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Derive the magnitude and direction of the linear acceleration

  1. Nov 21, 2008 #1
    [Solved] Centripetal Acceleration

    1. The problem statement, all variables and given/known data
    Derive the magnitude and direction of the linear acceleration at the tip of the stick, ignore the effect of gravity.

    2. Relevant equations
    r = length of stick
    a = dv/dt

    3. The attempt at a solution
    I first find the x and y components of v, then I took the derivative to find ax and ay:

    vx = rω cos(θ)
    vy = rω sin(θ)

    ax = rω ( -sin(θ) dθ/dt ) = rω ( -sin(θ) (-ω) ) = rω2 sin(θ)
    ay = rω ( cos(θ) dθ/dt ) = rω ( cos(θ) (-ω) ) = -rω2 cos(θ)

    I recognized that ω is -dθ/dt since it is in the opposite direction. This shows that the magnitude of a is rω2. If I were to draw the vector on the graph, it would also show that a is pointing toward the hinge.

    But how would I show that a is always pointing toward the hinge?

    Is this argument valid:

    Let q be the a vector from the tip to the center, then

    qx = sin(θ)
    qy = -cos(θ)

    Now, a • q = rω2cos(ф) = rω2, where ф is the angle between the two vectors and ф=0 in this case. Therefore a and q are in the same direction.

    Is there a simpler way to show this?

    - Thanks
    Last edited: Nov 21, 2008
  2. jcsd
  3. Nov 21, 2008 #2
    Your solution is perfect.

    Since the components of the acceleration vector are proportional to sin θ and -cosθ, you can also draw a right triangle to show that the acceleration vector points towards the center.
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