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Math Help - mechanics problem discussion

  1. #1
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    mechanics problem discussion

    Consider the following questionthis is not a homework problem,but it is one I want to discuss regarding its feature)

    A ball is falling onto a ground from a certain height and collides elastically and reaches the same height and the process continues.What is the averages force on the ground?

    There are plenty of methods available to do this.What I like most is to use the ROCKET EQUATION!!!

    m(dv/dt)=F-u(dm/dt) F is the external force to the system.

    It first may look strange.But if you think about it,there is nothing wrong in it.After all,it is meant for a system of particle and is quite general in application.


    [If you doubt,you apply it correctly to the famous problems asking the average force in water jet impinging on a wall/chain hanging on a flat platform suddenly starts to fall freely and falls a distance x/conveyer belt prblems etc...Just you will have to consider that LHS m(dv/dt) denotes the force on the concerned body...may be it is not accelerating---like rain falls on a roof and we want to find the average force exerted on the roof.It will begiven by the m(dv/dt) term but individually (dv/dt) will have no meaning here]

    Applying that in this problem you will see that the required average force is the external force Mg.And this is the correct answer.

    Now,the current problem that is haunting me.I refer to the same problem I quoted first in this thread,by assuming not perfectly elastic collision.That is I am given a co-eff. of restitution e<1.Suppose,I try to find the average force.My method still predicts it is Mg.But,the calculation works are not easy.Because,each time you have to find (del p/del t) where each of these are changing.Then,you have to take the average...and it is by no meanss easy to see what is the result...

    Can anyone tell me what is the correct result?By my tool,I am almost sure that it is Mg.But,is there any other means to verify that?
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  2. #2
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    Quote Originally Posted by kolahalb View Post
    Consider the following questionthis is not a homework problem,but it is one I want to discuss regarding its feature)

    A ball is falling onto a ground from a certain height and collides elastically and reaches the same height and the process continues.What is the averages force on the ground?

    There are plenty of methods available to do this.What I like most is to use the ROCKET EQUATION!!!

    m(dv/dt)=F-u(dm/dt) F is the external force to the system.

    It first may look strange.But if you think about it,there is nothing wrong in it.After all,it is meant for a system of particle and is quite general in application.


    [If you doubt,you apply it correctly to the famous problems asking the average force in water jet impinging on a wall/chain hanging on a flat platform suddenly starts to fall freely and falls a distance x/conveyer belt prblems etc...Just you will have to consider that LHS m(dv/dt) denotes the force on the concerned body...may be it is not accelerating---like rain falls on a roof and we want to find the average force exerted on the roof.It will begiven by the m(dv/dt) term but individually (dv/dt) will have no meaning here]

    Applying that in this problem you will see that the required average force is the external force Mg.And this is the correct answer.

    Now,the current problem that is haunting me.I refer to the same problem I quoted first in this thread,by assuming not perfectly elastic collision.That is I am given a co-eff. of restitution e<1.Suppose,I try to find the average force.My method still predicts it is Mg.But,the calculation works are not easy.Because,each time you have to find (del p/del t) where each of these are changing.Then,you have to take the average...and it is by no meanss easy to see what is the result...

    Can anyone tell me what is the correct result? I am almost sure that it is Mg.But,is there any other means to verify that?
    Hmmmm...

    On first blush it would seem to me that you are being too fancy about this: assuming that we are calling g a constant, the force on the ball due to gravity is a constant, thus the average force on the ground to stop it over time had better be that same mg. After all, it shouldn't matter whether the ball is bouncing or sitting on the floor if the force on the ball is constant.

    I'll have to think it through a bit more (my argument seems a bit too simplistic) and I'll get back to you if I think of something else.

    -Dan
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  3. #3
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    The average force the ball excerts on the ground is equal to the minus the
    average force the ground exerts on the ball. This is equal to the change in
    momentum per unit time.

    Drop the ball form a height h, then it takes t = sqrt(2h/g) s to hit the ground,
    and it hits the ground with speed v=sqrt(2 g h). Therefore the change in
    momentum per bounce is 2mv = 2m sqrt(2 g h), and the interval between
    bounces is 2t, so the average force is:

    2mv/2t = m sqrt(2gh)/sqrt(2h/g) = mg.

    RonL
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    CaptainBlack,you are assuming an elastic collision...I am talking of a collision that is partly elastic...

