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Engineering Dynamics problem

  1. Aug 27, 2013 #1
    1. The problem statement, all variables and given/known data
    A ball of 10kg mass is suspended by a cable that passes through a hole in a floor. The ball rotates in a circular orbit, A, that is in a plane that is 3m below the floor. If the orbit has a radius of 1m, calculate;

    a) Tension in the cable
    b) speed va of the ball in orbit A
    c) angular momentum about the vertical axis

    The motor then draws in the cable until the ball rotates in orbit B. If the radius of orbit B is 0.5m, calculate;

    d)The new speed vb of the ball
    e) The angle between the cable and vertical axis
    f) The height, h, gained by the ball


    2. Relevant equations
    θa - tan-1(1/3) = 18.43


    3. The attempt at a solution
    I am able to a-c by working out the tension as T = 10x9.81/cos18.43 = 103.4 N
    the velocity √1x103.4xsin18.43/10 = 1.808 m/s
    the angular momentum as L = r x mv = 1 x 10 x 1.808 = 18.08

    When I attempted to do the rest of the question I got the new speed to be half of the original speed and angle B to be smaller. However in the question the image for angle B looks to be greater than angle A so I am not sure whether its just the question or if I am doing something wrong to calculate the angle. (I find the sum of the forces in the x and y direction and make them equal to each other and then find angle B). For part f I am using the conservation of energy however when I make h the subject I find that it comes out as 3 which doesn't make sense.

    Can someone please help me. I have been stuck on this question for a long time and cannot seem to find where I have gone wrong.
     
  2. jcsd
  3. Aug 27, 2013 #2
    How did you compute the new speed of the ball?
     
  4. Aug 27, 2013 #3
    I assumed that the angular momentum was constant so 18.08 = 0.5 x 10v and v as 0.904
     
  5. Aug 27, 2013 #4
    If the new radius is half the original radius and angular momentum is conserved, how can the new speed be half the original speed? Note that if you substitute v = 0.904 into 18.08 = 0.5 x 10v, you get 18.08 = 4.52, which can't be right.
     
  6. Aug 27, 2013 #5
    Yes you're right sorry. So assuming angular momentum is conserved then from that equation v = 3.616. so for part e to get the angle I used mv^2/rsintheta = mg/costheta to get tantheta = v^2/rg and theta as 69.43°. Does that sound correct?
     
  7. Aug 27, 2013 #6
    Yes, that looks good.
     
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