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If a body moves such that its net acceleration always points towards a particular point does this body have constant angular velocity around this point?

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When the acceleration always points towards a particular point, then

  • the angular velocity $\frac{\mathrm{d}\theta}{\mathrm{d}t}$ is not necessarily constant,
  • but the areal velocity $\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{1}{2}r^2\frac{\mathrm{d}\theta}{\mathrm{d}t}$ is constant, as shown in this animated image from Kepler's laws of planetary motion - Second law of Kepler.
    Kepler-second-law
    The same (blue) area is swept out in a fixed time period. The green arrow is velocity. The purple arrows are acceleration and its components.
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  • $\begingroup$ My compliments for the choose also of the picture. $\endgroup$ – Sebastiano Jul 18 at 15:56
  • $\begingroup$ I would like to add that the areal velocity is equal to the magnitude of the angular momentum with respect to the point the body is being pulled toward, divided by twice the mass. The fact that it is constant can be seen as a consequence of conservation of angular momentum. $\endgroup$ – Puk Jul 19 at 2:03
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Lets look at the equations of motion:

The position vector $\vec{R}$ in polar coordinate is:

$$\vec{R}=\left[ \begin {array}{c} r\cos \left( \varphi \right) \\ r\sin \left( \varphi \right) \end {array} \right] $$

with the velocity $\vec{v}=\dot{\vec{R}}$ is the kinetic energy:

$$T=\frac{1}{2}\,m\,\vec{v}^T\,\vec{v}=\frac{1}{2}\,m\left(\dot{r}^2+r^2\dot{\varphi}^2\right)$$

the components of the gravitational force $F$ are :

$$F=\frac{G\,M\,m}{r^2}\left[ \begin {array}{c} \cos \left( \varphi \right) \\ \sin \left( \varphi \right) \end {array} \right] $$

with EL methode we get the equations of motion:

$$\ddot{r}=\frac{G\,M}{r^2}+\dot{\varphi}^2\,r\tag 1$$ $$\ddot{\varphi}+\frac{2\dot{\varphi}\,\dot{r}}{r}=0\tag 2$$

form equation (2) we get

$$\frac{d}{dt}(r^2\,\dot{\varphi})=0\,,\quad \rightarrow\quad r^2\,\dot{\varphi}=h\,,\dot{\varphi}=\frac{h}{r^2}$$ where h a constant.

conclusion:

$\frac{d\varphi}{dt}$ is constant only if $r$ is constant (circular path)

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You are thinking of uniform circular motion. However, you are missing some requirements.

Net acceleration/force has to be constant. You can see this from $a=\omega^2R$, where $\omega$ is the angular frequency. So if this needs to be constant, the acceleration has to be too.
Another requirement from this equation is a constant radius - this is why elliptical orbits aren't with constant angular velocity.
                                                 enter image description here (aboutscience.net)

The other requirement is linear velocity that is perpendicular to acceleration. This is derived from a vector diagram. If there is no velocity, the body would move towards the center. If it isn't perpendicular, then it would not be circular and so wouldn't have constant angular velocity.

Given these requirements, and what you mentioned about acceleration pointing inwards (called centripetal acceleration): yes, there will be constant angular velocity in theory.

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Consider the solar planetary motion, it can also be elliptical.

More detailed answer: Thomas Fritsch, Alaz, and Eli has provided detailed answer.

However, just for the fun:

For a constrained case where the acceleration was always perpendicular to the motion:

Consider a pendulum of a finite size, being attracted by a massive point object of finite distance, like the earth. Moving the planet away while increase it's mass $r^2$. Eventually at certain distance, you can see that, though the force is towards a distant point, it's can be treated uniform "downwards". Connecting the arm of the highest point, it's evidential that the angular velocity was not constant.

For a case of your interest: The comet question that has been popular addressed in classical mechanics.

Think about the comet that went around the planet but could not be attracted, at long distance, it's angular velocity approach $0$, while around the planet it's angular velocity was relatively faster.

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  • $\begingroup$ Yes but the angular velocity about the sun is constant $\endgroup$ – Rituraj Tripathy Jul 18 at 12:53
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    $\begingroup$ @RiturajTripathy, When an astronomical body follows an eccentric, elliptical orbit around a much more massive body, it's angular velocity is not constant. You should be able to easily see that if you consider Kepler's second law. $\endgroup$ – Solomon Slow Jul 18 at 13:20
  • $\begingroup$ Bodies in orbit are not necessarily accelerating towards any particular point. $\endgroup$ – Adrian Howard Jul 18 at 14:08

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