# Radius of curvature of the trajectory of points A and B

#### Davidllerenav

1. The problem statement, all variables and given/known data
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.

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

3. 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?

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#### Henryk

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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

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3. 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

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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

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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

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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 realised a problem with my thinking. Proceed with the other advice while I try to fix it.

#### haruspex

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That's not what I meant. But I just realised 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

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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

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

#### Davidllerenav

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

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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, wich 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

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I think I got it. I can draw the velocity with respect to the center, wich 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 wich has the radius?

#### haruspex

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But can't I find the angular velocity wich 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

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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

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BC is 2⋅OC
No, and I am really puzzled that you could think so. Are you looking at the right points?

"Radius of curvature of the trajectory of points A and B"

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