Velocity Vectors in Circular Motion: Understanding Acceleration

[Peter G.]

Hi,

I was drawing the velocity vectors in circular motion to show that, the difference between them would yield an acceleration with direction towards the center of the circle. The problem I am having though is understanding from which point that accelerating takes place. I.e:

A ball moves from point A to B.

I do the vector subtraction and get my change in velocity, my acceleration. Is that acceleration the acceleration at A, at B or at halfway through?

Thanks,
Peter G.

Oh, and also:

I learned from a website that, in circular motion:

Speed = Distance / Time, therefore, Speed = 2πr / t

But the book says the Velocity is equal to 2πr / t, which, applying the same logic as the website did, makes no sense, since, in one time period, the displacement would be 0, not 2πr...

 

[Doc Al]

When going from A to B, subtracting the displacements (and dividing by time) will give you the average velocity. Which of course is zero for the round trip. Using the distance traveled will give you the average speed. (Sometimes, velocity as speed are used interchangeably... just to confuse you!)

Similarly, subtracting the velocities (and dividing by time) will give you the average acceleration. What you probably want is the instantaneous acceleration, which will be the limit as A and B get closer.

 

[Peter G.]

Ok, so,

in the case of the equation: Speed = Distance / Time to yield: 2πr / t, I can use velocity and speed interchangeably?

And, I'm not sure if I got this. So, the resultant in my attachment (the difference between the velocity vectors) are my average acceleration between A and B? Like, if I got the length of the arc length between A and B and halved it, the acceleration I got would be there? (It seems to make sense, because, if I do so, my resultant points to the center of the circle)

 

[Doc Al]

Ok, so,

in the case of the equation: Speed = Distance / Time to yield: 2πr / t, I can use velocity and speed interchangeably?

That equation will give you the speed, which will be the magnitude of the instantaneous velocity. (Assuming uniform circular motion--constant speed.)

And, I'm not sure if I got this. So, the resultant in my attachment (the difference between the velocity vectors) are my average acceleration between A and B? Like, if I got the length of the arc length between A and B and halved it, the acceleration I got would be there? (It seems to make sense, because, if I do so, my resultant points to the center of the circle)

That would give you the average acceleration between those points, but not the acceleration in the middle. In the limit as A and B are arbitrarily close, you'll get the instantaneous acceleration. (Which will point towards the center.)

 

[Peter G.]

Ok, got it thanks! I used to kind of ignore the "infinitesimally small..." descriptions the books give but I have just recognized how they are important! Stupid me