Centrifugal forces don't exist in reality?

[Aeronautic Freek]

We can often hear that centrifugal force don't exist in reality...Helicopter mi-26 can lift 56tons ,it has 8 blades,so each blade hold 7 tons of force!
Do you know if you put 7tons at blade when blade is not rotating(static) ,bending moment will be way too much and blade will broke at root.

But when blade is rotating centrifugal force "straighten" blade so centifugal force basicaly "reducing" bending moment and this is reason why so tiny blades can hold so much load on it..
So how you can say,that centrifugal force don't exist?

centifugal force has outward direction so it cause tension in blade..

 

[Nugatory]

So how you can say,that centrifugal force don't exist?

Because it doesn't. The force on the blades is centripetal force, an inwards force applied by the fastenings at the hub. This force is pulling the blades into a circular path; withou that force they'd go flying off in a straight line at a constant speed just like Newton's first law says they should.

But you are right that the tension in the blades is what straightens them out - it's just that the tension comes from the inwards-directed centripetal force and not the fictitious centripetal force.

 

[Ibix]

This has come up before. There's a seventeen page argument about it here if you want to read. Its link is broken, but appears to reference this Wikipedia page.

I think the OP is referring to the "reactive centrifugal force". In the case of a ball on the end of a string being swung around, the string exerts a centripetal force on the ball and the ball exerts a "reactive centrifugal force" on the string. That's a real force (if you accept the analysis - I haven't finished reading the links myself).

The centrifugal force that emerges from transforming to a rotating frame is an inertial (also called fictitious) force and is sometimes said not to exist (although I don't think that's helpful). Either way, it's a different thing.

 

[Aeronautic Freek]

Because it doesn't. The force on the blades is centripetal force, an inwards force applied by the fastenings at the hub. This force is pulling the blades into a circular path; withou that force they'd go flying off in a straight line at a constant speed just like Newton's first law says they should.

But you are right that the tension in the blades is what straightens them out - it's just that the tension comes from the inwards-directed centripetal force and not the fictitious centripetal force.

force with inward direction (so called centripetal)will cause compression of blade,in reality blade is in tension so force has outward direction..
if you put load-cell between hub and blade it will show tension not compression..

 

[Dale]

force with inward direction (so called centripetal)will cause compression of blade

This is only true if the blade is not accelerating. A rotating blade is accelerating

 

[Ibix]

force with inward direction (so called centripetal)will cause compression of blade,in reality blade is in tension so force has outward direction..
if you put load-cell between hub and blade it will show tension not compression..

That's not measuring the centrifugal force, at least not in the sense of the words "centrifugal force" people mean when they say it doesn't exist. That would be measuring either the centripetal force or the reactive centrifugal force, depending on how you wish to interpret it.

 

[Aeronautic Freek]

This is only true if the blade is not accelerating. A rotating blade is accelerating

load cell will show tension when blade is rotating,never compression.
when blade is rotating it is little bit longer than when is not rotating,because centrifugal force cause tension in blade structure..

 

[Ibix]

load cell will show tension when blade is rotating,never compression.
when blade is rotating it is little bit longer than when is not rotating,because centrifugal force cause tension in blade structure..

Nobody is disagreeing with this, except for your assertion that this is due to the centrifugal force. There are two things you could mean by that. One meaning is the "reactive centrifugal force", and you are correct that this is a real force. This is not the force that people say doesn't exist. The other meaning of centrifugal force is a coordinate dependent effect that arguably "isn't real", and certainly doesn't appear in the inertial frame you appear to be using. You would be wrong to attribute anything to this.

You seem to be arguing against a straw man. Your example of "centrifugal force" is not an example of the "centrifugal force" that is said to be not real.

 

[Dale]

load cell will show tension when blade is rotating,never compression.
when blade is rotating it is little bit longer than when is not rotating,because centrifugal force cause tension in blade structure..

Agreed, but that fact is not incompatible with what @Nugatory and @Ibix said. We all agree that the blade is in tension.

What you said that is incorrect is that an inward force implies compression. That is not correct.

If you imagine cutting the blade at any point along its length you wind up with two surfaces. One surface is the outward facing surface of the part of the blade attached to the center, and the other surface is the inward facing surface of the part of the blade that is “free”. In tension the second surface has a force pointing inward, this is the real centripetal force and it is an inward force that is present in tension.

If you had an inward pointing force on the first surface then that would be compression.

 

[A.T.]

force with inward direction (so called centripetal)will cause compression of blade,in reality blade is in tension so force has outward direction.

Every part of the blade (except the last bit) has a inwards force (from the next inner part) and a smaller outwards force (from the next outer part), The net force from these two real forces is inwards, to allow circular motion in the inertial frame.

The outwards force, that people say doesn't exist, comes into play in the co-rotating frame where the blade is at rest. It cancels the net real force inwards, to make Newton's 2nd Law work in that frame. It's only used to explain the acceleration (or the lack of it) in the non-inertial frame. It has nothing to do with the measurable tensions, which are frame invariant.

 

[FactChecker]

centifugal force has outward direction so it cause tension in blade..
View attachment 264167

How and why are you distinguishing between centrifugal force and linear inertia? Why aren't you just calling this linear inertia? The blade mass wants to go straight due to linear inertia, but the connection to the rotor forces it to curve. As @Nugatory said, the only true force is the centripital force.

 

[Halc]

Centrifugal force is real in a rotating reference frame, causing stationary objects to accelerate without external force being applied, or to have weight where they are prevented from accelerating. Coriolis force is a similar effect where angular acceleration can be achieved without application of torque. The discussion above is all using inertial reference frame in which yes, centrifugal force is a psuedo-force.

 

[Nugatory]

Centrifugal force is real in a rotating reference frame...

It's not exactly wrong to say this (that depends on what you mean by "real") but if you're going to take this position you are also accepting that fictitious forces are exactly those that are "real" in a non-inertal frame... so a force can be fictitious and real at the same time and we're left wondering what the value of the word "real" is.

It is much more helpful and consistent with mainstream practice to consider "real" things to be those that cannot be made to go away by the mathematical trick of a coordinate transform.

 

[Dale]

Or just avoid the word “real” altogether.

 

[Nugatory]

Or just avoid the word “real” altogether.

Which is why we have the word "invariant", of course.

 

[Halc]

Well what adjective do we use to label a non-pseudo force?
Centrifugal is only meaningful in a rotating frame? I'm not trying to make any ontological claim here.

so a force can be fictitious and real at the same time

But not in the same way, so it isn't a contradiction.

 

[Herman Trivilino]

if you put load-cell between hub and blade it will show tension not compression..

Of course it will. That's why the blade is fastened to the hub.

But that is not relevant. The hub exerts a centripetal force on the blade and the blade exerts a centrifugal force on the hub. Newton's Third Law!

Note, though, that the net force on the hub is centripetal. It's made of a material strong enough to withstand the centrifugal exerted on the hub by the blade yet still remain intact.

 

[weirdoguy]

The discussion above is all using inertial reference frame in which yes, centrifugal force is a psuedo-force.

In inertial frame centrifugal force is non-existent, so I don't see any point in calling it a "pseudo-force in inertial frame".

 

[A.T.]

Well what adjective do we use to label a non-pseudo force?

Using less overloaded terms helps to avoid that whole philosophical mess about what is real/fictitious/pseudo:

Interaction force : frame invariant, obeys Newtons 3rd Law
Inertial force : exist only in non-inertial frames, frame dependent, doesn't obey Newtons 3rd Law

 

[Aeronautic Freek]

Every part of the blade (except the last bit) has a inwards force (from the next inner part) and a smaller outwards force (from the next outer part), The net force from these two real forces is inwards, to allow circular motion in the inertial frame.

The outwards force, that people say doesn't exist, comes into play in the co-rotating frame where the blade is at rest. It cancels the net real force inwards, to make Newton's 2nd Law work in that frame. It's only used to explain the acceleration (or the lack of it) in the non-inertial frame. It has nothing to do with the measurable tensions, which are frame invariant.

