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Angular Momentum of Multiple Objects with Collision

  1. Dec 6, 2015 #1
    1. The problem statement, all variables and given/known data
    A device consists of eight balls each of mass 0.6kg attached to the ends of low-mass spokes of length 2.1m, so that the radius of rotation of the balls is 1.05m. The device is mounted in the vertical plane. The axle is held up by supports that are not shown, and the wheel is free to rotate on the nearly frictionless axle. A lump of clay with mass 0.36kg falls and sticks to one of the balls at the location shown, when the spoke attached to that ball is at 45∘ to the horizontal. Just before the impact the clay has a speed 2m/s, and the wheel is rotating counterclockwise with angular speed 0.5rad/s.

    Diagram is http://imgur.com/d5iJsH9

    1) (Complete) Which of the following statements are true about the device and the clay, for angular momentum relative to the axle of the device?

    True: The angular momentum of the device and clay system just after the collision is equal to the angular momentum of the device and clay system just before the collision.
    False: The angular momentum of the falling clay is zero because the clay is moving in a straight line.
    False: The angular momentum of the device is the same before and after the collision
    False: Just before the collision the angular momentum of the wheel is zero.
    True: The angular momentum of the device is the sum of the angular momenta of all eight balls.

    2) Just before the impact, what is the angular momentum of the combined system of device plus clay about the center? (As usual, x^ is to the right, y^ is up, and z^ is out of the page.) *Note: should be a vector answer*

    3) Just after the impact, what is the angular momentum of the combined system of device plus clay about the center? *Note: should be a vector answer*

    4) Just after impact, what is the angular velocity of this device? *Note: should be a vector answer*

    5) (Complete) Qualitiatively, what happens to the total linear momentum of the combined system?

    Correct: The downward linear momentum decreases because the axle exerts an upward force.
    Incorrect There is no change because linear momentum is always conserved.
    Incorrect Some of the linear momentum is changed into energy.
    Incorrect Some of the linear momentum is changed into angular momentum.
    You are correct.

    6) (Complete) Qualitatively, what happens to the total kinetic energy of the combined system?

    Incorrect There is no change because kinetic energy is always conserved.
    Incorrect Some of the kinetic energy is changed into angular momentum.
    Correct: The total kinetic energy decreases because there is an increase of thermal energy in this inelastic collision.
    Incorrect Some of the kinetic energy is changed into linear momentum.

    2. Relevant equations
    Translational angular momentum=p⋅r⋅sin(θ)
    Rotational Angular Momentum=Iω
    Moment of inertia of a ball attached by a massless rod=mr^2

    3. The attempt at a solution

    So I need help with 2, 3 and 4. I know from question 1 that 2 and 3 must have the same answer. I tried to do it by adding up the angular momenta of the clay and the device with respect to the center of the device like this:

    Clay: Translational angular momentum is p⋅r⋅sin(θ). p is <0, -.72, 0>. r⋅sin(θ) is also known as the component of r in the direction perpendicular to the momentum and is .89345. Multiply these and we get .6433. Since the clay is falling down on the ball, it would cause it to rotate in the clockwise direction. Therefore, the angular momentum of the clay is into the page or in the -z direction. So, we get <0, 0, -0.6433>.

    Second, the device: Rotational angular momentum is Iω. Question 1 tells us that the angular momentum of the device is the sum of the angular momenta of the balls. So we take the angular momentum of 1 ball and multiply by 8. One ball attached by a massless rod has a moment of inertia of mr^2. In this case, m=0.6kg and r=1.05m. So I=0.6615. ω=.5, so the angular momentum of one ball is 0.33075. Multiply by eight balls and we have 2.646. Since the device rotates in the counterclockwise direction, it has angular momentum out of the page or in the +z direction. Therefore, the angular momentum is <0, 0, 2.646>.

    I figured I should just add these together to get the total, which would then be <0, 0, 2.0027> but I'm told this is wrong. Can anyone see where I'm going wrong? Thanks!
     
  2. jcsd
  3. Dec 6, 2015 #2

    haruspex

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    I don't get the same numbers for the angular momentum of the clay before impact. Please write out your steps there in more detail.
     
  4. Dec 6, 2015 #3
    My calculator was in radians. I feel like an idiot, thank you so much.

    Going to give steps in case anyone needs to use this thread in the future:
    for angular momentum of the clay, I used p⋅r⋅sin(θ).
    p=m⋅v, or 0.36⋅-2=0.72
    r⋅sin(θ) is the component of r in the direction perpendicular to p. In this case, it is 1.05*sin(45) = 1.05⋅.7071 = 0.7425.
    Multiply together: Angular momentum of clay = p⋅r⋅sin(θ) = 0.72⋅.07425 = 0.5346
     
    Last edited: Dec 6, 2015
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