, for point $A$, the best-fit circle to the cycloid trajectory is not a circle of radius $OA$. You will see this when you work it out.

 

[Davidllerenav]

Yes, you gave the correct formula $R = v^2/a_n$ in your first post.

The radius of curvature is the radius of the circle that "best fits" the trajectory curve. However, in the case of point $B$, this best-fit circle is not a circle that has radius $OB$. Likewise, for point $A$, the best-fit circle to the cycloid trajectory is not a circle of radius $OA$. You will see this when you work it out.

Oh, I think I see it, If I draw the trayectory of the points, the radius is bigger than the radius of the circle, right?

 

[TSny]

Oh, I think I see it, If I draw the trayectory of the points, the radius is bigger than the radius of the circle, right?

Yes, the radius of curvature of the trajectory at $A$ is bigger than $OA$. Similarly, for $B$.

 

[haruspex]

Yes, the radius of curvature of the trajectory at $A$ is bigger than $OA$. Similarly, for $B$.

Damn, I worried about that (post #10) but convinced myself it was ok.
Thanks for jumping in.

 

[Davidllerenav]

Damn, I worried about that (post #10) but convinced myself it was ok.
Thanks for jumping in.

Ok, so I'll have to do it with relative velocity as @TSny suggested.I need to find the expression of velocity of A, the use that on the expression of acceleration and then find the radius, right? The same for B.

 

[TSny]

Damn, I worried about that (post #10) but convinced myself it was ok.
Thanks for jumping in.

Yes, it's tricky. I've gotten bit by this stuff before.

 

[Davidllerenav]

If $\vec v_C$ is the velocity of C relative to the table, then you can find $\vec v_B$ by using the relative velocity formula $\vec v_B = \vec v_C + \vec v_{B/C}$.

Likewise, you can find $\vec a_B$ from the relative acceleration formula $\vec a_B = \vec a_C + \vec a_{B/C}$.

You can get $\vec v_A$ by using the relative velocity fomula $\vec v_A = \vec v_C + \vec v_{A/C}$

Why $\vec v_A = \vec v_C + \vec v_{A/C}$ and $\vec a_B = \vec a_C + \vec a_{B/C}$? I tried to understand that but I don't get it.

 

[haruspex]

Why $\vec v_A = \vec v_C + \vec v_{A/C}$ and $\vec a_B = \vec a_C + \vec a_{B/C}$? I tried to understand that but I don't get it.

Just standard equations of relative motion. If C moves a distance x_C in some direction and B moves a distance x_B in the same direction then B's motion relative to C is x_B/C=x_B-x_C. Rearranging and differentiating yields the velocity and acceleration equations.

 

[Henryk]

I'm really disappointed with this discussion. So many voices, none contributing anything useful. This forum should be by physicists who know math.
What we have is a curvature of a cycloid, a purely mathematical problem. Here is how to deal with it.
To begin, pick a frame of reference. The geometry is purely 2-dimensional and we need only x and y coordinates.
I choose the origin at the centre of the cylinder and the x-direction along the direction of the movement of the cylinder.
Let me denote the position of the axis of the cylinder as u and the radius of the cylinder as R. Since the cylinder rotates, the angle of the cylinder (measured from its starting position) will be u/R.
Therefore, the coordinates of the point A (top of the cylinder) will be (u+R sin(u/R), R cos(u/R)). The coordinates of the point B will be (u +R cos(u/R), -R sin(u/R)).
Now, we have x, and y coordinates of the points, It is time to apply the curvature formula as given in my previous post (see the link https://en.wikipedia.org/wiki/Curvature#Precise_definition)
For the point A, the X-coordinate is u + R sin(u/R), therefore, the derivative (with respect to u) is x' = 1 + cos(u/R) and the second derivative x" = - sin(u/R)/R.
Similarity, the Y-coordinate y = R cos(u/R) can be differentiated to give y' = - sin(u/R) and y" = - cos(u/R)/R.
Now, all we have to do is to substitute the above expression into the formula for the curvature. Denoting the radius of the curvature as r (lower case) we have
$\frac 1 r = \kappa = \frac {[ x'y"-y'x"]}{(x' ^2 + y'^2)^{3/2}}$.
At this point I could substitute the expression for the first and second derivatives into the above formula, then see the value of u to 0 and evaluate. This would be too complex, I will set u = 0, evaluate the derivatives and substitute the values into the expression for the curvature.
Here we go, if I set u = u, x' = 2, x" =0, y' =0, y" = -1/R, therefore, the curvature is
$\frac 1 r = \kappa = \frac {[ 2(-1/R) - 0*0]}{(2^2 + 0^2)^{3/2}} = \frac 1 {4R}$
Therefore, the radius of curvature of the point A is 4 times the radius of the cylinder !.
Similar calculations can be done for the point B.

 

[Davidllerenav]

Just standard equations of relative motion. If C moves a distance x_C in some direction and B moves a distance x_B in the same direction then B's motion relative to C is x_B/C=x_B-x_C. Rearranging and differentiating yields the velocity and acceleration equations.

But isn't B rotating arround C? A does the same.

 

[Davidllerenav]

I'm really disappointed with this discussion. So many voices, none contributing anything useful. This forum should be by physicists who know math.
What we have is a curvature of a cycloid, a purely mathematical problem. Here is how to deal with it.
To begin, pick a frame of reference. The geometry is purely 2-dimensional and we need only x and y coordinates.
I choose the origin at the centre of the cylinder and the x-direction along the direction of the movement of the cylinder.
Let me denote the position of the axis of the cylinder as u and the radius of the cylinder as R. Since the cylinder rotates, the angle of the cylinder (measured from its starting position) will be u/R.
Therefore, the coordinates of the point A (top of the cylinder) will be (u+R sin(u/R), R cos(u/R)). The coordinates of the point B will be (u +R cos(u/R), -R sin(u/R)).
Now, we have x, and y coordinates of the points, It is time to apply the curvature formula as given in my previous post (see the link https://en.wikipedia.org/wiki/Curvature#Precise_definition)
For the point A, the X-coordinate is u + R sin(u/R), therefore, the derivative (with respect to u) is x' = 1 + cos(u/R) and the second derivative x" = - sin(u/R)/R.
Similarity, the Y-coordinate y = R cos(u/R) can be differentiated to give y' = - sin(u/R) and y" = - cos(u/R)/R.
Now, all we have to do is to substitute the above expression into the formula for the curvature. Denoting the radius of the curvature as r (lower case) we have
$\frac 1 r = \kappa = \frac {[ x'y"-y'x"]}{(x' ^2 + y'^2)^{3/2}}$.
At this point I could substitute the expression for the first and second derivatives into the above formula, then see the value of u to 0 and evaluate. This would be too complex, I will set u = 0, evaluate the derivatives and substitute the values into the expression for the curvature.
Here we go, if I set u = u, x' = 2, x" =0, y' =0, y" = -1/R, therefore, the curvature is
$\frac 1 r = \kappa = \frac {[ 2(-1/R) - 0*0]}{(2^2 + 0^2)^{3/2}} = \frac 1 {4R}$
Therefore, the radius of curvature of the point A is 4 times the radius of the cylinder !.
Similar calculations can be done for the point B.

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. Even though, I find this way interesnting. How did you manage to define the coordinate poins of A and B like that?

 

[haruspex]

none contributing anything useful.

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.

 

[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 $\vec v_{A/O}=\vec v_{A/C}+\vec v_{C/O}=\vec v_{A/C}-\vec v_{O/C}$