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For circular motion to happen we need a centripetal force or acceleration i.e. a force perpendicular to the direction of motion at all instants. So without any perpendicular component of force circular motion will not occur.

Does the above statements mean that the circular path is a resultant path of the given velocity and the velocity due to the centripetal acceleration ?

1 : If yes then why is this resultant path always closer to the direction of given velocity and not towards the velocity due to centripetal acceleration ? Also If it is the resultant path then shouldn't the body fall spiralling towards the center as shown in the two figures below ?

2 : If not then why doesn't the particle gain any velocity in the centripetal direction although it has some acceleration in that direction? A horizontally projected body gains some velocity in the direction of $mg$, so a body in a circular motion should also gain some velocity in the centripetal direction.

Assume that all these things (shown in the picture) are happening within an infinitesimal distance . Here $v'$ shows the velocity due to centripetal acceleration and the dot at the center shows the cause of centripetal acceleration and dotted lines represent the path it would have taken with no force on it.

enter image description here

After joining the paths I get this picture

enter image description here

Hope the question is clear.

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To start off, it seems like you are thinking about uniform circular motion, as you are fixated on the centripetal acceleration and are not mentioning anything about tangential acceleration. Therefore, for now let's assume we are talking purely about uniform circular motion.

For circular motion to happen we need a centripetal force or acceleration i.e. a force perpendicular to the direction of motion at all instants. So without any perpendicular component of force circular motion will not occur.

Does the above statements mean that the circular path is a resultant path of the given velocity and the velocity due to the centripetal acceleration?

If by "velocity due to the centripetal acceleration" you mean $\text d\mathbf v=\mathbf a\,\text dt$, then yes; this is just applying the definition of acceleration $\mathbf a=\text d\mathbf v/\text dt$ to the velocity: $$\mathbf v(t+\text dt)=\mathbf v(t)+\mathbf a\,\text dt$$

This is true for all motion, not just circular motion.

Why doesn't the particle gain any velocity in the centripetal direction although it has some acceleration in that direction? A horizontally projected body gains some velocity in the direction of mg, so a body in a circular motion should also gain some velocity in the centripetal direction.

Something to remember is that the centripetal direction changes as the particle goes around the circle. The particle does gain velocity in the centripetal direction, but since the velocity was along circular path the instant before, once the object does pick up this velocity component, that component is no longer completely centripetal. The velocity does change to be more "aligned with" the acceleration vector, but since the acceleration vector is always changing directions the velocity vector will be constantly trying to align with different directions, and so we get the circular motion you describe.

This is different from the projectile case you give, where the vertical direction is a constant direction.

I am adding a picture to show what I think about circular motion. Imagine the picture as a magnified image of very small distance.

Here $v′$ shows the velocity due to centripetal acceleration and the dot at the center shows the cause of centripetal acceleration and dotted lines represent the path it would have taken with no force on it.

There are two issues with your diagram:

1)It looks like your centripetal acceleration is too large

2)It looks like your $\Delta t$ is too big

In order for circular motion to occur, the centripetal acceleration has to be exactly equal to $v^2/r$. It is not sufficient for the acceleration to just have a component perpendicular to the velocity at all points in time. From your diagram, it is obvious that $v'=a\text dt$ is too large.

To explore this more, let's numerically solve the differential equations with a larger than required centripetal acceleration. For planar motion in polar coordinates, we normally have the differential equations $$a_r=\ddot r-r\dot\theta^2$$ $$a_\theta=r\ddot\theta+2\dot r\dot\theta$$

Since we are assuming no tangential forces, let's set $a_\theta=0$. Now, if we correctly made $a_r=-r\dot\theta^2$, then we would be left with $\ddot r=0$, which would give us our uniform circular motion for $\dot r(t=0)=0$. However, let's impose a centripetal acceleration to be just a little larger than this (still dependent on the velocity) $a_c=-1.1r\dot\theta^2$, so we have the differential equation $\ddot r+0.1\cdot r\dot\theta^2=0$.

Solving the system of differential equations $$\ddot r+0.1\,r\dot\theta^2=0$$ $$r\ddot\theta+2\,\dot r\dot\theta=0$$ with initial conditions (dropping units) $r(0)=1$, $\dot r(0)=0$, $\theta(0)=0$, $\dot\theta(0)=1$, we get the trajectory

enter image description here

And we see that we get spiraling inward

Similarly, for $a_c=-.9\,r\dot\theta^2$, we get a trajectory that goes away from the origin

enter image description here

Of course, this is not exactly like your diagram since now the acceleration is not exactly perpendicular to the velocity, but if we had a spiral trajectory where the acceleration was always perpendicular to the velocity then we would have to abandon the assumption of a non-tangential acceleration.

