Centrifugal forces don't exist in reality?

[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."

 

[Mister T]

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."

 

[Mister T]

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?

 

[Dale]

isn't that contradictory

No. I am not sure why you think it would be 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?

No, this is definitely wrong. You do not need to have outward radial acceleration. You only need to have inward radial acceleration that is less than the usual circular centripetal acceleration at that radius.

Also, note that the tangential acceleration plays a role. If you have a circular path then your outward velocity is a fixed constant 0. If you also accelerate tangentially at a constant rate then your inward acceleration will be continuously increasing.

I encourage you to work a few of these problems for yourself. Start with some standard spiral equations and simply calculate the outward components of velocity and acceleration.

 

[PeterDonis]

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?

You can't decompose things this way; "radial" is not always the same direction, because we're working in an inertial frame. If you want to have a workable separation into "radial" and "tangential", you need to work in a non-inertial frame with time-dependent angular velocity relative to an inertial frame. In such a frame, as I said in post #46, there is an outward coordinate acceleration. But not in an inertial frame.

In an inertial frame, the way to think about it is this: "no acceleration" corresponds to a straight line, but the path in @Dale's diagrams is never a straight line. It is always, as @Dale pointed out, concave inward. That means there must always be an inward acceleration, causing the path to deviate inward from a straight line. The inward acceleration simply has a magnitude which is not quite large enough to keep the path an exact circle; instead it's a spiral.

 

[Delta2]

Thanks for your replies @Dale and @PeterDonis but there is something i don't understand

The calculations that @Dale did are made in an inertial frame right?
In that inertial frame we see that the radial component $v\cdot\hat r$ of velocity is increasing. Hence there must be a radial component of acceleration (in the same direction as velocity), hence there must be a net radial force(which should be a real or interaction force because we are in an inertial frame) in the same direction as velocity. Simple as that.

 

[A.T.]

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?

Consider a mass on a string in circular motion, then you cut the string and the mass flies off tangentially in a straight line. The acceleration is zero but the radial velocity increases.

 

[A.T.]

In that inertial frame we see that the radial component $v\cdot\hat r$ of velocity is increasing. Hence there must be a radial component of acceleration...

No, because the radial vector is rotating, not fixed in the inertial frame.

 

[Delta2]

Ok thanks @A.T. i think i understand now, by decomposing the velocity into radial component is like we "silently working" into a rotating non-inertial frame.

 

[DrStupid]

Their acceleration is inwards at all times.

That could erroneously be read as inward radial acceleration only. Maybe it's just me, but that would have confused me if I wouldn't know how the device works.

 

[Delta2]

That could erroneously be read as inward radial acceleration only. Maybe it's just me, but that would have confused me if I wouldn't know how the device works.

There is also tangential acceleration (in the inertial frame) if that's what you mean. This tangential acceleration is rotating in the inertial frame, (while it is not rotating in the rotating frame) and i believe is responsible for the increasing radial velocity. That's at least how i understand it. In the rotating (non inertial) frame it is the fictitious centrifugal force that is responsible for this.

 

[A.T.]

This tangential acceleration is rotating in the inertial frame, (while it is not rotating in the rotating frame) and i believe is responsible for the increasing radial velocity.

See my simple example with an inertially moving mass in post#54. There is no acceleration at all, and yet the radial velocity is increasing. So I don't think you can blame the tangential acceleration for that.

 

[Delta2]

See my simple example with an inertially moving mass in post#54. There is no acceleration at all, and yet the radial velocity is increasing. So I don't think you can blame the tangential acceleration for that.

Yes you right in post #54,however i have another example in my mind , i ll tell you what exactly i have in my mind and you tell me where i go wrong:

I am thinking of the ball inside a pipe problem:

Suppose we have a pipe of some diameter and we insert inside the pipe a ball with the same diameter. Assume that there are no frictions and that the ball is sitting stationary at the center of the pipe. Then we rotate the pipe with some angular acceleration with respect to one end of the pipe. Then the ball gets ejected from the other end with some radial velocity right? The ball starts from rest and ends up with some velocity, and hence kinetic energy. All of these in the inertial reference frame. An the question is : What force is responsible for the increase in kinetic energy of the ball? My answer is that it is the normal force from the walls of pipe that acts always in the tangential direction. Where am i wrong?

