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How to calculate accelerationby Eagle9
Tags: acceleration 
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#19
Jan911, 03:53 PM

#20
Jan911, 07:11 PM

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EDIT:
Your graph in Post #19 looks pretty good. You might try calculating the acceleration at point A, using the (ac^{2}+at^{2})^{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. Note, the contribution from a_{t}=9.42 m/s^{2} is so small, that your 1775 m/s^{2} 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}. 


#21
Jan1011, 08:07 AM

P: 134

Redbelly98
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 speedlet’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 itare the same In other wordsif 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. 


#22
Jan1011, 09:00 PM

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


#23
Jan1111, 08:01 AM

P: 134

….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 


#24
Jan1111, 02:50 PM

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P: 11,618

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
Jan1111, 03:40 PM

P: 134

gneill



#26
Jan1111, 03:54 PM

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P: 11,618

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! 


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