But if I called centrifugal force "reactive centrifugal force" than everything is OK?

Imagine ball connect to the blade in the way that can slide in and out .When blade start rotate ball will moves outward.
What is force which push ball outward in inertial frame of reference?
Or how centrifugal clutch works if centrifugal force "dont exist"(whatever that mean)?

 

[A.T.]

But if I called centrifugal force "reactive centrifugal force" than everything is OK?

There are two different outward forces mentioned in my post. They are not the same thing and must not be conflated, so you need two different names for them. I don't care which you use, as long you have two different ones.

Imagine ball connect to the blade in the way that can slide in and out .When blade start rotate ball will moves outward.
What is force which push ball outward in inertial frame of reference?

Moving in a straight line also brings you away from the center (outward), but you don't need a force to move in a straight line in an inertial frame.

 

[Dale]

But if I called centrufigal force "reactive centrifugal force" than everything is OK?

The inertial centrifugal force is a fictitious force. The reactive centrifugal force is a real force. They are different things, but if you use each correctly then both are “OK”

Imagine ball connect to the blade in the way that can slide in and out .When blade start rotate ball will moves outward.
What is force which push ball outward in inertial frame of reference?

In an inertial frame the ball never accelerates outward. Don’t confuse “move” with “accelerate”. Forces are required to accelerate, not to move. There is nothing inconsistent with Newton’s laws in moving outward without an outward force as long as there is never any outward acceleration.

 

[Aeronautic Freek]

In an inertial frame the ball never accelerates outward. Don’t confuse “move” with “accelerate”. Forces are required to accelerate, not to move. There is nothing inconsistent with Newton’s laws in moving outward without an outward force as long as there is never any outward acceleration.

Do you know how centrifugal clutch works?

if i put ball close to blade root when blade is not rotating,than start rotate blade,ball will accelerate outward,so which force push ball outward?

 

[A.T.]

...wich force push ball outward?

In which frame? What is the acceleration of the ball in that frame that needs to be explained by a force?

 

[Aeronautic Freek]

In which frame? What is the acceleration of the ball in that frame that needs to be explained by a force?

i know that in non-inertial frame is centrifugal force..
so i ask for inertial frame..

 

[A.T.]

so i ask for inertial frame..

What is the acceleration of the ball in that frame that needs to be explained by a force?

 

[PeroK]

Do you know how centrifugal clutch works?

if i put ball close to blade root when blade is not rotating,than start rotate blade,ball will accelerate outward,so which force push ball outward?

The contact force with the blade pushes the ball outwards. If you put a ball at rest on a frictionless turntable, then the ball stays at rest (in an inertial frame) when the turntable rotates. If there is friction, the friction force pushes the ball as the turntable rotates.

 

[Dale]

if i put ball close to blade root when blade is not rotating,than start rotate blade,ball will accelerate outward,so which force push ball outward?

The ball never accelerates outward. It only accelerates tangentially.

And no, I don’t know how a centrifugal clutch works. The fact that it has the word “centrifugal” in the name doesn’t change anything I said above.

 

[PeroK]

The ball never accelerates outward. It only accelerates tangentially.

... the ball is constrained to move in the radial direction (relative to the rotating blade). To remain at rest (relative to the blade) would require a centripetal acceleration. If there is insufficient friction for this, then the ball has no option other to accelerate outwards relative to the blade.

The real force is a tangential force in the rotating frame, but this tangential direction is constantly changing in an inertial frame, which results in constrained radially outward motion.

 

[Nugatory]

Do you know how centrifugal clutch works?

I'm not @Dale whom you are asking, but I can say with confidence that the the answer to that question will be "Yes".

if i put ball close to blade root when blade is not rotating,than start rotate blade,ball will accelerate outward,so which force push ball outward?

If the ball is unconstrained (no friction, not in a trough that forces it to move along the length of the blade, no mechancal connection to the blade, ...) it won't accelerate outwards. It will stay put (Newton's first law) while the blade moves out from under it, and then it will fall straight to the ground.

