1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the acceleration of a variable crank mechanism.

  1. Dec 5, 2012 #1
    I've attempted multiple times, but don't really know which way to go and none of them are getting to the answer. I've got the question (below), got it wrong and been told the correct answer.

    A variable crank mechanism consists of a slider in a slot which is rotated around an axis at constant speed 120 rev/min. The slider is moving outwards relative to the slot at constant velocity 9 m/s. What is the magnitude of its acceleration (in m/s^2) at the instant when its radius is 0.3 m?


    tangential velocity v = ωr

    maybe the equations of motion

    v^2=u^2 + 2 x a x s



    I know i have to convert the speed 120rev/min into rad/s by (120*2∏)/60 = 4∏.
    Then a=W^2 x r.
    so a = 16 x ∏^2 x 0.3=47.374...

    The answer i'm supposed to be getting is 231.1 m/s^2
     

    Attached Files:

  2. jcsd
  3. Dec 5, 2012 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Welcome to PF!

    You've calculated the centripetal component of the acceleration. There is also a tangential component that you will need to find.

    Polar coordinates are appropriate for this problem. There are well-known expressions for the components of acceleration in polar coordinates. Are you familiar with them?
     
  4. Dec 6, 2012 #3
    Thanks.
    I realized there was a tangential acceleration but didn't know the equations for the components of acceleration in polar coordinates.
     
  5. Dec 6, 2012 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    A derivation of acceleration in polar coordinates is here.
     
  6. Dec 9, 2012 #5
    ok i've solver it now thank you
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook