Radius of curvature of the trajectory of points A and B

[Davidllerenav]

Homework Statement

A cylinder rolls without slippage on a horizontal plane. The radius of the cylinder is equal to r. Find the radious of curvature of the trajectory of points A and B.

Homework Equations

• Ciruclar motion equations.
• $R=\frac{1}{C}$

The Attempt at a Solution

First I drew the velocity of A and B, as well as the angular acceleration of both like this:

After that I used the formula for the radious of cuvature $R=\frac{1}{C}=\frac{ds}{d\varphi}\Rightarrow ds=Rd\varphi \Rightarrow d\varphi=\frac{ds}{R}=\frac{vdt}{R}\Rightarrow \frac{d\varphi}{dt}=\frac{v}{R}=\omega$. Then, I used the angular acceleration formula $a_{n}=\frac{v^2}{R}=\omega^2 R$.

After that I attempted to find the radious of A:
$v_A=\omega R_A$
$a_{nA}=\frac{v_A^2}{R_A}\Rightarrow R_A=\frac{v_A^2}{a_{nA}}=\frac{\omega^2R_A^2}{a_{nA}}$
After that I don't know whta to do, is that it? Or do I have to do something else?

 

[Henryk]

I can give you a hint
You have to write the equation of a trajectory of a point of the cylinder. The trajectory is know in mathematics as cycloid.
Once you have an equation of the trajectory, use the formula for the curvature https://en.wikipedia.org/wiki/Curvature
Good luck

 

[TSny]

The Attempt at a Solution

First I drew the velocity of A and B

Your vector $\vec v_B$ in your drawing is not correct if it is meant to represent the velocity of B relative to the table. The direction you have indicated would be the direction of the velocity of B relative to the moving center C, $\vec v_{B/C}$.

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 $\vec a_B = \vec a_C + \vec a_{B/C}$.

After that I used the formula for the radious of cuvature $R=\frac{1}{C}=\frac{ds}{d\varphi}$

Here, $d\varphi$ represent the change in angle of the tangent line to the trajectory when moving along the trajectory by $ds$. $\frac{d \varphi}{dt}$ is not equal to the angular speed $\omega$ of the cylinder.

After that I attempted to find the radious of A:
$v_A=\omega R_A$

This relation between $v_A$ and the radius of curvature at A is not correct. You can get $\vec v_A$ by using the relative velocity fomula $\vec v_A = \vec v_C + \vec v_{A/C}$

$a_{nA}=\frac{v_A^2}{R_A}\Rightarrow R_A=\frac{v_A^2}{a_{nA}}$

This looks good. But, you cannot replace $v_A$ by $\omega R_A$.

In summary, you can find the radius of curvature at A by using $R_A=\frac{v_A^2}{a_{nA}}$. But you have to be careful in obtaining $v_A$ and $a_{nA}$.

With this approach, you do not need to work with the explicit formula for a cycloid.

Alternately, as @Henryk suggests, you can get the answer using the equation for a cycloid and the general formula for the radius of curvature of a curve.

 

[Davidllerenav]

I can give you a hint
You have to write the equation of a trajectory of a point of the cylinder. The trajectory is know in mathematics as cycloid.
Once you have an equation of the trajectory, use the formula for the curvature https://en.wikipedia.org/wiki/Curvature
Good luck

Thanks, I'll give it a try.

 

[Davidllerenav]

dφdtdφdt\frac{d \varphi}{dt} is not equal to the angular speed ωω\omega of the cylinder.

Why? Isn't it the change of the angle as time pases?

This expression for vAvAv_A is not correct.

That is not correct beacuse I need to set the coordinate system in the table, right? It would only be correct if ti is at the center.

 

[haruspex]

You have to write the equation of a trajectory of a point of the cylinder.

No, it's much simpler than that.
@Davidllerenav , think about which part of the cylinder is, momentarily, stationary.

 

[Davidllerenav]

No, it's much simpler than that.
@Davidllerenav , think about which part of the cylinder is, momentarily, stationary.

