Of Balls and Flies: CR4 Challenge (06/10/08)

A fly follows the exact trajectory of the ball, but with a constant speed equal to v .

There seems to be some confusion here... Acceleration can only occur if there is a change in Velocity...

You say constant speed...presumably you are distinguising between velocity and speed....?

As the only difference between the two is direction...then presumably we need to know in what direction you propose to measure the acceleration?

Del

It doesn't say that the fly traveled at the same time as the ball, but that it travelled the same trajectory.

My thought is that the question askes for the angular acceleration that the fly experiences while moving at a constant speed along the arched flight path.

Dwight

The parabolic trajectory at its peak could be aproximated with a cyrcle of radius h(max).

Therefore the radial acceleration would be

a = (v**2)/ h(max) = 2*g/ (cos 45)**2

Hi NaturalU,

You had the same idea for calculation as I did but you have made a slight mistake somewhere:-

If you assume, at any instant in time, the fly to be travelling in a circle, whose radius r creates an equivalent centripetal acceleration, then for any arbitrary speed u, the value of this acceleration a will be u^2/r .

Then the radius of such a circle is given by r = u²/a.

Thus at the top of the projectile's curve the magnitude of the velocity is

v.cos(45) = v/√2

Hence the equivalent circle at this point for the projectile will have a radius,

r = v²/(2.g).

Now since the fly is travelling at speed v all the time we can also say that his acceleration a_fly on this equivalent circle is,

a_fly = v²/r = v² / (v²/(2.g)) = 2g (Same result as Fyz)

Hi Mat Phys Maniac, you are right, I made a mistake taking the vertical component as v* cos(45) instead of v*sin(45).

Your reasoning ignores the ball paradigm presented initially; at the peak of ballistic trajectory the height is h(max) = [v*sin(45)]**2 / 2*g. If at this point the fly trajectory is approximated with a circle of radius = h(max) then

a= (v**2)/h(max) = 2g*(v**2)/[v*sin(45)]**2 = 2g / [sin(45)]**2 = g

I did it again: the answer is actually 4g! since [sin (45)]**2 = 1/2.

General solution is a = 2g/ [sin(alpha)]**2. The graph below shows number of gs vs ball throw angle.

Fine - except for the factor of √2 difference between your approximation to the radius of the circles and the actual value.

The parabola and the circle are tangent at the apex where h(max) = radius. The acceleration value is momentary and specific to this point only. What approximation factor are you suggesting. Could you be more specific?

Thank you.

As the error factor should be 2 rather than √2, I thought it appropriate to insert my oar (again).

Let us consider a simple parabola: y=1/2-x²/2. The derivative (dy/dx) is -2.x, so the slope at y=0 (x=+/-1) is 45^O. The second derivative (d²y/dx²) is everywhere 1; when the gradient is zero (i.e. at x=0, y=1/2), this corresponds to a circle of unit radius. However, the maximum height is 1/2, which is only half this radius.

Thus the radius of curvature at the top of a parabola with a 45-degree launch angle is twice its height. Insert that in your answer, and it aligns with the others.

P.S You could alternatively calculate the radius of curvature at the top based on the horizontal velocity of the ball (=v.cos(θ)), and the known acceleration being g. That also would give r=(v.cos(θ))²/g - i.e. v²/(2.g) for a 45^O launch angle.

Remember from dynamics that "SPEED" is a scalar value and carriers no vector information like velocity does. So whether the fly is moving at a constant speed or the ball, which without air resistance will be at a constant speed, the acceleration of both will be G. Of course this assumes that acceleration is a vector quantity, which it would have to be for the question to make any sense.

You seem to be missing that the ball does not travel at constant speed.

The horizontal velocity will be constant as there is no air resistance. For the fly to match the trajectory, it must match the vertical acceleration of the ball, thus the acceleration is the same.

Well done Fyz,

Nice easy solution without having to worry about fitting an equivalent circle into a parabola as in NaturalU's solution (Although, I now see he/she has made a simple mistake but too late as I awarded that solution a GA too).

However, just to clarify a slightly misleading statement in your answer:-

The speed of the ball or projectile is not constant at the top of the trajectoty - it is in fact a minimum, which you quite rightly point out is vCos45 = v/√2, which is the horizontal component of the projectile at the beginning. The vertical component of the velocity of the projectile at the top is of course zero.

Also, one further point to satisfy Del_The_Cat, The direction of the acceleration of the fly, at all times, is inwards into the parabolic curve and at a right angle to the tangential velocity vector( This centripetal acceleration is what keeps the fly on a curved path). The vector tangential velocity, (as opposed to the speed which is a scalar value) at the top of the curve is obviously horizontal, therefore the acceleration direction of the fly is vertically downwards at that point.

Finally, the acceleration on the projectile is always equal to g and is vertically downwards at all times.

