Bouncing Ball Position, Velocity and Acceleration Graphs.


by habibclan
Tags: acceleration, ball, bouncing, graphs, position, velocity
habibclan
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#1
Jun14-08, 07:09 PM
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Hi!
I've attempted to draw the position, velocity and acceleration graphs for a bouncing ball. This is the link for it:

http://i196.photobucket.com/albums/a...ncing_ball.jpg

What i'm wondering is, the instant in which the ball is in contact with the ground, is the acceleration zero, which is why there are circles? Just wanted to confirm that! [as there is no acceleration due to gravity?] Thanks in advance!
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G01
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Jun14-08, 09:57 PM
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Even though the velocity is zero at the point it hits the ground the acceleration is not, since the velocity is still changing (from negative to positive). Thus, there has to be an acceleration.
habibclan
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Jun14-08, 10:42 PM
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Does it not change from the "momentum" of the collision rather than the acceleration. I just remembered reading in my textbook that a horizontal surface has an angle of 0 therefore the parallel component of acceleration is 0. So the instant that it bounces off from the ground, its a horizontal surface, making the instantaneous acceleration 0? Am i right?

habibclan
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#4
Jun18-08, 12:21 PM
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Bouncing Ball Position, Velocity and Acceleration Graphs.


Can someone please help me this! Thanks in advance =)
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Jun18-08, 12:50 PM
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Quote Quote by habibclan View Post
Does it not change from the "momentum" of the collision rather than the acceleration. I just remembered reading in my textbook that a horizontal surface has an angle of 0 therefore the parallel component of acceleration is 0. So the instant that it bounces off from the ground, its a horizontal surface, making the instantaneous acceleration 0? Am i right?
No. The ball still has a vertical component of acceleration. While looking at the problem in terms of momentum is correct, it does not mean there is no acceleration. There is still acceleration. The velocity is changing. That is the definition of an acceleration. The momentum picture is one way to look at this situation, but it doesn't mean that is the only way. You can look at this scenario from the point of view of momentum conservation and also from the point of view of kinematics and acceleration. Both are valid here.
habibclan
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#6
Jun18-08, 12:56 PM
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Quote Quote by G01 View Post
No. The ball still has a vertical component of acceleration. While looking at the problem in terms of momentum is correct, it does not mean there is no acceleration. There is still acceleration. The velocity is changing. That is the definition of an acceleration. The momentum picture is one way to look at this situation, but it doesn't mean that is the only way. You can look at this scenario from the point of view of momentum conservation and also from the point of view of kinematics and acceleration. Both are valid here.
If there is acceleration, then there should be no empty circles in the acceleration graph. Why is it that the acceleration is undefined at the point of reflection from the ground [ derivatives of a cusp are undefined].
tiny-tim
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Jun18-08, 01:59 PM
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Quote Quote by habibclan View Post
If there is acceleration, then there should be no empty circles in the acceleration graph. Why is it that the acceleration is undefined at the point of reflection from the ground [ derivatives of a cusp are undefined].
Hi habibclan!

The vertical acceleration is minus g while the ball is in the air.

The vertical acceleration is hugely poisitive when the ball is in contact with the ground.

The shorter the ball is in contact with the ground, the larger the vertical acceleration.

Vertical acceleration is change in vertical momentum so it has to be in the opposite direction to g when the ball bounces, and all the downward momentum that it got during the quite long time in the air has to be reversed in the very short time on the ground.

The graph of the vertical acceleration should really have a huge positive spike at each bounce, not just a tiny little circle!
habibclan
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#8
Jun18-08, 04:56 PM
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Quote Quote by tiny-tim View Post
Hi habibclan!

The vertical acceleration is minus g while the ball is in the air.

The vertical acceleration is hugely poisitive when the ball is in contact with the ground.

The shorter the ball is in contact with the ground, the larger the vertical acceleration.

Vertical acceleration is change in vertical momentum so it has to be in the opposite direction to g when the ball bounces, and all the downward momentum that it got during the quite long time in the air has to be reversed in the very short time on the ground.

The graph of the vertical acceleration should really have a huge positive spike at each bounce, not just a tiny little circle!

Thank you so much! That makes a lot of sense to me now =).


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