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Velocity and acceleration of a particle (polar coords)

  1. Nov 2, 2013 #1
    1. The problem statement, all variables and given/known data
    I am asked to find the radial and transverse velocity and acceleration for a particle with polar coordinates [itex]r=e^t[/itex] and [itex]\theta=t[/itex]

    3. The attempt at a solution
    Therefore let the position of the particle be [itex]\underline{r}=\underline{\hat{r}}e^t[/itex]

    $$\therefore \dot{\underline{r}}=\frac{\mathrm{d} }{\mathrm{d} t}(\underline{\hat{r}})e^t+\frac{\mathrm{d} }{\mathrm{d} t}(e^t)\underline{\hat{r}}$$

    But [itex]\underline{\hat{r}}=\cos\theta\underline{\hat{i}}+\sin\theta\underline{\hat{j}}[/itex]

    $$\therefore\dot{\underline{r}}=\frac{\mathrm{d} \theta}{\mathrm{d} t}\frac{\mathrm{d} }{\mathrm{d} \theta}(\cos\theta\underline{\hat{i}}+\sin\theta\underline{\hat{j}})e^t+e^t\underline{\hat{r}}$$

    $$=(-\sin\theta\underline{\hat{i}}+\cos\theta\underline{\hat{j}})e^t +e^t\underline{\hat{r}}$$

    [itex]\therefore \underline{\hat{r}}=e^t\underline{\hat{\theta}}+e^t\underline{\hat{r}}[/itex]

    This is the particles velocity, the acceleration is given by:
    $$\ddot{\underline{r}}=\frac{\mathrm{d} }{\mathrm{d} t}(\underline{\hat{\theta}}e^t+\underline{\hat{r}}e^t)$$

    $$=\frac{\mathrm{d} }{\mathrm{d} t}(\underline{\hat{\theta}})e^t+\frac{\mathrm{d} }{\mathrm{d} t}(e^t)\underline{\hat{\theta}}+\frac{\mathrm{d} }{\mathrm{d} t}(\underline{\hat{r}})e^t+\frac{\mathrm{d} }{\mathrm{d} t}(e^t)\underline{\hat{r}}$$

    $$=\frac{\mathrm{d} \theta}{\mathrm{d} t}\frac{\mathrm{d} }{\mathrm{d} \theta}(-\sin\theta\underline{\hat{i}}+\cos\theta\underline{\hat{j}})e^t+e^t \underline{\hat{\theta}}+e^t\underline{\hat{\theta}}+e^t\underline{\hat{r}}$$

    $$=-e^t\underline{\hat{r}}+e^t\underline{\hat{\theta}}+e^t \underline{\hat{\theta}}+e^t\underline{\hat{r}}$$

    $$=2e^t\underline{\hat{\theta}}$$

    This is an acceleration in the transverse direction. But if [itex]\theta=t[/itex] and [itex]r=e^t[/itex] then why do I get an acceleration in the transverse direction, and not the radial direction? Is the transverse velocity not constant (I also got [itex]e^t\underline{\hat{\theta}}[/itex] for transverse velocity, why is this?) Help is very much appreciated
     
  2. jcsd
  3. Nov 2, 2013 #2

    verty

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    Homework Helper

    I agree with you that something just doesn't seem right about all this. The first thing is that it seems, because ##{dθ \over dt} = 1##, that the angle between the velocity and the radius should go to zero as t gets large. But the formula suggests the angle is a constant ##θ + {\pi \over 4}##. So this looks highly suspect.

    The second odd thing is that the acceleration is perpendicular to the radius. This seems to imply that the radial velocity is constant, that the acceleration is only adding to the rotational velocity of the vector, but as ##{dθ \over dt} = 1##, this looks doubly wrong.

    After a lot of thought, I am convinced that these calculations are correct. The tangential velocity is proportional to the radius (why?), and the radial velocity is proportional to the radius (why?), so the angle of the velocity must be ##θ + {\pi \over 4}##.

    And the acceleration, being not perpendicular to the velocity, lengthens and rotates it. And because the velocity is at a constant angle to the radius, the radial velocity is lengthened when the velocity is lengthened.

    So the strangeness can be explained way.
     
  4. Nov 2, 2013 #3

    vanhees71

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    Science Advisor
    2016 Award

    I just checked that the calculations done are right by doing everything in Cartesian coordinates. Then the motion is given by
    [tex]\vec{r}(t)=\exp t [\cos t \hat{x}+\sin t \hat{y}].[/tex]
    Since [itex]\hat{x}[/itex] and [itex]\hat{y}[/itex] are constant, you can easily do the derivatives wrt. [itex]t[/itex], and then you can find the components of velocity and acceleration simply by taking the dot products with the polar-coordinates orthonormal basis
    [tex]\hat{r}=\cos t \hat{x} + \sin t \hat{y}, \quad \hat{\theta}=-\sin t \hat{x} + \cos t \hat{y}[/tex]
    along the trajectory. I admit, the solution is not so intuitive but correct :-).
     
  5. Nov 2, 2013 #4
    This makes sense after thinking about it, thanks for your help! It is kinda simple now that you explained it too :p
     
    Last edited: Nov 2, 2013
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