The acceleration of the fly when it reaches the highest point woould be same as that of thing thrown. i.e. 'g" the garvitational acceleration i.e approximately- 9.8 m/sec2.towards the ground. 'g' is same irrespectve of mass. The fly will also follow the trajetcory assuming there is no air resistance.

shivaganti

OK, I admit it, only constant in the sense that d(speed)/dt=0, and that for zero time.

Hi Fyz.

Your calcs are right, except for one interpretation problem. If the accelerations of the ball and the fly would be measured by onboard accelerometers, you have to subtract 1g; hence the ball would register zero g and the fly 1g.

From a straight geometrical point of view, you are right of course, but that interpretation is reference frame dependent. My interpretation is reference frame independent and is called the "proper acceleration".

Jorrie

Hi Jorrie

Very droll. But for those not so au fait with custom and practice:
My accelerations are w.r.t. to the same frame of reference as taken for the speed of the fly. Of course, "Proper Acceleration" has its proper place, but if a clear frame of reference already exists, that would normally be the one to use - unless otherwise requested, or there are strong and apparent reasons for using P.A., in which case I would of course still have to say what I was using [relativistic calculations being a case in point]

Fyz

All accelerometers I have used are really measuring a force on a mass and not really an acceleration. An accelerometer in free fall does measure zero "acceleration" but that is because the instrumented mass and the base it is attached to are both accelerating at the same rate. The mass and the base are really both accelerating at one g relative to the earth. When you are in freefall, you don't experience any sense of acceleration but you are accelerating towards the earth. This is the reason the vomit comet follows the path it does. Both the airplane and the people inside are accelerating at the same rate relative to the earth but have zero acceleration relative to each other. The earth would seem to be a reasonable reference for the ball and the fly.

The figure below shows the balls flight as taken using a strobe. Since there is no air friction or gravity acting in the horizontal direction, the horizontal velocity is constant or √2/2v. No air friction but gravity act in the vertical direction so the vertical velocity is √2/2v - g*t where g is the acceleration of gravity.

At the top of the arc, the vertical velocity is zero so the speed of the ball is just the horizontal velocity or √2/2v. The vertical acceleration at the top is -g. For the fly, the speed at the top will be v since its speed is constant. The centrifugal acceleration is equal to v^2/r where r is the local radius of curvature. Since the speed of the fly is √2 times the speed of the ball at the top, the acceleration of the fly at the top is -2g.

IO don't think we can have centrifugalacceleration here.

Surely the acceleration is in the same direction as the gravitational acceleration? => 2.g

Alternatively, if g is taken as an unsigned scalar and the acceleration is measured by accelerometers on the fly (as per Jorrie) j.g (which assumes the fly is looking in the direction from which the world is coming at it)

Hi Fyz, you wrote: "Surely the acceleration is in the same direction as the gravitational acceleration? => 2.g"

Again, it depends on frame of reference. If you fly an aircraft normally (i.e., upright) and follow the ball's trajectory at the constant v specified, the craft's g-meter will read -1g at the top. If you fly it inverted, it will read +1g at the top.

Jorrie

It's doubtless all arguable, but:

I would have expected the aircraft values to be defined as the "vertical component"of acceleration, and the frame of reference for this "Proper Acceleration" defined by convention (which incidentally invalidates the claimed "frame of reference independence" of "Proper Acceleration" - except as regards its modulus - but I feel that is an argument for another forum).
In any event, this is a situation where the frame of reference is specifically defined, and g itself is implied as a scalar.

If we have a frame of reference where the directions are not specifically defined, then the sign becomes inappropriate unless g is understood to be a vector, the direction of which is downwards - so the acceleration of the ball is simply g (and that of the fly 2.g.)

Ha! A golf ball - did the challenge allow you to neglect lift as well as air resistance?
(but I gave a G.A. nonetheless)

-32.2ft/sec2

Silly Answer = 0.

A fly can not fly if there is no air resistance.

Very Good answer!! :-D

Very good answer.

So I've changed the challenge to be a very dense ball that is very small in diameter so that air resistance can be approximated as zero. How the fly knows where to fly to follow the balls path, I don't know. Surely the fly is not following just behind the ball because the fly would shortly over take the ball and then where would it know where to go.

I've generalize the challenge a little to have the ball make an initial angle of θ with the horizontal. Using the accleration proportional to the local radius of curvature and the square of the velocity ratio, I get the peak acceleration of the fly to be g/cosθ^2. The following figure show this relationship.

This means that the radius of curvature must be smaller at the top for the larger values of θ. Following is a figure showing the ball's (and fly's) path for initial angles of 15, 30. 45, 60, and 75 degrees. Interesting that the horizontal distance travelled when the ball returns to the original elevation for 15 and 75 degrees are the same as are those for 30 and 60 degrees. As expected, the carry distance for the 45 degree launch is largest.