This does however relate to the second issue in your diagram; you are only applying a perpendicular acceleration at set times rather over the entire trajectory. Now, I know we can always approximate the change in velocity as $\mathbf v(t+\Delta t)\approx\mathbf v(t)+a\Delta t$, but if $\Delta t$ is too large, then you are not going to get the correct trajectory.

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  • $\begingroup$ I think my words could not explain my doubt. I have edited the question . Please respond to it. Sorry for the inconveniences . $\endgroup$ – Ankit Sep 28 at 6:42
  • $\begingroup$ the second part is very clearly explained but can you consider the case in which it had no circular path followed on from earlier and just begins the circular motion ? In that case what happens to the velocity due to centripetal acceleration ? $\endgroup$ – Ankit Sep 28 at 6:46
  • $\begingroup$ @Ankit Your scenario isn't specific enough. Do you mean the particle starts at rest? Do you mean it's just moving through space and then you turn on only a centripetal acceleration? $\endgroup$ – BioPhysicist Sep 28 at 6:55
  • $\begingroup$ yes I meant the same which you wrote in your second part. Just like an electron entering a region of uniform magnetic field perpendicular to the direction of motion. $\endgroup$ – Ankit Sep 28 at 6:57
  • $\begingroup$ @Ankit Then it's what I describe in my answer. The velocity does start moving in the direction of the acceleration. But since the direction of the acceleration is then changing, then the direction the velocity is trying to move in will also be changing. Thus you get circular motion. $\endgroup$ – BioPhysicist Sep 28 at 7:00
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Your statement that, "A horizontally projected body gains some velocity in the direction of $mg$" is correct. And the direction of velocity gained is always downward since we have assumed that for short-ranged projectiles, gravity acts in one direction (usually $-\hat j$). Now, for circular motion this in not the cases because here the direction of centripetal acceleration is not fixed.

Also, acceleration is something which changes velocity. It does't mean only the magnitude but also the direction since velocity is a vector quantity. In circular motion, the centripetal acceleration is always perpendicular to the velocity of the particle i.e. no component of centripetal acceleration is along the velocity. Hence, the centripetal acceleration is responsible for only changing the direction of motion (velocity) of the body in the direction of centripetal acceleration.

But, since the direction of centripetal acceleration is not fixed, the direction of velocity of the particle is not fixed, too.

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Does the above statements mean that the circular path is a resultant path of the given velocity and the velocity due to the centripetal acceleration ?

Yes.

1: If yes then why is this resultant path always closer to the direction of given velocity and not towards the centripetal acceleration which will give velocity to the body in the centripetal direction ?

Because, the perpendicular velocity component caused by the centripetal acceleration is tiny, tiny, tiny. In fact, it is negligibly tiny and negligibly short-lived. It has the size of the mathematical ideal that only turns without causing any change in magnitude.

2 : If not then why doesn't the particle gain any velocity in the centripetal direction although it has some acceleration in that direction ? A horizontally projected body gains some velocity in the direction of mg , so body in circular motion should also gain some velocity in the centripetal direction.

The answer again is: because that new velocity component is tiny, tiny, tiny. So tiny and short-lived that as soon as the slightest change happens, which causes turning, this component turns away so as to not change the magnitude.

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  • $\begingroup$ for circle of radius 1m . The velocity after 1second will b $v^2$ which is greater than given velocity $v$. $\endgroup$ – Ankit Sep 28 at 6:47
  • $\begingroup$ That $v^2$ would be the centripetal acceleration (for a radius 1) so it is greater than v but it is not velocity it’s acceleration. $\endgroup$ – Dr jh Sep 28 at 7:09
  • $\begingroup$ I'm sorry, @Ankit, I don't understand your comment here. Under what conditions will the velocity after 1s be squared? And how can a squared velocity ever be an acceleration - the units don't match up. Where does this result come from? $\endgroup$ – Steeven Sep 28 at 7:09
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    $\begingroup$ @Ankit Your picture seems correct now that I can see the indicated accumulation of the paths into a spiralling motion. With non-negligible perpendicular components you will indeed see a spiralling motion as you seem to have drawn. Now imagine those components being smaller due to an acceleration only being present for a shorter time - then the spiralling becomes larger. And for even smaller components, it becomes even larger. Now imagine that the components are zero; you wouldn't see turning at all and ... $\endgroup$ – Steeven Sep 28 at 8:38
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    $\begingroup$ @Steeven it's fun to imagine this scenario . I have never thought that this could happen 😂 and neither my teachers told us about this . Thanks for your comments. I have learnt something new today 🎉🎉😁 . $\endgroup$ – Ankit Sep 28 at 9:20
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This answer has two parts. The first part does the leg work of introducing the complex number method of analyzing planar circular motion and the second shows how it can be used to give simple answers to all your questions.