    What I am getting:Force at a time on the floor,F(t)=dp(t)/dt+Mg
    Supposing that the initial time moment is t=0, the average force on the ground can be found by

    Average force <F>=Limit(T-->0) (1/T) integration(0 to T) F(t) dt
    =Limit(T-->0)[{p(t)-p(0)}/T +Mg]

    Since the motion of the ball is confined in a finite region, p(T)-p(0) is always bounded, and so the limit is zero
    Last edited by kolahalb; May 17th 2007 at 05:52 PM. Reason: As that was leading to my discussion with another forum regarding this question itself
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    Quote Originally Posted by kolahalb View Post
    CaptainBlack,you are assuming an elastic collision...I am talking of a collision that is partly elastic...

    What I am getting:Force at a time on the floor,F(t)=dp(t)/dt+Mg
    Supposing that the initial time moment is t=0, the average force on the ground can be found by

    Average force <F>=Limit(T-->0) (1/T) integration(0 to T) F(t) dt
    =Limit(T-->0)[{p(t)-p(0)}/T +Mg]

    Since the motion of the ball is confined in a finite region, p(T)-p(0) is always bounded, and so the limit is zero
    And for a coefficient of restitution e different from 1 the outgoing
    momentum is ep where p is the incoming momentum, so the change in
    momentum per "boumce" is (1+e)p rather than 2p, and the rest of the
    argument goes through, so the average force on a bounce is:

    [(1+e)/2] mg

    That is the detail of the individual bounce (the height of the bounce) cancels
    out.


    But now I think about this again I don't like it, as in the limit as the height
    of a bounce goes to zero, you should expect the average force to go to
    mg, which with this argument it does not

    RonL
    Last edited by CaptainBlack; May 17th 2007 at 08:01 PM.
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    I did a mistake...I wrote Limit T--->0.I meant that T--->infinity
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  7. #7
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    I said this as time of bounce is infinite w.r.t. the time of collision...However,I think I will have a more convincing arguement.
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  8. #8
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    Quote Originally Posted by kolahalb View Post
    CaptainBlack,you are assuming an elastic collision...I am talking of a collision that is partly elastic...

    What I am getting:Force at a time on the floor,F(t)=dp(t)/dt+Mg
    Supposing that the initial time moment is t=0, the average force on the ground can be found by

    Average force <F>=Limit(T-->0) (1/T) integration(0 to T) F(t) dt
    =Limit(T-->0)[{p(t)-p(0)}/T +Mg]

    Since the motion of the ball is confined in a finite region, p(T)-p(0) is always bounded, and so the limit is zero
    The average force:

    <F>=Limit(T--> infty) (1/T) integration(0 to T) F(t) dt

    If this exists, and then you conclude that the limit is mg.

    But now as T->infty this is dominated by the final state where
    the ball is at rest, and you have no guarantee (from this argument)
    that the short term average (say over a bounce) is stationary or not.

    RonL
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    CaptainBlack,I am sorry and shameful...
    I started correctly but then,possibly,I made a mess of everything...

    Please note my reasoning carefully...

    The formula for averaging any quantity is-
    <f(t)>=(1/T) integration (0 to T) [f(t) dt]

    now,what does T mean here?T is the "period".In our problem T may be identified as the time in which the ball falls and rises up.That is I am denoting the moment the ball starts to fall as t=0 and the moment the ball comes to rest after rebound as t=T.Then p(0)=p(T)=0

    So,(1/T)[p(T)-p(0)]=0

    So,my earlier reasoning was wrong and this time it is better and does the job as well.
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  10. #10
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    Quote Originally Posted by kolahalb View Post
    CaptainBlack,I am sorry and shameful...
    I started correctly but then,possibly,I made a mess of everything...

    Please note my reasoning carefully...

    The formula for averaging any quantity is-
    <f(t)>=(1/T) integration (0 to T) [f(t) dt]

    now,what does T mean here?T is the "period".In our problem T may be identified as the time in which the ball falls and rises up.That is I am denoting the moment the ball starts to fall as t=0 and the moment the ball comes to rest after rebound as t=T.Then p(0)=p(T)=0

    So,(1/T)[p(T)-p(0)]=0

    So,my earlier reasoning was wrong and this time it is better and does the job as well.
    This begins to look right, but I will think about it some more.

    RonL
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