1. Section 9.6 Momentum and Collisions in Two Dimensions (cont.)
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2. Example Problem
Two pucks of equal mass 100. g collide on an air hockey table. Neglect friction. Prior to the collision, puck A travels in a direction that can be considered the +x-axis at 1.0 m/s, and puck B travels in the –y-direction at 2.0 m/s prior to the collision. After the collision, puck B travels 30 degrees above the +x-direction (between +x and +y) at 0.80 m/s.
What is the velocity (direction and speed) of puck A after the collision?
Is the collision completely inelastic?
Answer: Solve for conservation of the x and y components of momentum independently. The initial x momentum is 0.1*1 = 0.1 kgm/s. After the collision puck 2 has an x momentum of 0.8*cos(30)*0.1 = 0.07 kgm/s. Thus puck 1 must have a final x momentum of 0.03 kgm/s. Thus the x component of the velocity of puck 1 after the collision is 0.3 m/s. The initial y momentum is -0.2 kg m/s. After the collision puck 2 has a y momentum of 0.8*sin(30)*0.1 = 0.04 kgm/s. Thus puck 1 must have a post-collision momentum of kgm/s and a y component of velocity of -1.6 m/s. The speed of puck 1 is thus sqrt(1.6^2+0.3^2) = 1.6 m/s. The direction of puck 1 after the collision is atan(.3/1.6)=10 degrees to the right of the –y axis. The initial kinetic energy is 0.25 joules and the final kinetic energy is 0.16 joules. Thus the collision is partially inelastic.
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3. QuickCheck
A stationary firecracker explodes into three pieces – one piece heads straight east, the other heads straight north. Which of the vectors below could be the velocity vector of the third piece?
Answer: B
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4. QuickCheck
A stationary firecracker explodes into three pieces – one piece heads straight east, the other heads straight north. Which of the vectors below could be the velocity vector of the third piece?
Answer: B
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5. Section 9.7 Angular Momentum
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6. Angular Momentum
Momentum is not conserved for a particle moving in a circle because the direction of motion keeps changing – the centripetal force exerts an impulse.
Still, just as there are rotational analogs of distance, velocity, mass, etc., there is a rotational analog of momentum.
This quantity is called angular momentum, and is usually represented by the symbol L.
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7. Angular Momentum We can calculate the angular momentum L.
Newton’s second law gives the angular acceleration:
We also know that the angular acceleration is defined as
If we set those equations equal and rearrange, we find
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8. Angular Momentum
For linear motion, the impulse-momentum theorem is written
The quantity Iω is the rotational equivalent of linear momentum, so it is reasonable to define angular momentum L as
The SI units of angular momentum are kg  m2/s.
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9. Angular Momentum
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10. Conservation of Angular Momentum
Angular momentum can be written
If the net external torque on an object is zero, then the change in angular momentum is zero as well:
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11. Conservation of Angular Momentum
The total angular momentum is the sum of the angular momenta of all the objects in the system.
If no net external torque acts on the system, then the law of conservation of angular momentum is written
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12. Example 9.11 Analyzing a spinning ice skater
An ice skater spins around on the tips of his blades while holding a 5.0 kg weight in each hand. He begins with his arms straight out from his body and his hands 140 cm apart. While spinning at 2.0 rev/s, he pulls the weights in and holds them 50. cm apart against his shoulders. If we neglect the moment of inertia of the skater, how fast is he spinning after pulling the weights in?
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13. Example 9.11 Analyzing a spinning ice skater (cont.)
prepare There is no external torque acting on the system consisting of the skater and the weights, so their angular momentum is conserved – the figure below shows a top- down view.
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14. Example 9.11 Analyzing a spinning ice skater (cont.)
solve The two weights have the same mass, move in circles with the same radius, and have the same angular velocity. Thus the total angular momentum is twice that of one weight. The mathematical statement of angular momentum conservation, If f = Iii, is
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15. Example 9.11 Analyzing a spinning ice skater (cont.)
Because the angular velocity is related to the rotation frequency f by  = 2f, this equation simplifies to When he pulls the weights in, his rotation frequency increases to
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16. Example 9.11 Analyzing a spinning ice skater (cont.)
assess Pulling in the weights increases the skater’s spin from 2 rev/s to 16 rev/s. This is somewhat high because we neglected the mass of the skater, but it illustrates how skaters do “spin up” by pulling their mass in toward the rotation axis.
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17. Example Problem
Bicycle riders can stay upright because a torque is required to change the direction of the angular momentum of the spinning wheels. A bike with wheels with a radius of 33 cm and a mass of 1.5 kg (each) travels at a speed of 10 mph. What is the magnitude of the angular momentum of the bike? Treat the wheels of the bike as though all the mass is at the rim.
Answer: The momentum of inertia of each wheel is simply m*r^2 = 1.5*0.33^2 = 0.16 kg-m^2 (each wheel). The linear velocity of the bike is 10 mph or 4.5 m/s. Thus the tangential velocity of the rim of the wheel is 4.5 m/s and the angular velocity is 4.5/0.33 = 14 rad/s. Thus each wheel has an angular momentum of 0.16*14 = 2.2 kg-m^2/s and the whole bike has an angular momentum of 4.4 kg-m^2/s.
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18. QuickCheck
If the Earth warms significantly, the polar ice caps will melt, and water will move from near the poles (near the Earth’s axis of rotation), and spread out evenly around the globe. What effect will this have on the length of the day?
A. Increase slightly
B. Stay exactly the same
C. Decrease slightly
Answer: B
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19. QuickCheck
If the Earth warms significantly, the polar ice caps will melt, and water will move from near the poles (near the Earth’s axis of rotation), and spread out evenly around the globe. What effect will this have on the length of the day?
A. Increase slightly
B. Stay exactly the same
C. Decrease slightly
Answer: B
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20. Intermezzo: Practice with Torque and Statics
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21. Example Problem
A baseball bat has a mass of 0.82 kg and is 0.86 m long. It’s held vertically and then allowed to fall. What is the bat’s angular acceleration when it has reached 20° from the vertical? (Model the bat as a uniform cylinder).
Answer: The bat tips due to a torque due to gravity which acts at the center of the cylinder. The torque at this instant is mg*sin(20)*L/2 where L is the length of the bat. The acceleration is torque/I where I is the moment of inertia. For a uniform cylinder rotating about the end I = m*L^2/3 so alpha 6 = rad/s^2. Note that the mass of the rod is superfluous information, but not the length.
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22. Example Problem
An 80. kg construction worker sits down 2.0 m from the end of a 1450 kg steel beam, as shown. The cable supporting the beam is rated for a maximum tension of 15 kN. Should the worker be worried?
(8.44) Answer:
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