The set up:

The most transparent and simplest explanation can be done using complex numbers. Consider the following function:

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

This should be a self-explanatory formula if you've done the polar form of complex numbers. Simply consider the polar form but with the magnitude and angle changing as some functions of time. To find the acceleration of the body, we have to twice differentiate this position function.

$$ v(t) = \frac{d}{dt} \big( r e^{ i \theta } \big) = \bigg[ \dot{r} e^{i \theta} +i r \dot{\theta} e^{i \theta } \bigg] $$

And differentiating once more,

$$ a(t) = \ddot{r}e^{i \theta } + i \dot{r} \dot{\theta} e^{ i \theta } + i [\dot{r} \dot{\theta}e^{i \theta } + r \ddot{\theta} e^{i \theta} +i r \dot{\theta}^2e^{ i \theta} ]= \ddot{r}e^{i \theta} - r \dot{\theta}^2e^{i \theta} +i [2 \dot{r} \dot{\theta}e^{i \theta } + r \ddot{\theta} e^{i \theta} ]$$

Or,

$$ a(t)= \ddot{r}e^{i \theta} - r \dot{\theta}^2e^{i \theta} +i [2 \dot{r} \dot{\theta}e^{i \theta } + r \ddot{\theta} e^{i \theta} ]$$


Now with the most general equation in hand, we can address your questions:

  1. Do the above statements mean that the circular path is a resultant path of the given velocity and the velocity due to the centripetal acceleration?

For a circular path, the only condition is that $ \dot{r} = 0$. Applying this consideration onto our formula for acceleration:

$$a(t) = -r \dot{\theta}^2 e^{i \theta} + i[ r \ddot{\theta} e^{i \theta} ]$$

It can be seen that the first term is in the same direction as $r(t)$ and the second term is perpendicular to $r(t)$ [ multiplication by i]. The first term denotes the centripetal 'pull' and the second term term denotes the tangential acceleration.

Also notice that since the path is constant , the $r(t)$ doesn't change and hence velocity is given by:

$$ v(t) = [ i r \dot{\theta} e^{i \theta}] = i \dot{\theta} z(t)$$

Even in the case of non-uniform circular motion, it is easy to find that the velocity is always perpendicular to the position! This is because we want the particle is constrained to move in the curve. You can easily convince yourself that the particle will fall of the curve if it's velocity is not tangent to the curve at all points.

Another way to think about it is, an acceleration at an instant of time will only have its effect play out in the next instant. If the particle has a tangent velocity vector at a point, the centripetal acceleration at that point will 'turn' the velocity as it moves a small arc length of the curve.. but now at this new point, the centripetal acceleration's direction as changed for accommodating the next turn!

To get the case of the spiraling motion which you described, simply stray away from imposing the condition that the distance from the origin is constant! ( i.e $ \frac{dr}{dt} \neq 0 $)


Note: $r(t)$ and $ \theta(t)$ are purely real!!

You can find more of these complex number techniques in Tristan Needham: Visual Complex Analysis

Hope this helps!

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why doesn't the particle gain any velocity in the centripetal direction although it has some acceleration in that direction?

Because every centripetal acceleration is accompanied with centrifugal resisting acceleration :

$$ {\vec a_{~\text{centrifugal}}}=\omega ^{2}{\vec {r}} $$

A horizontally projected body gains some velocity in the direction of $mg$

Yes, because angular velocity component is negligible, so no centrifugal force = falling back to Earth, unless you put so much kinetic energy equaling to gravitational potential energy, so it can escape gravitational field influence and become Earth satellite or a free-walking object in a universe, subject to be captured by other body gravitational field.

EDIT

BTW, your pictures of representation of speed due to centripetal acceleration are wrong. If body, like you pictured, is moving in a spiral trajectory, then this means that centripetal force and acceleration is varying, i.e. not constant. Just try to draw relative curve segment center, and you will see that it will be different,- depending on what segment you will chose :

enter image description here

So it means that with time particle is attracted to different "temporary places" until it converges somewhere (or not).

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  • $\begingroup$ if there is a centrifuge partner then doesn't that mean that the net force on the body is zero and hence it should follow a straight line ? $\endgroup$ – Ankit Sep 28 at 8:33
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    $\begingroup$ You can't add these forces, because they acts on different bodies. Centripetal force is rotating system COM force which attracts particle towards center and reactive centrifugal force is a particle force which induces tension to system COM directed outwards from it. So because these forces forms action–reaction Newton pair, you can't add them. $\endgroup$ – Agnius Vasiliauskas Sep 28 at 9:41
  • $\begingroup$ and the downvote is for... ? $\endgroup$ – Agnius Vasiliauskas Sep 28 at 14:01
  • $\begingroup$ I don't know who downvoted ? $\endgroup$ – Ankit Sep 28 at 14:36

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