How to calculate acceleration

In summary, the conversation discusses calculating the acceleration of a rod as it rotates around a motionless black sphere. The rod's rotation starts at Position A and increases to Position B. The formula for calculating the acceleration when the rod rotates at a constant speed is given, but the conversation focuses on finding the formula for when the rotational speed changes. The formula for finding the centripetal acceleration with respect to time or angle is discussed, along with the formula for finding the linear acceleration at the end of the radius vector. The main question is finding the formula for the acceleration of a point moving on a circle when its speed is changing.
  • #1
Eagle9
238
10
On the picture below you see the motionless black sphere and the green rod rotating around it. At the Position A the rod is motionless, and then it begins rotating and increases this rotational speed up to Position B (you can notice it-the color gradually changes from light green to dark green). After Position B the rod rotates at constant speed :rolleyes:
I would like to know how I can calculate the acceleration between A and B positions when the rod increases the rotational speed. I know how to compute the acceleration when the rod rotates at the constant speed: a=v^2/r, but what about the situation that I would like to find out? In other words when the rod quickens :cool: let’s assume that length of the rod is 45 meters and it takes 200 seconds to reach Position B from A. So, is there any formula for this purpose?
http://img89.imageshack.us/img89/5485/backupoff.gif [Broken]
 
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  • #2
Of course there are formulae.
if F is the force applied to accelerate the rod at a distance r from the pivot, Frsin(th) is the torque applied , th is the angle between the direction of force and the rod.
divide this torque by the moment of inertia of the rod to find the angular acceleration.

for the rod I guess the MI is (ML^2)/12 , M the mass and L the length.
Multiply the angular acceleration by the radial distance of any point on the rod to find its acceleration.
this is the tangential acceleration, perpendicular to the centripetal acceleration rw^2. w is the ang.acc muliplied by time.
for the part with constant velocity, there isn't tangential acc. but cent.acc. and w will be ang.acc.*200
 
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  • #3
It looks like you want the centripetal acceleration with respect to time (or angle) as the rod rotates from A to B. Is that right? If so...

[tex]a_c = \omega^2 R[/tex]

where [tex]\omega[/tex] is the instantaneous angular velocity of the rod, R is its length (from the hub to its end).

So how do we find [tex]\omega(t)[/tex]? First find the angular acceleration from your given conditions. Start with the angular position, [tex]\theta[/tex]. The usual motion formula gives

[tex]\theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2 [/tex]

In your case [tex]\theta_0 = 0, \omega_0 = 0, \theta = \frac{\pi}{2}, t = 200 seconds[/tex]

So that

[tex]\alpha = (2 \pi /2)/(200^2 s^2) = 7.854 \cdot 10^-5 rad/sec^2 [/tex]

Now that you've got [tex]\alpha[/tex] you can find [tex]\omega[/tex] for any time between points A and B:

[tex]\omega(t) = \alpha t^2[/tex]

Then go back to your formula for ac.
 
  • #4
gneill
Well, I know that linear acceleration is calculated by means of this formula: at = (V - Vo)/t
and circular with this: ar=V^2/R
But how can I “combine” them? That is I want to calculate the circular acceleration when the rotational speed changes, increases :wink:
 
  • #5
Eagle9 said:
gneill
Well, I know that linear acceleration is calculated by means of this formula: at = (V - Vo)/t
and circular with this: ar=V^2/R
But how can I “combine” them? That is I want to calculate the circular acceleration when the rotational speed changes, increases :wink:

I'd like to be sure that we're using the same terms in the same way. What is your definition of "circular acceleration"? Is it the centripetal acceleration? The angular acceleration? Something else?

Given the angular acceleration [tex]\alpha[/tex], the linear acceleration at the end of the radius vector is given by [tex]a = r\;\alpha [/tex].
 
  • #6
gneill said:
I'd like to be sure that we're using the same terms in the same way. What is your definition of "circular acceleration"? Is it the centripetal acceleration? The angular acceleration? Something else?

Given the angular acceleration [tex]\alpha[/tex], the linear acceleration at the end of the radius vector is given by [tex]a = r\;\alpha [/tex].

Well, I am not physicist, so perhaps I did not use the correct words :)
So: "circular acceleration"-when some point moving on the circle (like in our case) increases its speed, that is it. For example its speed is equal 5 m/sec and it increases this speed up to 22 m/sec :)
 
  • #7
Eagle9 said:
Well, I am not physicist, so perhaps I did not use the correct words :)
So: "circular acceleration"-when some point moving on the circle (like in our case) increases its speed, that is it. For example its speed is equal 5 m/sec and it increases this speed up to 22 m/sec :)

That's the linear acceleration. It's equal to the angular acceleration (in radians per second squared) multiplied by the length of the radius. But this is not what you indicated you were looking for in your first post. There you implied that you wanted the centripetal acceleration
("...the acceleration when the rod rotates at the constant speed: a=v^2/r").

