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One can cover circular motion with different speeds at different positions, right? The only aim is to complete his circular motion, right?

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    $\begingroup$ Said where? Your question could use a reference and a quote from the source to clarify the context/origin. $\endgroup$ – user985366 Sep 22 at 12:35
  • $\begingroup$ Most highschool textbooks only deal with scenario which op has described above so $\endgroup$ – Buraian Sep 22 at 18:58
  • $\begingroup$ You can, but you tend to need calculus to do so. Once the system has reached steady state, at sufficiently high RPMs, it's a good approximation since inertia helps keep things constant. $\endgroup$ – DKNguyen Sep 22 at 22:08
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    $\begingroup$ No, it's not. Tangential speed is a vector quantity and vectors do have direction and magnitude. In every circle position speed direction changes, thus speed vector is not a constant. $\endgroup$ – Agnius Vasiliauskas Sep 23 at 9:19
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Circular motion at constant speed is just the simplest type of circular motion, so the first examples of circular motion that you come across often have constant speed to keep them simple. In more complicated examples of circular motion the speed can change with time and with position. And when you progress to orbital mechanics then you have examples of objects in elliptic or hyperbolic orbits where both the speed and the radius of the orbit change with time.

By analogy, the simplest type of linear motion has constant speed - but you can have linear motion with varying speed and constant acceleration or even with varying acceleration - an example of the latter is simple harmonic motion.

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  • $\begingroup$ 'Constant acceleration' If we consider centripetal acceleration ( magnitude is same, direction changes) as the direction is different the acceleration is always varying right ? can u give an example of constant acceleration? $\endgroup$ – Angeline varghese Sep 22 at 14:01
  • $\begingroup$ @Angelinevarghese An example of linear motion with constant acceleration would be an object falling under gravity, or a ball rolling down an inclined plane. $\endgroup$ – gandalf61 Sep 22 at 14:55
  • $\begingroup$ Actually i asked about circular motion $\endgroup$ – Angeline varghese Sep 22 at 16:33
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    $\begingroup$ "Circular motion with constant magnitude acceleration is circular motion at constant speed". This is not true. One could have acceleration with a constant magnitude where the vector always has a tangential component as well. $\endgroup$ – BioPhysicist Sep 22 at 18:36
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    $\begingroup$ You're assuming the tangential component stays constant. One could change the tangential and centripetal accelerations accordingly so that the acceleration maintains a constant magnitude while also maintaining circular motion. Holding the derivative of $a$ to be $0$ for circular motion would mean that $\dddot\theta=-2\dot\theta^3$. $\endgroup$ – BioPhysicist Sep 22 at 20:07
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In uniform circular motion, the velocity has a constant magnitude. In this case the acceleration always points towards the center of the circle (perpendicular to the velocity) and has a magnitude exactly equal to $v^2/r$.

In the more general sense, we can have acceleration that has a tangential component that points along the circle (parallel/anti-parallel to the velocity). In this case we will have a velocity that also changes in magnitude in addition to changing direction.

For the more mathematically inclined reader, the acceleration vector in polar coordinates is

$$\mathbf a=(\ddot r-r\dot\theta^2)\,\hat r+(r\ddot\theta+2\dot r\dot\theta)\hat\theta$$

However, for circular motion it is the case that $\dot r=0$ and $\ddot r=0$, so we have

$$\mathbf a=-r\dot\theta^2\,\hat r+r\ddot\theta\,\hat\theta$$

In uniform circular motion (constant speed), $\ddot\theta=0$ also, and so we get what was stated above, that the acceleration only has a centripetal component equal to $r\dot\theta^2=v^2/r$.

In general $\ddot\theta\neq 0$, and so we have an additional component to the acceleration that is tangent to the circle. In this case the speed is not constant because $\dot\theta=v/r$ is changing. The magnitude of the acceleration vector is

$$a=\sqrt{r^2\dot\theta^4+r^2\ddot\theta^2}$$

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I want to address what you said about the aim being to "complete his circular motion". I assume you mean that the particle is moving in a circle with a constant radius. Let $v$ be the speed of the particle, $a_{||}$ be the component of the acceleration in the direction of the velocity of the particle, and $a_{\perp}$ be the component of the acceleration perpendicular to the velocity.

The condition to not change the radius of an orbit is: $$ a_{\perp} = \frac{v^2}{ r}. $$ If $a_{\perp} < \frac{v^2}{r}$ then there isn't enough force to keep the object in orbit and the radius of the orbit will increase. If $a_{\perp} > \frac{v^2}{r}$ then there is so much force pulling the object in and the radius will decrease.

