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Moment of inertia at CoP

  1. Feb 7, 2015 #1
    I am trying to learn to calculate collisions on the center of percussion of a rod.

    Can you tell me how to find the moment of inertia?

    Suppose we have a rod of length 1m and mass M = 12 kg., we know that ##I = 12 * 1^2 / 3 = 4## , if a collision takes place at the CoP ( at 0.166666 m from the Center of mass and 0.66666 from the pivot), is still I = 4? .

    If a collision takes place at the CoM, isn't it just an ordinary collision with a body of ( I =) M = 12 ?
    If an incoming ball (m = 1, v = p = 20) hits the rod at the CoM it will rebound at v = 16.92 and the rod will move at v = 3.077, without any rotation, right?

    If it his the CoP, is its angular momentum = 20 * 0.6666 or 20 * 0.16666? And what is the angular momentum of the swinging rod?

    Thanks for your help
     
  2. jcsd
  3. Feb 7, 2015 #2
    the moment of intertia would change because the length changed, if the length from the pivot point increased there would be a greater moment of interia(I), if it decreased there would be a lower MOI(I) but a higher angular velocity, this is because of conservation of angular momentum, the equation for angular momentum is L=wI, where L=angular momentum in (kg*m^2/s), w=angular momentum in (rad/sec), and I=moment of intertia in (kg*m^2)
    Your second question needs to be more organized or it needs further specifications. As a general rule of thump a good way to solve these problems is to write down all the variables given and include the units of the variables, next right the basic equations and then you substitute for unknown values. This forum is not meant for homework questions..
     
    Last edited: Feb 7, 2015
  4. Feb 8, 2015 #3
    (I'll try to frame my post more clearly, it was not a HW question)
    -------------
    I have a couple of questions about the [CP](http://en.wikipedia.org/wiki/Center_of_percussion#Center_of_percussion_of_a_uniform_beam)

    The formula is :
    ##b=\frac{L^2}{12A}## where b is the distance from CoG, therefore if L = 1 (and M = 12),
    the distance from the center of gravity be 1/12*.5 = 1/6

    Now suppose a ball A (m =1, v = 20) hits the CP,

    1. will the rod react in the same way as if there were no U-bolt at the pivot?
    2. how do we calculate the angular momentum of the ball, (p = 1 *20 = 20), shall we consider r from the CoG (1/6, L = 10/3) or from the pivot (2/3, L = 40/3)?
    3. is the moment of inertia $I = 12/12 * 1*2 = 1$, or, considering the pivot at the U-bolt 12/3 = 4?

    Thanks for your attention
     
  5. Feb 8, 2015 #4
    1. No, it would not react the same way because a U-bolt is needed to let the rod move freely.
    2. Basic Equations: v=rw, L=wI, I=mr^2. I=(1)(40/3)^2=I=1600/9,w=r/v=>w=(40/3)/(20)=2/3, so L=2/3*1600/9=3200/7. by the way the distance of the pivot point is measured from the distance of where the center of gravity, the same procedure can be used for the other radius.here are the units for the above equations v=linear velocity in m/s, r=distance from center of gravity, w=angular velocity in rad/sec, I=moment of inertia in kg*m^2, L=angular momentum in kg*m^2/sec, and m=mass in kg..
    3. I am confused...are you sure this concept of center of percussion is required? the above equations I gave you should be able to answer most of your questions

    If I may ask for what purpose you are asking these questions, is it for fun or for class work specifically about center of percussion?I have never encountered CoP in general physics so understand that I am using my knowledge from angular momentum to try to help you, hopefully someone more experienced with CoP will help you.
     
    Last edited: Feb 8, 2015
  6. Feb 8, 2015 #5
    I do know how to use those equations: https://www.physicsforums.com/threads/is-angular-momentum-always-conserved.792401/#post-4977667 when the rod is really free.
    In the cited case if the tip is hit L= 20*.5 = 10, the ball rebounds at v = 10 (L= -5) , the rod (I = 1) moves at v = 2.5 and rotates with omega = L = 15

    The U-bolt lets the rod free to move or swing but not to rotate, that is the difference with a real pivot

    you did not answer my questions:
    I want to know if L and I are to be related to the the CoG or to the pivot
     
  7. Feb 8, 2015 #6
    "I want to know if L and I are to be related to the the CoG or to the pivot", they are related to both because L(angular momentum) is based off of I(moment of inertia), I is based off the R (the radius). The radius (R) is the distance between the center of gravity and the pivot point,. I bet I have been no help ;___; you can hurt me if you want.
     
  8. Feb 8, 2015 #7
    If the ball hits the rod at the CP, p = 20, what is its L? and I = 1 or 4?
     
  9. Feb 8, 2015 #8
    if you want me to solve a problem I can, but I need the variables and the units with them and tell me what you want to find. do you mean p as in momentum? CoP is the moment of inertia that is the most effective according to what I have read about CoP. if you right down all the variables and unknowns the basic equations anyone can solve physics problems with little knowledge of what is actually going on, after all science math problems are only substitution.
     
    Last edited: Feb 8, 2015
  10. Feb 8, 2015 #9
    You do not realize that what I am asking is just the value of that variable. (L or I)

    One last time:
    If a ball (m=1 v = 20, L = 10) hits a free rod (L = 1, M = 12), on the tip it will bounce back at v =10, the rod will translate at v = 2.5 and rotate at omega = 15. Since I = 1 , l = 15.

    now , the ball hits the rod not at b =.5 from CoG but at 0.1666666.The rod is free or fixed with a U-bolt
    What are the differences?
    If you can/ want to solve that, you are welcome, what I am asking is what how to determine L of the ball in both cases (and I)

    I hope that is clear. If you can give any concrete answer, I'll appreciate that.
     
  11. Feb 8, 2015 #10
    If I understand correctly you want to know how the radius of the collision affects the L and the I. Conservation of angular momentum still applies in collisions in all directions. So when the ball hits the rod it is transfering some of its angular momentum to the rod. therefore the ball would lose angular momentum but the beam would gain angular momentum, where the ball hits the rod determines if I or omega would increase or decrease to cause the angular momentum of the rod to be constant. After the collision the angular velocity of the ball would be less because the ball lost some of its momentum in hitting the rod. If the radius from the center of gravity increases the I would increase, this would cause the angular velocity to decrease in order to obey conservation of momentum so the net L would not change in either cases because conservation of momentum still applies in collisions.Try to remember that the sum of the angular momentums must equal the intiall angular momentum, in order for the law to occur some variables such as the angular velocity or moment of inertia must change. Furthermore try to remember the the angular momentum variables are relative, so they are different for the ball and the rod. Apply the law of conservation of angular momentum to all of these angular momentum problems, it states " the total momentum in a closed system is constant"; meaning that although the momentum can be transfered to other objects in that system it will always be the same, angular momentum is kept constant by either the angular velocity changing or the the moment of inertia changing. Overall this is an angular momentum collision problem, I believe that using energy to solve this type of problem would be a lot more easy.
    The 3 important equations for angular momentum are I=mr^2, L=omega&I, and V=r* omega
     
    Last edited: Feb 8, 2015
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