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Torque and angular momentum with ball shot into air

  1. Nov 21, 2005 #1
    A 0.380 kg ball is shot directly upward with an initial speed of 38.5 m/s. What is its angular momentum about point P, at a horizontal distance of 1.85 m from the launch point, when the ball is at its max height? Halfway back to the ground? What is the torque on the ball about point P due to the gravitational force when the ball is at its max height? Halfway back to the ground?

    Where should I begin?
     
  2. jcsd
  3. Nov 21, 2005 #2

    siddharth

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    The angular momentum of a particle wrt to a point would be
    [tex] l=r \times p [/tex].
    So, what do you need to find first to find the angular momentum?
     
  4. Nov 22, 2005 #3
    to find angular momentum when the object is halfway back to the ground, should i use the velocity at that point? and to find r, would i use the hypotenuse of the height of the object and the distance the object is from P?
     
  5. Jan 11, 2008 #4
    Sorry to resurrect this thread, but I have this exact same question on a problem due tonight. Only difference is my initial velocity is 41.0m/s and point P is 2.05m from the base of the launch point and the mass is 0.360kg.

    I used kinematics to find the maximum height:
    v^2 - v0^2 = 2*a*D
    0 - 41^2 = 2 * (-9.8) * D
    D = 85.7m

    I'm solving the torque one first.

    Torque = rxF = r * F * sin theta

    F in this case, in the problem they are defining it as the gravitational force which would be mg = 0.360 * 9.8 = 3.52N

    Now, we need to find r.

    That would be pythagorean theorem...sqrt ( 2.05^2 + 85.7^2 ) = 85.789m..

    So we have mg * r = 0.360 * 9.8 * 85.789 * sin theta

    I keep getting the problem wrong because I think I am getting the wrong value of theta?

    Theta is the tan inverse of (2.05 / 85.7), right?

    Here's an image:
    [​IMG]
     
  6. Jun 9, 2011 #5
    I believe the angle should be tan^(-1)(y/x) so its your max height over your distance to point p
     
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