How to calculate acceleration


by Eagle9
Tags: acceleration
Eagle9
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#1
Jan5-11, 11:00 AM
P: 129
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
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 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?
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vin300
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#2
Jan5-11, 11:34 AM
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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 appled , 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
gneill
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#3
Jan5-11, 01:23 PM
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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.

Eagle9
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#4
Jan8-11, 11:21 AM
P: 129

How to calculate acceleration


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
gneill
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#5
Jan8-11, 11:54 AM
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Quote Quote by Eagle9 View Post
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
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].
Eagle9
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#6
Jan8-11, 01:00 PM
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Quote Quote by gneill View Post
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 :)
gneill
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#7
Jan8-11, 02:00 PM
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Quote Quote by Eagle9 View Post
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?
Eagle9
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#8
Jan8-11, 02:48 PM
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Quote Quote by gneill View Post
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 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 so I need somehow to "combine" this two kinds of acceleration and find the formula
gneill
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#9
Jan8-11, 03:10 PM
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Quote Quote by Eagle9 View Post
Well, I will write one more 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 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]
Redbelly98
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#10
Jan8-11, 08:45 PM
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Quote Quote by gneill View Post
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.
Eagle9
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#11
Jan9-11, 08:14 AM
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Quote Quote by Redbelly98 View Post
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?
gneill
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#12
Jan9-11, 08:28 AM
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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]
Redbelly98
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#13
Jan9-11, 08:46 AM
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Quote Quote by Eagle9 View Post
So, the total acceleration that I seek is equal square root from sums of these accelerations in square, right? Is this formula correct?
The formula you posted is identical to the one I gave you, so I'd have to say I agree with it
Eagle9
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#14
Jan9-11, 12:01 PM
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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

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
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? 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?
And finally we calculate total acceleration for the purpose of which I opened this topic here:




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?
gneill
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#15
Jan9-11, 12:17 PM
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Quote Quote by Eagle9 View Post
so, the humans will have to withstand the 181 gs, right?
Right.
Eagle9
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#16
Jan9-11, 01:55 PM
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Quote Quote by gneill View Post
Right.
I would like to specify: 181 gs-this will be the g-load at any point between point/positions A and B? 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?
gneill
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#17
Jan9-11, 02:24 PM
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Quote Quote by Eagle9 View Post
I would like to specify: 181 gs-this will be the g-load at any point between point/positions A and B? 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!
Redbelly98
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#18
Jan9-11, 03:07 PM
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Quote Quote by Eagle9 View Post
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.


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