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Angular Momentum of a wheel at a Location

  1. Apr 8, 2009 #1
    1. The problem statement, all variables and given/known data
    A device consists of eight balls each of mass 0.5 kg attached to the ends of low-mass spokes of length 1.8 m, so the radius of rotation of the balls is 0.9 m. 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 0.25 kg falls and sticks to one of the balls at the location shown (First quadrant at a 45 degree angle), when the spoke attached to that ball is at 45 degrees to the horizontal. Just before the impact the clay has a speed 8 m/s, and the wheel is rotating counterclockwise with angular speed 0.35 radians/s.

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

    1.)The angular momentum of the device + clay just after the collision is equal to the angular momentum of the device + clay just before the collision.
    3.)The angular momentum of the device is the same before and after the collision.
    4.)The angular momentum of the falling clay is zero because the clay is moving in a straight line.
    5.)Just before the collision the angular momentum of the wheel is 0.
    6.)The angular momentum of the device is the sum of the angular momenta of all eight balls.


    (b) Just before the impact, what is the angular momentum of the combined system of device plus clay about the center C? (As usual, x is to the right, y is up, and z is out of the screen, toward you.)
    LC,i = < , , > kg · m2/s

    (c) Just after the impact, what is the angular momentum of the combined system of device plus clay about the center C?
    LC,f = < , , > kg · m2/s

    (d) Just after the impact, what is the angular velocity of the device?
    omega f = < , , > radians/s

    (e) Qualitatively, what happens to the total linear momentum of the combined system? Why?
    1.)The downward linear momentum decreases because the axle exerts an upward force.
    2.)Some of the linear momentum is changed into energy.
    3.)There is no change because linear momentum is always conserved.
    4.)Some of the linear momentum is changed into angular momentum.


    (f) Qualitatively, what happens to the total kinetic energy of the combined system? Why?
    1.)Some of the kinetic energy is changed into linear momentum.
    2.)There is no change because kinetic energy is always conserved.
    3.)The total kinetic energy decreases because there is an increase of thermal energy in this inelastic collision.
    4.)Some of the kinetic energy is changed into angular momentum.



    2. Relevant equations
    L=r*p
    L=r*p*sin(theta)
    I=mr^2
    v=2*pi*r/T
    v=omega*r
    L=I*omega
    K=1/2*I*omega^2
    torque=r*Fnet

    3. The attempt at a solution
    I'm not this smart, I have no idea.
     
  2. jcsd
  3. Apr 8, 2009 #2

    LowlyPion

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    Homework Helper

    Your angular momentum of the system is made of two parts.

    L of wheel = I * ω

    L of clay = r X P

    But the r here is not the R of the wheel. It will be R*√2/2 which is the distance from the center along the axis that is ⊥ to the downward direction of P of the clay. Draw a picture to satisfy yourself that this is the appropriate distance vector to use for the falling clay.
     
  4. Apr 9, 2009 #3
    im working on the same type of problem
    i found the L of clay, but would that be negative because it is falling??
    also on the
    L of wheel part
    how do you find I?
    i know I = M * r(perpendicular) ^ 2 but since there are 8 balls do we multiply by 8?
    also would the balls at the 45 degree angle have the same r(perp) as the others?
     
  5. Apr 9, 2009 #4

    LowlyPion

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    The sign is determined by the right hand rule in taking the vector cross product. Whether it is + or -, is determined then by which side of the axle the ⊥ to a radius is taken. For ω you would use the right hand curl rule. Curl the fingers of your right hand in the direction of the rotation, and where your thumb points is plus. (Counter clockwise as you look at it is plus.)

    Otherwise, your I would be as you have figured 8 times the mass of a ball times the square of the radius, .9m. With the moment of inertia though you aren't concerned with the angle, just the distance from the origin.

    Right hand rule for x X y :
    400px-Cartesian_coordinate_system_handedness.svg.png
     
  6. Apr 9, 2009 #5
    so the L for the wheel is I*omega where I = (8*.5)(.9)^2 and omega is just the .35rad/sec. Then you add that to the L of the clay which is R*p where R=1.8*(sqrt(2)/2)) and p = (.25*8). If so, i got it wrong.
     
  7. Apr 9, 2009 #6

    LowlyPion

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    It may add or subtract. I can't see the drawing to know which side of the axle the clay falls.
     
  8. Apr 9, 2009 #7
    I added L of the wheel and L of clay. I also subtracted them and they were both incorrect. Is it the calculations?
     
  9. Apr 9, 2009 #8

    LowlyPion

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    Wait a minute. The distance to the axle is .9*√2/2 not 1.8*√2/2.

    Of course the Lc is Z directed. So it would be of the form <0,0,Lc>
     
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