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Eagle9
Eagle9 is offline
#19
Jan9-11, 03:53 PM
P: 129
gneill
No. There are two components to that acceleration. One comes from the tangential acceleration due to the constant acceleration of the rod as it picks up speed. The other component comes from the *current* velocity of the rod at any given instant -- the centripetal acceleration.

The total acceleration is maximum at point B. It is zero at point A. Obviously it grows in betwixt A and B!
Well…….at first I was surprised when I read this :) but apparently you are right. Of course at the point A acceleration will be equal to zero, and then it will gradually increase to 1774.752 m/sec^2 and after position B it will be constant and equal to 1774.728 m/sec^2 as we have already calculated according this formula ar=V^2/R, right? By the way these two values of accelerations are very close to each other, I nearly confused them
So, I think that we can imagine this acceleration like this graph:


Redbelly98
There's really no such thing as "final, total acceleration". There is either the instantaneous acceleration, at a single point or instant of time -- or there is the average acceleration between two points. I'm guessing you mean the latter, so you would use the definition of average acceleration:
aavg = (V - Vo) / t
Actually I wanted to know what is the acceleration during these 10 second between A and B :) as I see it gradually increases.
I'm guessing you mean the latter, so you would use the definition of average acceleration:
aavg = (V - Vo) / t
Yes I know, but I have got one question: as I knew this formula is for linear acceleration, in other words when the point is moving at the straight line. But as I see I can use it on the circle also

You have to pay attention that the V's are vectors; don't simply subtract the two numbers, you have to account for the directions of V and Vo since they are vectors.
I am not very good at mathematics, simply tell me-my calculations are right or wrong?