The part that is in contact with the ground would be momentarily stationary. In this case O.

 

[haruspex]

Right.

The part that is in contact with the ground would be momentarily stationary. In this case O.

So what can you say about how quickly the distance from A to that point on the ground is changing? What does that suggest about the radius of curvature of its motion?

 

[Davidllerenav]

Right.

So what can you say about how quickly the distance from A to that point on the ground is changing? What does that suggest about the radius of curvature of its motion?

It would be a function of time a think, right? Because the radius changes with time.

 

[haruspex]

It would be a function of time a think, right? Because the radius changes with time.

That's not what I meant. But I just realized a problem with my thinking. Proceed with the other advice while I try to fix it.

 

[haruspex]

That's not what I meant. But I just realized a problem with my thinking. Proceed with the other advice while I try to fix it.

No, I think it's ok...
If point O of the cylinder is momentarily stationary, what are other parts of the cylinder doing in relation to it?

 

[Davidllerenav]

No, I think it's ok...
If point O of the cylinder is momentarily stationary, what are other parts of the cylinder doing in relation to it?

Rotating, right? It would become the center.

 

[haruspex]

Rotating, right? It would become the center.

Yes, and with what radius for A?

 

[Davidllerenav]

Yes, and with what radius for A?

I think that 2r, because it would be the diameter.

 

[haruspex]

I think that 2r, because it would be the diameter.

Right. What about B?

 

[Davidllerenav]

Right. What about B?

Well, I really don't know. It would be a line from O to B, and maybe I can try to make a triangle, but I don't know how.

 

[haruspex]

Well, I really don't know. It would be a line from O to B, and maybe I can try to make a triangle, but I don't know how.

I'm sure you'll see how with a bit more thought.

 

[Davidllerenav]

I'm sure you'll see how with a bit more thought.

I think I got it. I can draw the velocity with respect to the center, which is tanget to the cylinder and draw the velocity of the center, it seem sthat the sum of both is the velocity of the point B with respect to the point O.

 

[haruspex]

I think I got it. I can draw the velocity with respect to the center, which is tanget to the cylinder and draw the velocity of the center, it seem sthat the sum of both is the velocity of the point B with respect to the point O.

Yes, but the problem is to find the radius of curvature of the path, not the instantaneous velocity.

 

[Davidllerenav]

Yes, but the problem is to find the radius of curvature of the path, not the instantaneous velocity.

But can't I find the angular velocity which has the radius?

 

[haruspex]

But can't I find the angular velocity which has the radius?

Yes, if you find the angular velocity and the linear velocity that will tell you the radius, but it does seem a long way round. Isn't B's distance from O easy to determine?

 

[Davidllerenav]

Yes, if you find the angular velocity and the linear velocity that will tell you the radius, but it does seem a long way round. Isn't B's distance from O easy to determine?

Maybe I can do the same thing I did with velocity but with position?

 

[haruspex]

Maybe I can do the same thing I did with velocity but with position?

How far is OC? How far is BC? What is the angle between?

 

[Davidllerenav]

How far is OC? How far is BC? What is the angle between?

OC is $r$, because it goes to the center. BC is $2\cdot OC=2r$, am I right? The angle between them is 45 degrees.

 

[haruspex]

BC is 2⋅OC

No, and I am really puzzled that you could think so. Are you looking at the right points?

 

[Davidllerenav]

No, and I am really puzzled that you could think so. Are you looking at the right points?

Oh, sorry I thought about OB. Then BC has a length r too and between BC and OC the angle is of 90 degrees.

 

[haruspex]

Oh, sorry I thought about OB. Then BC has a length r too and between BC and OC the angle is of 90 degrees.

Right, so what is the distance BC?

 

[Davidllerenav]

Right, so what is the distance BC?

It would be the radius.

 

[haruspex]

It would be the radius.

Sorry, I meant OB.

 

[Davidllerenav]

Sorry, I meant OB.

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

 

[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.