How to calculate acceleration

Redbelly98

Uh, no, not quite. 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:

a_avg = (V - Vo) / t

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.

Eagle9

gneill
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
Actually I wanted to know what is the acceleration during these 10 second between A and B :) as I see it gradually increases.
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

I am not very good at mathematics, simply tell me-my calculations are right or wrong?

Redbelly98

EDIT:
Your graph in Post #19 looks pretty good. You might try calculating the acceleration at point A, using the (ac²+at²)^1/2 formula. Since v_A=0, we know a_c=0 as well, so the acceleration is simply a_t at point A.

EDIT #2:
Folks, my apologies for writing the following without reading posts 17 & 19 very carefully. It looks like the problem is pretty much solved now, apart from the acceleration not being zero at point A.

In Post #14, you have calculated the acceleration at point B correctly. Or more specifically, at point B while there is still an angular acceleration. You used the velocity at point B to calculate a_c=v²/r, so it does not apply to "between points A and B"; it applies to point B only.

Note, the contribution from a_t=9.42 m/s² is so small, that your 1775 m/s² is pretty close to the acceleration after point B as well. In fact, your approximation π=3.14 introduces more error into the calculation than does neglecting a_t.

I agree.
Well, the centripetal acceleration is 0 at A. But don't forget there is still a tangential acceleration component due to the 3/10 rev/s² angular acceleration.
Yes, agreed.

Eagle9

Redbelly98
I think that this is right when the rod reaches this point A again, after one complete revolution

Now, I would like to clarify one thing that did not let me to sleep last night during several hours
One the image below you can see the similar situation:




Again, the rod is motionless at the position A and gradually increases its speed up to position B the only difference is that in CASE 1 the point B is closer to A than in CASE 2. In other words the rod has to accelerate much faster in CASE 1 than in CASE 2. So, I think that in CASE 1 acceleration at point B SHOULD be more than in CASE 2 at point B, right? (all other conditions are same: radius of the rod and its final speed-let’s say 5 meters/sec) the logic indicates me this and if I am right I also think that in CASE 1 the acceleration at point B should be more than acceleration after point B (when the rod rotates with constant speed). Exactly this situation we had at my post N 14. In CASE 2 the accelerations at point B and after it-are the same
In other words-if the rod has got enough distance/time to increase its speed then its acceleration (during increasing the speed!) will always be less than final acceleration (after position B)-this happens in CASE 2.

Redbelly98

That would be incorrect, since by that time the angular acceleration is zero. Multiple that by r to get tangential acceleration, and it is zero.

But when it is initially at point A, and there is angular acceleration, there is a nonzero tangential acceleration as well.

I don't know why you would think this. As in case 1, the angular acceleration drops from something nonzero down to zero at point B, and that will cause a drop in the acceleration graph you are trying to draw.

Eagle9

Ok, I would restate my question in a bit different way: From A to B the speed and acceleration increases and I want (for some specific purpose) the acceleration between these points (and at B point also) to be always less than after point B when the rod rotates with some constant velocity. How can I achieve this? I thought that to achieve this, the point B should quite far from the point A (CASE 2), in this case the rod may increase its speed much slower…….

….or: acceleration always will be the same at point B? Not depending how quickly the rod increases its speed from A to B? I wish to find this out

gneill

Perhaps an analogy is in order. Suppose you're riding in a car that's accelerating from a standing start with the goal of reaching 100km/hr. If the car accelerates constantly all the way and then suddenly stops accelerating when it reaches 100km/hr, there will be a jerk as the acceleration suddenly stops (constant pressure backwards into your seat while accelerating, suddenly no pressure when the acceleration stops).

For a smooth ride, the driver presses slowly on the gas at first (small acceleration), then builds the acceleration to a peak, then as the final speed is approached, backs off on the acceleration to transition smoothly to that speed.

Now extend the analogy slightly. The car is driving on circular track with a small radius. As the car accelerates, that acceleration pushes you back into your seat as before, plus the centripetal acceleration due to turning the curve presses you to the side, against the passenger door. The latter acceleration builds as the velocity increases. The former remains constant as long as the car's forward acceleration is constant. As before, the pressure into your seat will suddenly disappear (jerk) if the acceleration suddenly stops. The centripetal acceleration pressing you into the door will remain as long as the car's velocity remains.

So, what do you suggest you do for a smooth transition to the final angular speed?

Eagle9

gneill
Slowly picking the speed up

gneill

Or, more precisely, gradually decreasing the tangential acceleration on approach to point B.

Theoretically, I suppose you could arrange it so that the Total Acceleration magnitude remained constant. Probably easier to accomplish with a physical feedback mechanism than to figure out the required function with respect to time!