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I Central force and acceleration in the polar direction

  1. Apr 13, 2016 #1
    Consider a central force. The central force is radial by definition, so ##\vec{F}=f(r) \hat{r}##. Therefore, by definition, the acceleration caused by the force, in the direction of ##\hat{\theta}## must be zero, ##\vec{a_{\theta}}=0##.

    In presence of central force angular momentum is conserved, in particular its magnitude, which is ##|\vec{L}|=m r^2 v_{\theta}##, where ##v_{\theta}## is the magnitude of the velocity of the body in the direction of ##\hat{\theta}##.
    That means that the product ##r^2 v_{\theta}## stays constant and, if ##r## gets smaller, ##v_{\theta}## must increase.

    Since ##v_{\theta}## is the magnitude of the velocity of the body in the direction of ##\hat{\theta}##, this means that there must be and acceleration in this direction, necessarily.

    But how can this be possible, since, as said, the force is completely radial?

    I'm ok with the fact that the force is not orthogonal to the velocity, hence the magnitude of the velocity can increase, but, in particular, I really don't see how the magnitude of the component ##v_{\theta}## can change, provided the fact that the force is radial.

    In this picture it is clear that the magnitude of the velocity (which is tangential to the ellipse, of course) changes, but I can't understand the change in ##v_{\theta}##.

    View attachment 99029
     
  2. jcsd
  3. Apr 13, 2016 #2

    BiGyElLoWhAt

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    No, it doesn't mean that there is a theta component. It means that the initial velocity is v_theta. If there was an acceleration along theta, angular momentum wouldn't be conserved. There would be a non-zero net torque.
     
    Last edited by a moderator: Apr 13, 2016
  4. Apr 13, 2016 #3

    BiGyElLoWhAt

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    What does v equal in polar coordinates?
     
  5. Apr 13, 2016 #4

    Nugatory

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    Before you take on elliptical orbits in a central force, consider the simpler case of no force at all. What is the equation of motion of a particle that starts at the point ##(\theta=0,r=1))## with a speed ##v## in the tangential direction? It's not subject to any force or acceleration at all, yet neither the ##\theta## nor the ##r## components of its velocity are constant.

    Cartesian coordinates have the nice property that the basis vectors are functions of neither position nor time, so when you rewrite ##\vec{F}=m\vec{a}## as the differential equation ##\frac{d^2\vec{r}(t)}{dt^2}=\frac{\vec{F}}{m}## it simplifies into ##\frac{d^2A_x(t)}{dt^2}=\frac{F_x}{m}## (and likewise for the y and z components) and you can safely conclude that if ##F_x=0## then ##v_x## will be a constant. More general coordinate systems don't necessarily work that way.
     
  6. Apr 13, 2016 #5
    vθ not being constant does not mean that the acceleration in the theta direction is not zero.
     
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