1. Unit 5: Conservation of Momentum
Force and Momentum and Conservation of Momentum (9-1,9-2, 9-3)
Collisions (9-4, 9-5, 9-6)
Collisions, Center of Mass (9-7,9-8)
Center of Mass, Rocket Propulsion, and Quiz Three Review (9-9,9-10)
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2. Schedule Today Wednesday Friday
A bit more on collisions and then center of mass
Wednesday
Finish Conservation of Momentum
Review of Conservation of Energy and Momentum
Friday
Test Conservation of Energy & Momentum
Crib sheet: single 3”x5” note card.
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3. Collisions in 2 or 3 Dimensions
Head-on collisions are special, more generally a collision results in a final state with momentum in two or three dimensions.
Examples abound, on the field (billiards, football…) and in the laboratory (atomic, nuclear, particle physics...)
Since the Law of Conservation of Momentum is a vector statement it can still be gainfully applied to such cases.
Let’s consider a fixed target again….
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6. Scattering: Proton Scattering Revisited
A proton traveling at 8.2x105m/s elastically collides with a stationary proton (in a hydrogen target.) An outgoing proton is observed at a 60o scattering angle.
What are the final velocities of the protons and what is the scattering angle of the second proton?
m1=m2=m
60o
p1=8.2x105m/s
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7. Another Dread Derivation!
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10. Center of Mass
To date we’ve simplified and considered only translational motion.
General motion includes both translational and rotational motion.
Even if motion is general and many objects involved, there is always one point that moves in the same path that a single particle would if subjected to the same force. This point is the center-of-mass (cm or c-o-m).
The general motion of an extended body (or system of bodies) can be considered as the sum of the translational motion of the CM plus rotational motion about the CM.
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11. Translational and rotational
Translational only
Translational and rotational
but the same parabola
is followed by the CM
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12. Definition of CM for Two Objects
Consider a system of two objects of mass m1 and m2 at positions x1 and x2.
By definition:
That is, the CM can be found at the weighted average of the two positions:
If one mass is zero, at the position of the other mass.
If equal weight, at the midpoint
If one mass is greater, nearer the more massive object.
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13. Definition of CM for n Particles
Now consider xCM for n particles where n is large. If these are located along the x-axis:
Or more generally if each is at vector position ri.
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14. Example: 1-Dimensional
Three beads of equal mass are spaced on a string at positions 1.0m, 5.0m and 6.0m. Find the position of the CM.
We start with the definition of CM in one dimension and substitute one mass and the positions:
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15. A Bit More on Multi-dimensional CM
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16. Example: Particles in 2D
Three particles each with mass 2.50kg are located at the corners of a triangle as shown in the figure.
Where is the center-of-mass?
The (x,y) coordinates of the three masses are: (0,0), (2.0m,0), and (2.0m, 1.5m)
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18. CM for Continuous Distributions
An extended object can be described as a continuous distribution of matter
Imagine a collection of n particles each with a tiny mass Dmi in a small volume around a point xi, yi, and zi.
If we let n approach infinity then the mass element becomes an infinitesimal dmi at point x,y,z.
The summations then “go over” to the continuum or become integrals:
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20. Example: CM of a Thin Rod.
Show that the CM of a uniform thin rod of length L and mass M is at its center.
Where is the CM if the linear mass density l varies linearly from l=lo from the left end to l=2lo to the other?
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21. Here’s where our choice of axes can really help: Put the rod on the x-axis with the left end at x=0. Immediately we see that ycm=zcm=0.
For the first question regarding a uniform rod the mass per unit length is constant and l=M/L.
Divide the rod into infinitesimal elements of length dx with mass dm=ldx. Using the integrals for cm we have:
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23. Symmetry and CM
For symmetric objects the CM can often be found by inspection.
A sphere with uniform mass density is pretty clear.
For every mass element on one side of the center, there is an equivalent element on the opposite side of the center of geometry.
The CM must then lie at the center of geometry
The argument identical for a uniform cylindrical disk.
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24. Example: CM Outside of an Object!
What is the CM of the brace?
We can consider an object as a collection of it’s parts. For this object consider the rectangles A and B.
By symmetry
CM of A is (1.03m, 0.10m)
CM of B is (1.96m, -0.74m)
Assuming the brace has thickness t and uniform mass density r.
MA = (2.06m)(0.20m)tr = 0.412m2(tr)
MB = (1.48m)(0.20m)tr = 0.296m2(tr)
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26. Center of Gravity
The center of gravity of an extended object is defined as the average position of the weight, precisely defined its:
It’s easy to show at the surface of the earth this is the same as CM
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27. Experimentally Determining CG or CM
Since CG=CM we have a nifty way of determining the CM.
If an object is suspended from any point it will swing until the CG lies below the point.
If we pick a second suspension point and the CG will also lie below the point.
The intersection of the two lines will be the CG and the CM!
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28. To finish On Wednesday we conclude Conservation of Momentum.
This involves a fuller exploration of the behavior of the center of mass and of equation:
We also discuss the material on Quiz #3. Remember a crib sheet on a 3x5 card is allowed!
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