 

[DrStupid]

My answer is that it is the normal force from the walls of pipe that acts always in the tangential direction.

That's what I mean above. It doesn't work without the tangential component of the acceleration. The inward radial acceleration can't do the job. It just prevents the ball (or the weights in the centrifugal clutch) from moving outside even faster.

 

[Dale]

the radial component v⋅r^ of velocity is increasing. Hence there must be a radial component of acceleration (in the same direction as velocity),

This is simply wrong. I don’t know how it can be more clear. You should work several of these problems yourself. Sometimes there is simply no substitute for working it out.

 

[Delta2]

This is simply wrong. I don’t know how it can be more clear. You should work several of these problems yourself. Sometimes there is simply no substitute for working it out.

Yes it is wrong and @A.T. made me realized it with his posts #54 and #55.

 

[Dale]

An the question is : What force is responsible for the increase in kinetic energy of the ball? My answer is that it is the normal force from the walls of pipe that acts always in the tangential direction. Where am i wrong?

Nothing wrong with that.

 

[etotheipi]

Suppose we have a pipe of some diameter and we insert inside the pipe a ball with the same diameter. Assume that there are no frictions and that the ball is sitting stationary at the centre of the pipe. Then we rotate the pipe with some angular acceleration with respect to one end of the pipe. Then the ball gets ejected from the other end with some radial velocity right? The ball starts from rest and ends up with some velocity, and hence kinetic energy. All of these in the inertial reference frame. And the question is : What force is responsible for the increase in kinetic energy of the ball? My answer is that it is the normal force from the walls of pipe that acts always in the tangential direction. Where am i wrong?

If the force is purely tangential, then one has $\vec{F} = F_{\theta} \hat{\theta}$. Into Newton II: $$F_{\theta} = mr\ddot{\theta} + 2m\dot{r}\dot{\theta}$$But strangely, I can also write $$(\vec{\tau})_z = (\vec{r} \times \vec{F})_z = rF_{\theta} = mr^2 \ddot{\theta}$$which seems to suggest $F_{\theta} = mr\ddot{\theta}$ (if we assume the particle starts at $r>0$). This is odd, because $\dot{r}\dot{\theta}=0$ follows; and since $\dot{\theta} > 0$, we must have $\dot{r} = 0$, which implies the ball's radial coordinate will remain fixed...

I've made a mistake somewhere (because for circular motion we evidently need a radial component of force) but I can't see where

 

[jbriggs444]

My answer is that it is the normal force from the walls of pipe that acts always in the tangential direction. Where am i wrong?

The normal force always acts in a tangential direction, yes. But the tangential direction keeps changing. As it does, tangential velocity becomes radial velocity.

That's the Coriolis effect.

 

[vanhees71]

Consider a mass on a string in circular motion, then you cut the string and the mass flies off tangentially in a straight line. The acceleration is zero but the radial velocity increases.

The velocity after the cut is constant due to Newton's 1st (and 2nd) Law. It doesn't increase (what ever this means for a vector ;-)).

 

[jbriggs444]

The velocity after the cut is constant due to Newton's 1st (and 2nd) Law. It doesn't increase (what ever this means for a vector ;-)).

The radial component of the velocity increases. Which, of course, does not contradict your correct statement.

 

[A.T.]

The velocity after the cut is constant due to Newton's 1st (and 2nd) Law. It doesn't increase (what ever this means for a vector ;-)).

I wrote "radial velocity", not just "velocity ". This is the context:

...the radial component $v\cdot\hat r$ of velocity is increasing...

 

[vanhees71]

The particle goes on with the velocity at the moment you cut the string. At this moment the velocity was perpendicular to the radius vector since the tangent of a circle is always perpendicular to the radius...