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Tangential and centripetal acceleration

  1. Nov 15, 2014 #1
    Hello.

    Sorry for this very silly question, but I struggled myself too long on this (it was on the test and fortunataly I could answer correctly, but have not understood how to solve that).

    I know a particle, in a certain instant, has position ##(x_0,y_0,z_0)##
    I know the velocity magnitude ##\vec{v} = v_p \hat{\tau}##

    where tau is the tangent unit vector.
    And I know also the acceleration vector:
    ##\vec{a}=a_0\hat{x}+a_0\hat{y}##


    I have to find the tangential and centripetal acceleration of the particle at that instant.


    Thanks.
     
  2. jcsd
  3. Nov 15, 2014 #2

    td21

    User Avatar
    Gold Member

    tangential acc is a times tau.
    you can then figure out centripetal acceleration.
     
  4. Nov 16, 2014 #3
    Ok, I think I should insert the numbers here, because I have to figure out the numerical value and I do know nothing about tau.
    position ##(x_0 = \sqrt 3 m , y_0=1 m , z_0 =2.12 m)##
    ##v_p = 4.78 m/s##
    ##a_0 = 6.01 m/s^2##

    I have only this, nothing less, nothing more.
     
  5. Nov 17, 2014 #4
    Please, it's my last unfinished business.
    This is the original text. It's in italian, but mathematics is a universal language. Capture.PNG
     
  6. Nov 17, 2014 #5

    td21

    User Avatar
    Gold Member

    ep = cos30 ex + sin30 ey
    then take the dot product of ep and a
    i.e.
    (cos30 ex + sin30 ey)⋅(6.01 ex + 6.01 ey)
     
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