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How to calculate accelerationby Eagle9
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#1
Jan511, 11:00 AM

#2
Jan511, 11:34 AM

P: 513

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 


#3
Jan511, 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 a_{c}. 


#4
Jan811, 11:21 AM

P: 138

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 


#5
Jan811, 11:54 AM

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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
Jan811, 01:00 PM

P: 138

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
Jan811, 02:00 PM

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("...the acceleration when the rod rotates at the constant speed: a=v^2/r"). So, what exactly is it you need to know? 


#8
Jan811, 02:48 PM

P: 138




#9
Jan811, 03:10 PM

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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 v^{2}/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
Jan811, 08:45 PM

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P: 12,074

a = (a_{c}^{2} + a_{t}^{2})^{1/2}where a_{t} is the tangential acceleration and a_{c} is the centripetal acceleration. 


#11
Jan911, 08:14 AM

P: 138



#12
Jan911, 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] 


#13
Jan911, 08:46 AM

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#14
Jan911, 12:01 PM

#15
Jan911, 12:17 PM

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#16
Jan911, 01:55 PM

P: 138




#17
Jan911, 02:24 PM

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The total acceleration is maximum at point B. It is zero at point A. Obviously it grows in betwixt A and B! 


#18
Jan911, 03:07 PM

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P: 12,074

a_{avg} = (V  Vo) / tYou 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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