Radius of curvature of the trajectory of points A and B

[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.