In any realistic situation, there will be some frictional force on the ball as it wants to stay put while the blade moves under it (this is how we can use flatbed trucks to move things around). This force will have a radial outwards component and that's what pushes the ball outwards.

 

[jbriggs444]

And no, I don’t know how a centrifugal clutch works.

A centrifugal clutch is found for instance on go-karts or small motorcycles. You have a rotating assembly that is attached to the motor, often directly to the motor shaft. It spins with the shaft. This assembly contains a pair of weights which are spring-loaded so that, at rest, both are together hugging the motor shaft.

Around this assembly is a steel drum which is attached to the drive mechanism. Often a chain ring is bolted on and a drive chain runs from here to the rear wheel.

When the motor is idling, the rotation rate of the clutch assembly is insufficient to drive the weights apart. The force of the spring is sufficient to keep them together. When the motor is revved up, the rotation rate increases, the weights move apart and contact the drum. The drum spins and the kart or bike takes off.

https://en.wikipedia.org/wiki/Centrifugal_clutch

It is a convenient design since it engages and disengages automatically with no risk of stalling the engine.

As has been noted, when viewed from the inertial frame the weights move in a near-circular path. There is some degree of outward spiral as the clutch engages, but the acceleration of both weights is always centripetal.

 

[Aeronautic Freek]

A centrifugal clutch is found for instance on go-karts or small motorcycles. You have a rotating assembly that is attached to the motor, often directly to the motor shaft. It spins with the shaft. This assembly contains a pair of weights which are spring-loaded so that, at rest, both are together hugging the motor shaft.

Around this assembly is a steel drum which is attached to the drive mechanism. Often a chain ring is bolted on and a drive chain runs from here to the rear wheel.

When the motor is idling, the rotation rate of the clutch assembly is insufficient to drive the weights apart. The force of the spring is sufficient to keep them together. When the motor is revved up, the rotation rate increases, the weights move apart and contact the drum. The drum spins and the kart or bike takes off.

View attachment 264278
https://en.wikipedia.org/wiki/Centrifugal_clutch

It is a convenient design since it engages and disengages automatically with no risk of stalling the engine.

As has been noted, when viewed from the inertial frame the weights move in a near-circular path. There is some degree of outward spiral as the clutch engages, but the acceleration of both weights is always centripetal.

and how you will explain someone what push weights outward every time when engine increase RPM?

 

[Dale]

and how you will explain someone what push weights outward every time when engine increase RPM?

In an inertial frame the weights are never pushed outwards. Their acceleration is inwards at all times.

 

[Aeronautic Freek]

In an inertial frame the weights are never pushed outwards. Their acceleration is inwards at all times.

How do you mean accelerate inwards?if weights accelarte inward,like you said than spring will compressed and clucth will never transmit power from engine to wheel?

every time when you increase enigne RPM,weights moves outward,stretch spring and transmit power from engine to wheel
i am talking about real life situation what happened with weights in clucth..

open in youtube..

 

[Dale]

if weights accelarte inward,like you said than spring will compressed
...
every time when you increase enigne RPM,weights moves outward

You are again confusing “moving” with “accelerating”. The clutch moves outward, but at all times the acceleration is inwards. There is never any outward push because there is never any outward acceleration.

 

[Nugatory]

How do you mean accelerate inwards?if weights accelarte inward,like you said than spring will compressed

Not so... whether the spring is stretched or compressed depends on the velocity of the weights relative to the shaft, not the acceleration. It’s easy to miss this distinction if you choose to use the rotating frame of reference because in that frame both the acceleration and velocity are directed radially. It’s clearer (as are most problems) when you use an inertial frame: the weights’ velocity is tangential so the spring stretches until the clutch engages while the acceleration is still radial.

 

[Aeronautic Freek]

You are again confusing “moving” with “accelerating”. The clutch moves outward, but at all times the acceleration is inwards. There is never any outward push because there is never any outward acceleration.

weight start moving outwards when RPM is increased but accelration is inward?

or you just want to tell that weights in cluth can not be analyze from inertial frame,becuase it results show that clucth will never works correct?

