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Inclined throw in a vertical loop

  1. Sep 25, 2014 #1
    1. The problem statement, all variables and given/known data
    A small ball rotates in a vertical loop of radius R. it's velocity is the minimal required to stay in the loop.
    Now the loop is cut at an angle ##\alpha## and the ball shoots out of it.
    Express ##\alpha## so that the ball falls back into the loop and continues turning.

    2. Relevant equations
    The horizontal distance in an inclined throw: ##x=\frac{V^2\sin(2\theta)}{g}##
    The acceleration in circular movement: ##a=\frac{V^2}{R}##

    3. The attempt at a solution
    The velocity at the highest point, so it won't fall: ##V=\sqrt{gR}##
    The difference between the highest point and the cut in the loop, point A: ##h=R\left(1-\cos\frac{\alpha}{2}\right)##
    Energies between the highest point and A:
    $$\frac{mgR}{2}=-mgR\left(1-\cos\frac{\alpha}{2}\right)+\frac{mV^2}{2}$$
    $$\rightarrow V^2=gR\left(3-2\cos\frac{\alpha}{2}\right)$$
    Our horizontal distance:
    $$x=2R\sin\left(\frac{\alpha}{2}\right)$$
    The throw angle ##\theta## is ##\frac{\alpha}{2}##
    The horizontal distance at inclined throw with our velocity and angle:
    $$x=\frac{gR\left(3-2\cos\frac{\alpha}{2}\right)\sin \alpha}{g}=R\left(3-2\cos\frac{\alpha}{2}\right)\sin \alpha$$
    This distance equals the distance between points A and B:
    $$R\left(3-2\cos\frac{\alpha}{2}\right)\sin \alpha=2R\sin\left(\frac{\alpha}{2}\right)$$
    $$4\cos^2\frac{\alpha}{2}-6cos\frac{\alpha}{2}+1=0$$
    And it's wrong, it should be ##\alpha=120^0##
     

    Attached Files:

  2. jcsd
  3. Sep 25, 2014 #2
    I made a mistake. the last formula should be:
    $$2\cos^2\frac{\alpha}{2}-3\cos\frac{\alpha}{2}+1=0$$
    $$\alpha=90^0$$
    And it should be 120
     
  4. Sep 25, 2014 #3

    ehild

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    There are two roots for cos(a/2). One is 1, but that corresponds to a vertical throw. The other is cos(a/2)=1/2, that is alpha/2 = 60°, alpha= 120°.

    ehild
     
  5. Sep 25, 2014 #4
    Thanks very much, a silly mistake
     
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