1. Limited time only! Sign up for a free 30min personal 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!

I Question about circular motion and acceleration

  1. Dec 6, 2017 #1
    In circular motion
    1) V = rw and ##\vec V## = r ω##\vec e_{tan}##
    2) a = rα and ##\vec a## = -##\frac{v^2}{r}####\vec e_{rad}## + rα##\vec e_{tan}##
    Where ##\vec e_{tan}## is the unit vector along the tangent in increasing direction of θ
    And ##\vec e_{rad}## is the unit vector along the radial outward.
    From 1) we see that rω is the magnitude of velocity of particle executing circular motion and its direction is along tangent
    But in 2) we see that magnitude of acceleration is rα but this is not the magnitude of total acceleration
    How could you explain that rα is not the magnitude of total α
     
  2. jcsd
  3. Dec 6, 2017 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    This is not correct. It only describes the tangential acceleration.
     
  4. Dec 6, 2017 #3
    I know that, this is not the magnitude of acceleration, it will be root of the sum of the square of components in both tangential and radical direction, but in v=rω its the total magnitude of the velocity and in a= rα its not the magnitude of total acceleration, its just the magnitude along the tangent, this is what not satisfying me, in the previous case total magnitude is same as v= rw, but this is not the case with accelaration, why is this so
     
  5. Dec 6, 2017 #4

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Take a general velocity ##\vec v##. Its magnitude, i.e., the speed, ##v## satisfies ##v^2 = \vec v^2## which means that the time derivative ##\dot v## of the speed is given by the time derivative of this expression as
    $$
    \frac{dv^2}{dt} = 2v\dot v = \frac{d\vec v^2}{dt} = 2\vec v \cdot \frac{d\vec v}{dt} = 2\vec v \cdot \vec a.
    \quad
    \Longrightarrow
    \quad
    \dot v = \frac{\vec v \cdot \vec a}{v^2}.
    $$
    Hence, only the acceleration in the direction parallel to ##\vec v## matters for the change in the speed. Any acceleration orthogonal to the velocity will only change the direction, but not the speed.
     
  6. Dec 6, 2017 #5
    Wonderfull explanation, just replace v^2 by v
    Thank you so much, for this explanation
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Question about circular motion and acceleration
Loading...