Acceleration in circular motion -- conceptual doubt

[PhysicsKid703]

Homework Statement

Purely a conceptual/ terminology question, a simple yes or no will suffice :)
In circular motion, will a point on its circumference will have a LINEAR acceleration which can be broken into two components- the RADIAL component which is v²/r and the TANGENTIAL component which is r*angular acceleration?
If so, in rolling without slipping of a body on a fixed-at-rest surface, do we use the TANGENTIAL acceleration and c.o.m acceleration vector sum and equate it to zero or LINEAR acceleration and c.o.m acceleration vector sum and equate it to zero? I'm 99% sure we use tangential and not linear but just want to confirm. Thanks![/B]

Homework Equations

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The Attempt at a Solution

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[Suraj M]

If so, in rolling without slipping of a body on a fixed-at-rest surface, do we use the TANGENTIAL acceleration and c.o.m acceleration vector sum and equate it to zero or LINEAR acceleration and c.o.m acceleration vector sum and equate it to zero?

At which point? Contact? Then i guess your right, tangential! because you're not going to get 0 for linear acceleration and c.o.m acceleration sum.

LINEAR acceleration which can be broken into two components- the RADIAL component which is v2/r and the TANGENTIAL component which is r*angular acceleration?

Don't you think its the other way around? I mean it's almost the same , but still! I'm asking!

 

[PhysicsKid703]

Haha I'm not sure, but in the end it's just terminology and personal preference I guess, switching up linear and tangential. I guess it would be best to say total(net) acceleration is the vector sum of its tangential and radial components.

 

[BvU]

If so, in rolling without slipping of a body on a fixed-at-rest surface, do we use the TANGENTIAL acceleration and c.o.m acceleration vector sum and equate it to zero or LINEAR acceleration and c.o.m acceleration vector sum and equate it to zero?

Is a question that can be answered with a simple

no.​

In case you are interested in a bit more than a simple no, the answer to your a or b question becomes a double no:

"do we use the TANGENTIAL acceleration and c.o.m acceleration vector sum and equate it to zero ?"

no.​

"do we use the LINEAR acceleration and c.o.m acceleration vector sum and equate it to zero ?"

no.​

In case you are interested in the physics and want to know what we do use:

we use the TANGENTIAL acceleration and equate it to zero
​

Tangential acceleration for a point on the rim, that is. And only at the moment it is in contact with the fixed surface. At that moment the motion is purely vertical (and changing direction from downwards to upwards). Google cycloid.

And it isn't all that simple to ask a question that can be answered with a simple yes or no...

 

[PhysicsKid703]

I see. My bad, I wasn't trying to be cocky or anything, I just said it without thinking. You're right, a simple yes or no question is difficult to ask.
So just to reiterate and confirm,
The tangential acceleration of the point in contact should be equated to zero, and that is the vector sum of acceleration of com and r*alpha, correct? Since that particle has both the acceleration due to com and the r*alpha acceleration?

 

[Chestermiller]

I'm a little confused. Are you saying that the cylinder is rolling at constant velocity without slipping at the surface, or are you saying that the cylinder is accelerating, but is not slipping at the surface?

Chet

 

[BvU]

Hello, I tried to be neutrally cocky
And this time I think a simple "yes but" seems in order.
Have to admit you made me think twice about what exactly you describe in what frame of reference.
My picture is that $\vec\alpha \times \vec r$ is wrt the rotation axis in the accelerating frame of reference and adding the acceleration of that axis (in a frame of reference where the ground doesn't move) should give zero.

I don't like the idea of calling the result a tangential acceleration (tangential to what, exactly), but maybe it's a matter of taste.

(I realize I'm contradicting my own post #4 "what we do use" -- can't be helped)

[edit] Hi Chet, what do you think ? For the case of an non-slipping, accelerating wheel.

 

[PhysicsKid703]

BvU
I have the same picture in mind. The vector sum of both the r*alpha acceleration and the acceleration of c.o.m, possessed by point of contact should give zero, assuming surface to be at rest. The terminology of tangential etc I presume is best used for pure rolling or circular motion, when we can just split up the linear acceleration into its radial and tangential components. Thanks for the help, I think I've got it :D
Mr. Chet, yes, that was my question; I was referring to a body accelerating and not slipping on a surface-at-rest.

 

[Chestermiller]

Hi Chet, what do you think ? For the case of an non-slipping, accelerating wheel.

I think it's easy to show that, at the instantaneous point of contact, the tangential velocity of the cylinder surface relative to the table surface is zero. It wasn't as obvious to me that at the instantaneous point of contact, the tangential acceleration of the cylinder surface is zero. However, if we look at the tangential acceleration at the contact point relative to a reference frame traveling with the center of mass, it is just -rα (since r is not changing). And since the acceleration of the center of mass is +rα if no slip is occurring, the tangential acceleration of the cylinder relative to the table at the contact point must be zero also. As I said, this was not obvious to me.

Chet