So, what exactly is it you need to know?
 
  • #8
gneill said:
That's the linear acceleration. It's equal to the angular acceleration (in radians per second squared) multiplied by the length of the radius. But this is not what you indicated you were looking for in your first post. There you implied that you wanted the centripetal acceleration
("...the acceleration when the rod rotates at the constant speed: a=v^2/r").

So, what exactly is it you need to know?

Well, I will write one more :smile: I want to find the formula (if it exists generally) for the acceleration of the point moving on the circle when its speed is changing. If it was not changing then the formula would be a=V^2/R, but it is changing. I cannot use other formula at = (V - Vo)/t because this is for linear acceleration when the point is moving on the straight line and I have got circle :smile: so I need somehow to "combine" this two kinds of acceleration and find the formula
 
  • #9
Eagle9 said:
Well, I will write one more :smile: I want to find the formula (if it exists generally) for the acceleration of the point moving on the circle when its speed is changing. If it was not changing then the formula would be a=V^2/R, but it is changing. I cannot use other formula at = (V - Vo)/t because this is for linear acceleration when the point is moving on the straight line and I have got circle :smile: so I need somehow to "combine" this two kinds of acceleration and find the formula

The angular velocity of the rod is [tex] \omega [/tex]

The angular acceleration of the rod is [tex]\alpha[/tex]

The linear acceleration at the end of the rod is [tex] a = \alpha \; r [/tex]

The centripetal acceleration at the end of the rod is [tex] \omega^2\;r[/tex]

That last one is the same as your v2/r. When the angular velocity [tex]\omega[/tex] is changing with time, you need to plug in the value of [tex]\omega[/tex] at that particular time.

I showed in an earlier post in the thread how to calculate [tex]\alpha[/tex] and [tex]\omega(t)[/tex]
 
  • #10
gneill said:
The angular velocity of the rod is [tex] \omega [/tex]

The angular acceleration of the rod is [tex]\alpha[/tex]

The linear acceleration at the end of the rod is [tex] a = \alpha \; r [/tex]
I am used to calling that the tangential acceleration.

The centripetal acceleration at the end of the rod is [tex] \omega^2\;r[/tex]

The tangential and centripetal accelerations are at right angles to one another, and may be combined using the Pythagorean Theorem for adding vectors that are at right angles:

a = (ac2 + at2)1/2

where at is the tangential acceleration and ac is the centripetal acceleration.
 
  • #11
Redbelly98 said:
I am used to calling that the tangential acceleration.



The tangential and centripetal accelerations are at right angles to one another, and may be combined using the Pythagorean Theorem for adding vectors that are at right angles:

a = (ac2 + at2)1/2

where at is the tangential acceleration and ac is the centripetal acceleration.

So, the total acceleration that I seek is equal square root from sums of these accelerations in square, right? Is this formula correct?
http://img411.imageshack.us/img411/1888/a457a603b244copy.gif [Broken]
 
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  • #12
If you go back to post #3 in this thread, I showed how to calculate the angular acceleration and thus the centripetal acceleration and linear (tangential) accelerations for any time t while the rod is accelerating.

[tex]a_c = \omega^2 \; R[/tex]
[tex]a_t = \alpha \; R[/tex]
 
  • #14
gneill
If you go back to post #3 in this thread, I showed how to calculate the angular acceleration and thus the centripetal acceleration and linear (tangential) accelerations for any time t while the rod is accelerating.
Of course I noticed this post, but you know I am not Physicist or Mathematician, I cannot derive formulas, I needed the “ready” formula that I need to use for some specific reason :wink:

Redbelly98
The formula you posted is identical to the one I gave you
Yes, of course I noticed when you wrote this:
a = (ac2 + at2)1/2
so I'd have to say I agree with it!
I am happy! You can’t imagine to how many people I have asked about this :shy:
Now I will try to make some basis computations, I want to be sure that I calculate everything correctly:
So, here are the initial data:
1. Radius-5 meters
2. Angular velocity at position B and after it-3 revolutions/sec.
3. Time needed from A to B-10 seconds
So, the question is: what acceleration will be on this rod’s end when the rod is increasing the speed from A to B?
First we calculate ar. It is equal to ar=V^2/R. First we need to know the speed V, and for this purpose we need to know the circumference of this circle and it is equal to=2*r*3,14=31.4 meters. We multiply this on 3 revolutions and we receive 94.2 meters/sec velocity on the circle, right? :wink: Now we put these data in this formula: ar=V^2/R and we get: ar=94.2^2/5=8873.64/5=1774.728 m/sec^2

