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Beer and Johnston Dynamics 9th 11.139 angular acceleration

  1. Oct 8, 2011 #1
    1. The problem statement, all variables and given/known data

    An outdoor track is 420 ft. in diameter. A runner increases her speed at a constant rate from 14 to 24 ft./s over a distance of 95 ft. Determine the total acceleration of the runner 2 s after she begins to increase her speed.


    2. Relevant equations

    Vr = dr/dt
    Vθ = r*dθ/dt
    Ar = d2θ/dt2
    Aθ = r*d2θ/dt2 + 2 dr/dt*dθ/dt
    V = r*dθ/dt eθ
    A = -r*(dθ/dt)2er + r*d2θ/dteθ
    An = v2
    At = dv/dt


    3. The attempt at a solution

    diameter = 420 ft. therefore ρ = .5*420 ft. or ρ = 210 ft.

    An1 = (14 ft./s)2/(210 ft.)
    An1 = 0.933 ft./s2

    An2 = (24 ft./s)2/(210 ft.)
    An2 = 2.74 ft./s2

    I am not sure where to go from here. I know I can't use equations from rectilinear motion since this is angular. If I could find the time it takes her to run the 95 ft. I think I could use that to find an average tangential acceleration by At = Δv/Δt. If I could also find the speed at 2 s, use An = v2/ρ and find the normal component of acceleration. Taking the magnitude of the two would give me the total acceleration and α = tan-1(An/At).

    I'm just not sure of the next step. I'm also not sure if that is the right approach.. Thanks for your time and any help ahead of time.
     
    Last edited: Oct 8, 2011
  2. jcsd
  3. Oct 8, 2011 #2

    gneill

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    Staff: Mentor

    Angular velocities and accelerations are given in rad/sec and rad/sec2 respectively. Presumably your An1 and An2 are meant to be angular velocities.

    You can 'co-opt' the equations from linear motion if you change all the variables to their angular equivalents. So for example, v = a*t becomes ω = [itex]\alpha[/itex]*t. You can probably think of a linear equation that relates initial velocity, final velocity, acceleration, and distance. :wink:
     
  4. Oct 8, 2011 #3
    I'm confused when you say presumably your An1 and An2 are meant to be angular velocities. I thought those were the normal components of acceleration. The units work out in ft./s2 as well.

    Also, you are saying that I can use the equations for uniformly accelerated rectilinear motion for angular acceleration? I would assume that if I were able to use them they would give me the tangential component of acceleration correct?
     
  5. Oct 8, 2011 #4

    gneill

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    Staff: Mentor

    My apologies. I didn't see the 'squares' (must be going 'terminal blind'). Yes, you've got the centripetal accelerations there.
    Yes, and correct. Use the same equations but with angular measures.
     
  6. Oct 8, 2011 #5
    Got it! From v2 = v2o + 2a(x - xo)

    Substituting in to get (24 ft./s)2 = (14 ft./s)2 + 2At (95 ft. - 0 ft.)

    solving for At we get At = 2 ft./s2

    Using v = vo + at substituting in for vo = 14 ft./s a = At = 2 ft./s2 and the given t = 2 s

    v = 14 ft./s + 2 ft./s2*(2 s)
    v = 18 ft./s

    Using An = v2/ρ substituting in the v we just found and ρ = 210 ft.
    An = (18 ft./s2)/210 ft.
    An = 1.54 ft./s2

    To find total A take the magnitude of At and An
    A = sqrt((2ft./s2)2 + (1.54 ft./s2)2)
    A = 2.53 ft./s2

    EDIT: And thanks a ton for your helping along. I didn't think I could use those equations since the equations for finding the other values since the equations for finding velocity and acceleration changed for angular. However now it makes sense why I can.
     
  7. Oct 8, 2011 #6

    gneill

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    Staff: Mentor

    Well done! :smile:
     
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