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Finding centripetal acceleration

  1. Sep 30, 2008 #1
    1.The blade of a windshield wiper moves through an angle of 90 degrees in .28 . The tip of the blade moves on the arc of a circle that has a radius .76m. What is the magnitude of the centripetal acceleration at the tip of the blade.



    2. Ac= v^2/r



    3. okay, the accleration should equal r(theta)/t

    but whenever I plug in my variables and solve I get an answer in the 7000s which is not reasonable at all.

    The actual answer is 24 m/s^2 but I have no idea how to get there.
     
  2. jcsd
  3. Sep 30, 2008 #2
    This is not correct. This expression would give you the linear velocity at the end of the wiper blade.
     
  4. Sep 30, 2008 #3
    really? that is an equation my physics teacher and i derived together.

    V = Δs/Δt = (rΔ(theta)/Δt)^2

    then, plugging that into Ac

    v^2/r = (rΔ(theta)/Δt)^2 / r = rΔ(theta)/Δt
     
  5. Sep 30, 2008 #4

    Mentallic

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    Homework Helper

    Using [tex]A_{c}=\frac{v^{2}}{r}[/tex] It is clear you need to find the velocity since you already are given the radius.

    I will assume that it is 0.28 seconds. The question states that it moves through an angle of 90o. Therefore, the distance the tip of the blade moves is 90o through a circle.

    Using the circumference for a circle formula: [tex]C=2r\pi[/tex] where C=circumference, r=radius; This is the distance travelled around an entire circle, but we only want the distance covered in 0.25 of the circle.

    Therefore, [tex]\frac{C}{4}=\frac{1}{2}\pi(0.76)[/tex]

    The distance covered for the tip of the wiper is [tex]d=0.38\pi . m[/tex]

    Using [tex]v=\frac{d}{t}[/tex] where v=velocity, d=distance (not displacement), t=time

    Plugging in distance and time - [tex]v=\frac{0.38\pi}{0.28}[/tex]

    [tex]v=\frac{19}{14}\pi . ms^{-1}[/tex]

    Plugging back into the centripetal acceleration formula - [tex]A_{c}=\frac{(\frac{19}{14}\pi)^{2}}{0.76}[/tex]

    Hence, [tex]A_{c}\approx24ms^{-2}[/tex]
     
  6. Sep 30, 2008 #5
    Thanks! It seems simple once I get to look at all the work.

    Now I just really wonder what my physics teacher was going on about...the equation certainly didn't get me the right answer.
     
  7. Sep 30, 2008 #6
    v = 0.38pi/.28
    ...and this is the linear velovity, r(theta)/t, not the acceleration.


    Here is where you are making an error...
    you did not square the theta or the t
     
  8. Sep 30, 2008 #7

    Mentallic

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    Homework Helper

    Well if you are looking for an equation that can find the centripetal acceleration straight from this kind of information given, then: (oh I will be skipping a few of the simpler steps since you seem more intelligent than I first suspected :smile:)

    [tex]C=2r\pi[/tex]

    Hence, [tex]d=\frac{C\Theta}{360}=\frac{r\pi\Theta}{180}[/tex]

    Therefore, [tex]v=\frac{r\pi\Theta}{180t}[/tex]

    Therefore, [tex]A_{c}=\frac{(\frac{r\pi\Theta}{180t})^{2}}{r}[/tex]

    Finally, the equation you were looking for to plug and chug instantly is:

    [tex]A_{c}=\frac{r(\pi\Theta)^{2}}{(180t)^{2}}[/tex]

    where,
    [tex]r =radius (metres)[/tex]
    [tex]\Theta =Angle of turn (degrees)[/tex]
    [tex]t =time (seconds)[/tex]
     
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