Now we calculate at and this is equal to at=(V-V0)/t. at=(94.2-0)/10sec=9.42 m/sec^2, right? :wink:
And finally we calculate total acceleration for the purpose of which I opened this topic here:
http://img151.imageshack.us/img151/6043/18837187.gif [Broken]



So, the final, total acceleration from A to B point is equal to 1774.7529998239191594461301199631 meters/sec^2 right? If we intend to know the g-loads (let’s assume that the human is placed on the end of the rod) we should divide this number at 9.8 meters/sec^2, right? So we receive: 1774.7529998239191594461301199631/9.8=181.09724487999175096389082856765 :) so, the humans will have to withstand the 181 gs, right?
 
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  • #15
Eagle9 said:
so, the humans will have to withstand the 181 gs, right?

Right.
 
  • #16
gneill said:
Right.
I would like to specify: 181 gs-this will be the g-load at any point between point/positions A and B? :shy: I mean this: imagine that I am standing at the end of this rod, and the rod began rotating and increasing the speed from A to B. Will I have to withstand this 181 g continuously during these 10 seconds?
 
  • #17
Eagle9 said:
I would like to specify: 181 gs-this will be the g-load at any point between point/positions A and B? :shy: I mean this: imagine that I am standing at the end of this rod, and the rod began rotating and increasing the speed from A to B. Will I have to withstand this 181 g continuously during these 10 seconds?

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!
 
  • #18
Eagle9 said:
So, the final, total acceleration from A to B point is equal to 1774.7529998239191594461301199631 meters/sec^2 right?
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:
aavg = (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.
 
  • #19
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 :wink:
So, I think that we can imagine this acceleration like this graph:
http://img407.imageshack.us/img407/5514/11825360.gif [Broken]

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 :wink:

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?
 
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  • #20
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 vA=0, we know ac=0 as well, so the acceleration is simply at 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.

Eagle9 said:
Redbelly98Actually 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 :wink:I am not very good at mathematics, simply tell me-my calculations are right or wrong?
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 ac=v2/r, so it does not apply to "between points A and B"; it applies to point B only.

Note, the contribution from at=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 at.

gneill said:
The total acceleration is maximum at point B.
I agree.
It is zero at point A.
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.
Obviously it grows in betwixt A and B!
Yes, agreed.
 
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  • #21
Redbelly98
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.
I think that this is right when the rod reaches this point A again, after one complete revolution :smile:

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

http://img132.imageshack.us/img132/4603/19223238.gif [Broken]


Again, the rod is motionless at the position A and gradually increases its speed up to position B :smile: 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) :smile: the logic indicates me this :smile: 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 :smile:
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.
 
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  • #22
Eagle9 said:
Redbelly98

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

.
.
.
In CASE 2 the accelerations at point B and after it-are the same :smile:
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.
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.
 
  • #23
Redbelly98 said:
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.

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 :smile:
 
  • #24
Eagle9 said:
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 :smile:

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?
 
  • #25
gneill
So, what do you suggest you do for a smooth transition to the final angular speed?
Slowly picking the speed up :rolleyes:
 
  • #26
Eagle9 said:
gneill
Slowly picking the speed up :rolleyes:

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!
 

1. What is acceleration?

Acceleration is the rate of change of velocity over time. In simpler terms, it is how much an object's speed increases or decreases over a certain amount of time.

2. How is acceleration calculated?

Acceleration is calculated by dividing the change in velocity by the change in time. The formula for acceleration is a = (vf - vi) / t, where vf is the final velocity, vi is the initial velocity, and t is the time interval.

3. Can acceleration be negative?

Yes, acceleration can be negative. This occurs when an object is slowing down, or decelerating. Negative acceleration is also known as deceleration or retardation.

4. What are the units of acceleration?

The units of acceleration are typically meters per second squared (m/s²) in the SI system. Other common units include feet per second squared (ft/s²) and kilometers per hour squared (km/h²).

5. How does gravity affect acceleration?

Gravity plays a significant role in acceleration, as it is the force that causes objects to accelerate towards the Earth's surface. On Earth, the acceleration due to gravity is approximately 9.8 m/s². This means that any object in free fall will accelerate at a rate of 9.8 m/s² until it reaches terminal velocity.

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