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Ball rolling down smooth curve

  1. Apr 16, 2010 #1
    Ball sliding down smooth curve

    1. The problem statement, all variables and given/known data
    A mass of 2 kg is dropped vertically into a frictionless slide located in the x-y plane. The mass enters with zero velocity at (-5,5) and exits traveling horizontally at (0,0).

    Assuming the slide to be perfectly circular in shape construct a model of the forces acting on the mass and hence simulate its motion.

    2. Relevant equations
    Unknown.


    3. The attempt at a solution
    How is it done?

    This is the question written so it's easier to understand.......

    Quarter-Circle radius 5 starting at (-5,5), ending at (0,0), centre at (0,5).
    Ball mass 2.1kg is let go vertically at (-5,5) and left to slide down the slope.
    The ball can be assumed to be a sliding point mass.

    So,
    s_0 = (-5,5)
    v_0 = (0,0)
    a_0 = (0,-9.8)

    v_f = (sqrt(2*9.81*5),0) = (9.90, 0)
    s_f = (0,0)

    But, how do you model it?
     
    Last edited: Apr 17, 2010
  2. jcsd
  3. Apr 17, 2010 #2

    lanedance

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    the first thing to do is describe the force on the ball at each point on the circle

    to do this, you should parametise equation of the cricle in terms of a single variable, angle probably being a good choice
     
  4. Apr 17, 2010 #3

    HallsofIvy

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    By the way, you titled this "ball rolling down smooth curve". It is an important simplification in this problem that the slide is "frictionless" and so the ball does NOT roll! That means that you can consider the ball as a single point (its center of mass) and find the motion of that.
     
  5. Apr 17, 2010 #4
    ok, so the parametric equation for a circle radius 5 origin 0,5 is
    [tex]y = 5cos( \theta )+5; x = 5sin( \theta )[/tex]

    or in Cartesian form

    [tex](x)^2 + (y-5)^2 = 25[/tex]

    would the velocity at any given y be
    [tex]v^2 = 2 g (y-5)[/tex]
    or would that just be the y component of the velocity?
     
  6. Apr 17, 2010 #5

    lanedance

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    Halls makes a good point & as you haven't siad anything about ball mooment of inertia or radius, i will just assume it is a sliding point mass

    first, the circle centre is actually (5,0) as i read it

    if you want to use energy balance, the difference between total kinetic energy & gravitational potential will be conserved, so that should help you decide what v represents in the equation you gave
     
  7. Apr 17, 2010 #6
    centre is 0,5
    kby4b8.png

    but yeah, it's a sliding point mass.
    the difference between the two is the same through out? how? becuase it would have 0 GPE at 0,0 :S
     
    Last edited by a moderator: Apr 18, 2017
  8. Apr 17, 2010 #7

    lanedance

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    sorry i meant sum
     
  9. Apr 17, 2010 #8
    ok, so v is tangential velocity.
    so how do I go from tangential velocity to position (x,y) at time t?
     
  10. Apr 17, 2010 #9

    lanedance

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    so kinetic energy is based on the magnitude of velocity

    so for a given point on the curve you now know velocity magnitude & direction...
     
  11. Apr 17, 2010 #10
    I had another go at it, this is as far as I got:
    Thanks so far btw! :)



    A mass of 2 kg is dropped vertically into a frictionless slide located in the x-y plane. The mass enters with zero velocity at (-5,5) and exits traveling horizontally at (0,0).

    Assuming the slide to be perfectly circular in shape construct a model of the forces acting on the mass and hence simulate its motion.


    See http://imgur.com/EutWV.png

    Parametric form of the path:
    [tex]\begin{align*}x & = 5 Sin[\theta] \\
    y & = 5 Cos[\theta]+5
    \end{align*}[/tex]
    Since [tex]\phi = \theta - \pi[/tex] and [tex]\phi[/tex] is the angle we're interested in, the equation can be rewritten as:

    [tex]\begin{align*}x & = 5 Sin[\phi + \pi] \\
    y & = 5 Cos[\phi + \pi]+5
    \end{align*}[/tex]

    Since the curve is smooth and it is assumed there is no air resistance we say that all GPE is converted to KE. That is,
    [tex]m g h = \frac{1}{2}m v_t^2[/tex]
    So we can calculate the tangential velocity [tex]v_t[/tex] at any [tex]h[/tex].
    [tex]v_t = \sqrt{2gh}[/tex]
    Since, in this case, [tex]h[/tex] is just our [tex]y[/tex] co-ordinate we can get [tex]v_t[/tex] in terms of [tex]\phi[/tex].
    [tex]\begin{align*}v_t & = \sqrt{2gh} \\
    & = \sqrt{2gy} \\
    & = \sqrt{2g(5 Cos[\phi + \pi]+5)} \\
    v_t & = \sqrt{10g(1+Cos[\phi + \pi])}
    \end{align*}[/tex]
    Since [tex]v_t = r \dot{\phi} = r\omega[/tex], where [tex]r[/tex] is the radius of the circle's path, then
    [tex]\begin{align*}5 \dot{\phi} & = \sqrt{10g(1+Cos[\phi + \pi])} \\
    \dot{\phi} & = \frac{1}{5}\sqrt{10g(1+Cos[\phi + \pi])}
    \end{align*}[/tex]
    So the angular velocity, [tex]\omega[/tex], at any given [tex]\phi[/tex] is;
    [tex]\frac{d\phi}{dt} & = \frac{1}{5}\sqrt{10g(1+Cos[\phi + \pi])}[/tex]
     
  12. Apr 18, 2010 #11
    I've got this now

    http://mathbin.net/45802 [Broken]
     
    Last edited by a moderator: May 4, 2017
  13. Apr 18, 2010 #12

    lanedance

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    no worries, method looks ok, though i haven't checked working,

    angles are a little confusing i would just start with angle from horizontal, moving in direction of the motion then [itex] (x,y) = 5(-cos(\phi), 1-sin(\phi)) [/itex] but dond;t let this defintion confuse things.

    one other point, the velocity of any particle constrained to a path will always be tangential to that path (you used transverse somehere as well which doesn't quite make sense)
     
  14. Apr 18, 2010 #13

    lanedance

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    also note, that as you used energy balance & angluar acceleration, you haven;t actually solved for the forces anywhere if that is required...
     
  15. Apr 18, 2010 #14
    I've done it :D

    Using Runge-Kutta method..

    but now I need to work out vectors for the forces...
     
  16. Apr 18, 2010 #15

    lanedance

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    should be just trig problem
     
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