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Parametrize trajectory of a hocjey puck

  1. Nov 16, 2009 #1
    1. The problem statement, all variables and given/known data
    A hockey puck of radius 1 slides along the ice at a speed 10 in the direction of the vector (1,1). As it slides, it spins in a counterclockwise direction at 2 revolutions per unit time. At time t = 0, the puck’s center is at the origin (0,0).

    Find the parametric equations for the trajectory of the point P on the edge of the puck initially at (1,0).

    2. Relevant equations
    general eqn: (Rcosθ, Rsinθ), where R is the radius of the puck

    3. The attempt at a solution
    radius = R = 1
    frequency = f = 2
    angular frequency = w = 2πf = 4π
    θ = wt

    (cos4πt + 10cos(π/2)t, sin4πt + 10sin(π/2)t)
    Is this answer right?
    Last edited: Nov 16, 2009
  2. jcsd
  3. Nov 16, 2009 #2


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    Re: Parametrize

    No, it isn't.

    First, follow the motion of the center of the puck. It is, at time t= 0, at (0, 0) and moves in the direction of vector <1, 1> with speed 10: (x, y)= (vt, vt) which has speed [itex]\sqrt{v^2+ v^2}= v\sqrt{2}= 10[/itex]. [itex]v= 10/\sqrt{2}= 5\sqrt{2}[/itex]. The center of the puck, at time t, is at (5\sqrt{2}t, 5/sqrt{2}t).

    Now look at the rotation. Since it makes two revolutions per unit time, it makes one revolution when t= 1/2. [itex](x, y)= (cos(\omega t), sin(\omega t))[/itex] and has period 1/2: [itex]\omega (1/2)= 2\pi[/itex] so [itex]\omega= 4\pi[/itex]. [itex](x,y)= (cos(4\pi t), sin(4\pi t)).

    Add those two motions.
  4. Nov 16, 2009 #3
    Re: Parametrize

    Woops, I had a typo. Instead of (cos4πt + 10cos(π/2)t, sin4πt + 10sin(π/2)t), I meant:
    (cos4πt + 10cos(π/4)t, sin4πt + 10sin(π/4)t)

    Here, 10cos(π/4)t and 10sin(π/4)t are equivalent to 5sqrt2.
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