 

[Nugatory]

weight start moving outwards when RPM is increased but accelration is inward?

or you just want to tell that weights in cluth can not be analyze from inertial frame,becuase it results show that clucth will never works correct?

No, when you analyze it an inertial frame the springs stretch and the clutch works.

 

[Lnewqban]

How do you mean accelerate inwards?if weights accelarte inward,like you said than spring will compressed and clucth will never transmit power from engine to wheel?

every time when you increase enigne RPM,weights moves outward,stretch spring and transmit power from engine to wheel
i am talking about real life situation what happened with weights in clucth..

You have to force those weights to go around the shaft of the clutch.

If you liberate the weights from their springs and pivots, and remove the wheel, when you accelerate the engine, each of those will go flying away from the center.
That is exactly what would happen to our planet without having the gravitational force that keeps it rotating around the Sun.

Copied from
https://www.britannica.com/biography/Isaac-Newton/The-Principia

"The mechanics of the Principia was an exact quantitative description of the motions of visible bodies. It rested on Newton’s three laws of motion:
(1) that a body remains in its state of rest unless it is compelled to change that state by a force impressed on it;
(2) that the change of motion (the change of velocity times the mass of the body) is proportional to the force impressed;
(3) that to every action there is an equal and opposite reaction.

The analysis of circular motion in terms of these laws yielded a formula of the quantitative measure, in terms of a body’s velocity and mass, of the centripetal force necessary to divert a body from its rectilinear path into a given circle. When Newton substituted this formula into Kepler’s third law, he found that the centripetal force holding the planets in their given orbits about the Sun must decrease with the square of the planets’ distances from the Sun."

 

[Herman Trivilino]

weight start moving outwards when RPM is increased but accelration is inward?

Suppose you are driving north in your car when you apply the brakes. The motion of the car continues to be northward, but the direction of the acceleration is southward.

In Newtonian physics the direction of the net force is always the same as the direction of the acceleration. You need a net force in the centripetal direction to get a centripetal acceleration.

Here's a scenario for you. During the spin cycle, water is separated from the clothes in the washing machine. There's a centripetal force exerted on the clothes by the spinning drum, but each of the small holes in the drum is a place where the drum cannot exert a centripetal force on the water, so the water goes through the hole in the drum. But the water is not pushed outward by the spinning drum. It's not even in contact with the drum. It moves off in a tangential direction because that is the direction it was moving in before it escaped through the hole.

 

[A.T.]

or you just want to tell that weights in cluth can not be analyze from inertial

You cannot analyze them from any frame, until you have learned the basics, like the difference between velocity and acceleration, or between Newtons 2nd and 3rd law.

 

[Aeronautic Freek]

Suppose you are driving north in your car when you apply the brakes. The motion of the car continues to be northward, but the direction of the acceleration is southward.

but weight A don't brakes ,it accelarate outward..

when engine idle, t1 weight A stay at 2cm from center
now we push throttle and engine increase RPM
so in t2 weight is at 15cm from center.

weight A accelarate from 2cm to 15cm in time t2-t1,so weight A accelarate outward and velocity is also outward
isnt it?

 

[Dale]

weight start moving outwards when RPM is increased but accelration is inward?

Yes.

or you just want to tell that weights in cluth can not be analyze from inertial frame,becuase it results show that clucth will never works correct?

You can always use any frame for analysis. All physical observables are independent of the reference frame.

 

[Lnewqban]

Copied from
https://www.etymonline.com/word/centrifugal

"Centrifugal (adj.):
"Flying off or proceeding out from a center" 1690s, with adjectival suffix -al + Modern Latin centrifugus, 1687, coined by Sir Isaac Newton in "Principia" (which is written in Latin), from Latin centri-, alternative combining form of centrum "center" (see center (n.)) + fugere "to flee" (see fugitive (adj.)). Centrifugal force is Newton's vis centrifuga."

 

[Herman Trivilino]

weight start moving outwards when RPM is increased but accelration is inward?

No, but it spends all its time thereafter with the acceleration pointing inward. Note that it moves outward but it's slowing down, therefore the acceleration is inward.