If we demand the radius $r$ to be constant, then $a_{\perp}$ only depends on the speed of the particle $v$, and the easiest situation is having both of them to be constant. Which is why "they" always say a circular motion keeps the speed constant. However we also see that we can keep the particle with constant radius if we increased $a_{\perp}$ together with $v$. We need to increase them together in this very particular way though such that when $v$ is doubled, $a_{\perp}$ is quadrupled. Note that increasing $a_{\perp}$ means increasing the pull (the force) on the particle in the direction perpendicular to the velocity, so this might not always be possible. For example, increasing the force might be possible if the particle is held in orbit by a string, but will not be possible if the particle is held in orbit by gravity.

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The "aim" of circular motion is indeed to complete to circulate at a constant radius to the source of the force. The speed doesn't have to be constant.

Take a piece of string with a mass attached in your hand. Now give the mass on the string (while its length stays constant, which is a rather idealized piece of string) a circular motion. You can also attach the string to a rotating motor. You can make the circular speed vary by adjusting it with your hand, or even better by varying the speed of the rotation of the motor. But it's easier to use your hand.

When the speed of the mass is getting higher you feel the centrifugal force pulling at your hand getting bigger and vice-versa.

Another example. A mass attached to string (again with constant length). Attach the string to a nail on a table. Give the mass a push and due to the friction force (which is a non-conservative force, though this is not of real importance here), the mass will make a decelerating circular motion.

In both cases, the force on the mass is not directed towards the center of the circle on which the mass moves.

So, it is not always said that the speed in a circular motion is constant. Mostly though this is said in the context of astronomy and the math simplifies for describing a circular motion, but not in the case I mentioned, which is a non-uniform circular motion.

Note that exact circular motion exists only in theory. For the math, see the answer given above by @BioPhysicist.

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The equation modelling the most general kind of spiralling motion in a plane is:

$$ r(t) = A(t) e^{i \theta(t)}$$

Where $ A(t)$ is the radius of circle as a function of time, $ \theta(t)$ is angle of circle as a function of time. Now,

$$ \frac{dr}{dt} = v(t) = \frac{dA}{dt} e^{ i \theta(t) } + iA \frac{d \theta}{dt} e^{ i \theta(t) }$$

Now, the first term in the expression denotes radial velocity $ v_{rdl} (t)$ and the second denotes $ v_{tngt} (t)$.

$$v_{rdl} = \frac{dA}{dt} e^{ i \theta(t) }$$

$$ v_{tngt} = i A \frac{ d \theta}{dt} e^{ i \theta(t) } = i \frac{ d \theta}{dt} r(t)$$

Now, suppose we thought of complex numbers as vectors, we could think of a multiplication of 'i' to analogous to rotating the vector ninety degrees. So, if $ r(t)$ denotes the position vector, multiplying it with 'i' gives a vector perpendicular to it. In other words this denotes tangential component of the velocity.

So, the angle the rate at which the particle traverses the angle can be any function of time. Maybe it can start spinning faster and faster as times goes and could be given as:

$$ \frac{ d \theta}{dt} = e^{t}$$

So, like an exponential increase of angle speed, and hence lead to large tangential velocities as you've noted.

Hope this helps


In the regular circular motion, we say that $ A(t) = A$ and $ \theta(t) = \omega t$ where $ \omega $ is some constant


Reference:

Page-10,20 of Tristam Needham's visual complex analysis book

This website

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    $\begingroup$ Although your solution might be technically right I don't see how introducing complex numbers is appropriate to answer this question. $\endgroup$ – ZeroTheHero Sep 22 at 15:19
  • $\begingroup$ It models the physical situation very well and each part catches all elements of the motion. After op sees this, I can not imagine him having any more cases on the different kind of spiral motion. As for why I just introduced complex numbers, I thought this answer was an opportunity to showcase the use of complex numbers in modelling physics situations.. $\endgroup$ – Buraian Sep 22 at 16:20
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    $\begingroup$ radius of circle ... angle of circle Spirals are not circles. $\endgroup$ – G. Smith Sep 22 at 17:47
  • $\begingroup$ Not a precise explanation but what I meant is a spiral can be thought as a circle with increasing varies as the angle changes $\endgroup$ – Buraian Sep 22 at 17:55

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