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Q. Mechanics3 - Review exercise2 (circular motion) watch

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    Right, I have been having an absolute torrid time with some of these review questions for whatever reason, this question in particular, I always seem to get just off the answer for the last part of the question, qould like to know where I am going wrong, and the method for answering the question.


    23. A smooth sphere, with centre O and radius a, is fixed to the upper surface of a table. A smooth particle P is placed on the surface of the sphere, at a point A, where OA makes an angle Φ with the upward verticle
    (0 < θ <½π ).

    The particle is released from rest. Show that when OP makes an angle θ with the upward verticle, and that the particle is in contact with the sphere,

    a (∂θ/∂t)² = 2g(cosΦ - cosθ )

    Determine the value of θ when the particle leaves the sphere.

    Given that cosΦ = 3/4, show that the particle hits the table at a distance (3/4)√3 a from the verticle diameter


    I can tell you that the value of θ when the particle leaves the sphere:

    cosθ = 2/3 cosΦ

    But the last bit has really got me curfuddled, any help greatly appreciated
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    You know that when the particles leaves the surface:
    a²w² = 2ga(3/4 - 1/2) = ga/2
    => v = aw = sqrt[ga/2], which is the linear speed of the particle when it leaves the sphere.

    You also know that the vertical distance of the particle from the table at the last point of contact is a+acos60=1.5a.

    Since the particle leaves at a direction tangent to the sphere, then using some trig you can find that the intial angle of projection is 60. Assuming the particle falls freely under gravity after it leaves the sphere, then use some projectile analysis.

    Vertical motion:
    s = ut - 0.5gt²
    -1.5a = sqrt[ga/2]sin60 t - 0.5gt²
    0.5gt² - 0.5sqrt[1.5ga]t - 1.5a = 0
    t = (0.5sqrt[1.5ga] ± sqrt[(27/8)ga])/2g

    So, the time until the particle strikes the table is:
    (sqrt[1.5ga] + sqrt[13.5ga])/4g

    Horizontal motion:
    s = vt
    = sqrt[0.5ga]cos60 . (sqrt[1.5ga] + sqrt[13.5ga])/4g
    = (sqrt[3]/4) a

    This is the distance from the vertical at the last point of contact. Now we want to find the distance of this vertical from the vertical diameter, which is:
    a sin60 = (sqrt[3]/2) a

    So the total distance is:
    (sqrt[3]/2) a + (sqrt[3]/4) a = (3/4) sqrt[3] a
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    Danke Schon, I forgot to put 1.5a as a negative distance, I keep on doing that :p: ...anyway when doing v.t I was struggling with all the wierd square roots, but sorted them out now.
 
 
 

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