You are aware that when the velocity and acceleration are in opposite directions, the object slows down?

 

[PeterDonis]

weight start moving outwards when RPM is increased but accelration is inward?

Before this question can even be answered, you need to get clear about two distinctions that you appear to be failing to make. First, you need to distinguish inertial from non-inertial frames; second, you need to distinguish coordinate acceleration from proper acceleration.

The second distinction is easier to see conceptually. Coordinate acceleration is simply the second time derivative of position with respect to whatever frame (inertial or non-inertial) you are using. Thus, coordinate acceleration is frame-dependent. Proper acceleration is acceleration that is actually felt as "weight" (or measured by an accelerometer). Thus, proper acceleration is frame-independent.

Now for the first distinction. In an inertial frame, analysis is easier because coordinate acceleration can only be caused by proper acceleration; i.e., an object will only have coordinate acceleration if it is being acted on by a force that it actually feels as weight--proper acceleration. So the two always go together in an inertial frame.

In the case of the so-called centrifugal clutch, in an inertial frame, when the engine RPM increases, the weight's tangential speed increases, which means it has coordinate/proper acceleration in the tangential direction (from the engine itself, transmitted by the clutch assembly that is constraining the weight to stay in its channel). This increase in tangential speed causes the radius of the circular path the weight follows to increase; but there is no outward acceleration in the radial direction in an inertial frame, because the change in the radius of the weight's circular path is entirely accounted for by its increase in tangential speed and hence its tangential acceleration. The only radial acceleration is from the force applied to the weight by the spring, which is inward.

You appear to be analyzing this scenario in a rotating, non-inertial frame, and thinking of the weight's motion as purely radial; however, that actually cannot be the case for a rotating frame with fixed angular velocity, because the RPM, and hence the angular velocity of the weight relative to an inertial frame, is not constant. So, for example, if we use a rotating frame with angular velocity equal to the starting (slower) RPM, the weight starts out at rest in this frame; but when the RPM speeds up, the weight moves tangentially, not just radially, because the weight's angular velocity, relative to an inertial frame, is now faster than the angular velocity of the rotating frame, relative to an inertial frame. So this analysis ends up looking much like the analysis done above for the inertial frame: the weight's tangential acceleration accounts for the increase in its radial coordinate in this non-rotating frame.

If, instead, we use a rotating frame with angular velocity equal to the ending (faster) RPM, the weight does not start out at rest in this frame; it is moving tangentially, in a retrograde direction (i.e., opposite to the direction of rotation of the frame relative to a non-inertial frame), and the tangential acceleration induced by the RPM change of the engine now causes that retrograde tangential speed to decrease, which again accounts for the increase in the weight's radial coordinate.

However, using non-rotating frames does require observing the distinction between coordinate acceleration and proper acceleration, since they are no longer the same. Even with constant RPM, the weight experiences an inward proper acceleration, due to the spring; and the spring is the only thing that is providing any force that is felt as weight/proper acceleration in the radial direction. And that is the case regardless of which frame we choose to do the analysis in. So in all frames, it is true that the only proper acceleration on the weight is inward. This is why @Dale answered "yes" to the question you posed in what I quoted above: he was taking "acceleration" to mean "proper acceleration", since that's the only kind of acceleration that is not frame-dependent and therefore is telling us about the actual physics of the situation, instead of about our choice of coordinates. (And it also tells us that the clutch works just fine no matter what frame we use, as @Dale said.)

You appear to not be using any of the above frames, but to be thinking of the scenario in terms of a rotating frame with time-varying angular velocity, so that the weight only moves radially in this frame. In such a frame, yes, the weight experiences an outward coordinate acceleration, and this can be thought of as due to the fact that, while the RPM is changing, the inward force of the spring and the outward centrifugal force do not exactly balance. However, it is still true that only the inward force of the spring is felt as weight/proper acceleration, so the proper acceleration is still always inward (thought its magnitude, of course, changes as the spring stretches).

In short, if you are going to say that there is an outward acceleration on the weight, you need to do two things that you have not been doing up to now in this discussion: (1) specify the particular non-inertial frame you are using, and (2) specify that you are talking about coordinate acceleration, not proper acceleration.

 

[Dale]

when engine idle, t1 weight A stay at 2cm from center
now we push throttle and engine increase RPM
so in t2 weight is at 15cm from center.

weight A accelarate from 2cm to 15cm in time t2-t1,so weight A accelarate outward and velocity is also outward
isnt it?

Let’s work this out mathematically in the inertial frame. To simplify things we will assume that there is enough dissipative effects so that the spring is always at its dynamic equilibrium length. Let $k$ be the stiffness of the spring, $m$ be the mass of the clutch, $r_0$ be the unstretched length of the spring, $r$ be the radial position (the length of the spring), $\alpha$ be the angular acceleration, $\omega$ be the angular velocity, and $\theta$ be the angular position. The unit vector in the outward pointing radial direction is $\hat r = (\cos(\theta),\sin(\theta))$.

Setting the centripetal force equal to the spring tension we get $m r \omega^2 = k (r-r_0)$ which immediately gives us $r=\frac{k r_0}{k-m\omega^2}$. For a constant angular acceleration starting from rest we have $\omega = \alpha t$ and $\theta=\frac{\alpha}{2}t^2$ which gives a position $$x=\frac{k r_0}{k-m t^2 \alpha^2}(\cos(t^2\alpha/2),\sin(t^2\alpha/2))$$ then $v=\dot x$ and $a=\dot v$ and finally we obtain $$v \cdot \hat r = \frac{2 k m r_0 t \alpha^2}{(k-mt^2\alpha^2)^2}$$ There is also a closed form expression for $a \cdot \hat r$ but it is too annoying to write.

Setting $r_0$ to 2 cm, $k$ to 100 N/m, $m$ to 1 kg and $\alpha$ to 1 radian/s^2 we can plot the equations graphically. Here is a parametric plot of the position from t=0 s to t=9 s:

Here is a plot of the outward component of the velocity $v \cdot \hat r$

And here is a plot of the outward component of the acceleration $a \cdot \hat r$

Note that the outward velocity is always positive and the outward acceleration is always negative. In other words at all times it is moving outwards and at all times it is accelerating inwards. Since it never accelerates outwards there is no force that pushes it outwards. The idea that it must be accelerating outwards simply because it is moving outwards is not correct in an inertial frame.

 

[Delta2]

ehm sorry for this @Dale, from the above diagrams i can see that the outward velocity is increasing in the positive direction, while the outward acceleration is increasing in the negative direction. How can that be? How can the velocity be increasing in the positive direction, without having an acceleration in the positive direction as well? I mean since the outward acceleration and the outward velocity are on the same line i can write an equation like $$v_r=v_{0r}+a_{r}t$$ right?
If $v_r$ is positive and increasing, then $a_r$ has to be positive as well...

 

[Dale]

ehm sorry for this @Dale, from the above diagrams i can see that the outward velocity is increasing in the positive direction, while the outward acceleration is increasing in the negative direction. How can that be? How can the velocity be increasing in the positive direction, without having an acceleration in the positive direction as well?

Visually, in the first plot, it is already clear that the acceleration is always inwards because the path is always concave inwards. A path with outward acceleration would be concave outward.

As long as a curve has increasing radius it has an outward component of velocity and as long as it is concave inwards it has an inwards component of acceleration. Basically all outgoing spirals share this shape.

As far as how it can be, basically a constant zero inward velocity corresponds to an inward acceleration (the centripetal acceleration), not a zero inward acceleration. An inward velocity is associated with an inward acceleration even larger than the centripetal acceleration and an outward velocity is associated with an inward acceleration less than the centripetal acceleration.

 

[Delta2]

But @Dale, the radial$v\cdot\hat r$ velocity is increasing in the positive (outwards) direction, while the radial acceleration $a\cdot\hat r$ is increasing in negative (inwards) direction, isn't that contradictory. In order for the radial velocity to increase we must have radial acceleration in the same direction as radial velocity, how else does the radial velocity increases?