Simple but very true unless the fly is equipped with some type of thrust.

Out side the Box

A constant speed is an acceleration of zero

I think they mean by constant speed that the magnitude of the velocity is not changing. The kinetic energy of the fly would stay constant. However, the direction of the fly will be changing so there will be lateral acceleration.

Of Balls and Flies:

Uh, you lost me somewhere between a fly following the ball AT CONSTANT SPEED and later WHAT IS THE ACCELERATION of the fly. Has physics changed in the past 60 years???

Of course physics has changed in the past 60 years. But not this aspect. Centripetal acceleration has been widely understood since Edmund Halley paid for the publication of Isaac Newton's "Principia".

The acceleration of the fly when it reaches the highest point of the trajectory of the ball is 0

Assuming also, this ball and fly are on earth, where the aceleration, due to gravity, is constant 32 ft/sec/sec. The acceleration is 32ft/sec/sec. The upward velocity would slow until the apex because of the downward pull by gravity. The speed would remain constant until another force was exerted.

I don't know if this is a trick question or not, but since you said that the fly travels at a constant speed, then the fly's acceleration is 0.

Hi blizzard, you wrote: "I don't know if this is a trick question or not, but since you said that the fly travels at a constant speed, then the fly's acceleration is 0."

I don't think it's a trick question and the acceleration of a fly following the same path as the ball at constant speed (not velocity) cannot be zero, because it follows a curved path (parabola) under its own propulsion. It's not in free fall, so there must at least be centrifugal forces.

An on-board accelerometer will measure an inward acceleration profile that starts at zero, peaks at 1g at the top and reduces to zero again (until it hits the ground, that is). In the coordinates of a static (ground) observer, the fly's acceleration will start at 1g, peaks at 2g at the top and reduces to 1g again. This is probably what the author has in mind.

Jorrie

In order for the fly to remain at constant speed while following a parabola or an arc it must have a directional acceleration. In an arc it is constant and towards the center of the circle or the arc(ie: along the radius of the circle). In a parabola, the top of the parabola is the only spot where symmetry prevails and the acceleration is vector points directly down-wards.

if you plotted the acceleration of the fly relative to the fly itself, the graph would also be a parabola. because the fly would initially be accelerating forward and down, as it reaches the top of the parabola the forward acceleration would decrease and the downward would increase, with the acceleration pointing directly down-wards. As the fly passes the top, a backwards acceleration would increase.

If you compare the speed of the ball to the speed of the fly at the top you would see that the balls speed would be (1/sqrt(2))V and the flys would be V

therefore in order for the fly to follow the path of the ball while going sqrt(2) times faster it must accelerate that ratio directly down-wards. Therefore the flys' acceleration would be:

(√2)g

and assuming that we are on the earth and the g= 32.2 ft/s/s then the flys' acceleration would be:

45.54 ft/s/s

This is my first post everyone I hope I did ok :D

Cheers

Hi, LP Ultima!

Very nice first post, and welcome to the engineers' place for mutual enjoyment as well as serious information exchange. Glad to see you in here.

Mark

Not a bad start. But going at √2 times the speed means your velocity change in a given distance is a factor of √2 larger. However, covering that distance takes 1/√2 of the time. Each of these contributes a factor of √2 to the acceleration - hence the acceleration is 2.g, and not the √2.g you proposed.

Abide a while with us...

I've been trying to figure out what you guys are talking about and I still don't get it. If something, a fly, a jet plane, a race car or a planet, whatever, is moving at a CONSTANT SPEED , then then there is no acceleration until that speed increases. As long as it is CONSTANT then there is no acceleration. I know, there is something I'm not getting,but as hard as I try I can't understand how something moving at constant speed is accelerating. Waiting anxiously for an explanation that someone besides a mathematician or a physicist can wrap his head around. Constant speed is constant speed.......isn't it? jack

Imagine you're driving your car at a steady speed on a level. Then you go up a hill. In order to maintain your steady speed you need to step on the gas. (acceleration)

Hopefully this explanation is helpful to you.

I don't understand either, and the answer from ddk didn't help me much. It sounded more like trying to define the energy used to maintain a constant speed. These guys are a couple of steps above what I used to think I knew! (I miss the old math days).

A fundamental concept about objects in motion is that in the absence of any forces an object in motion will continue to move at the same speed in the same direction, ie, in a straight line. In order for it to either change its speed or its direction of motion an external force must be applied which in turn will result in an acceleration. Now in this case while the fly's speed is constant it is not moving in a straight line so an external force must be acting on it and so it must also have an acceleration.

When you drive round a corner (curve) in a car at say 30 mph you maintain the same speed, but, the direction changes. Speed is a scalar quantity (no direction involved) velocity is a vector (measuring both speed and direction).

If you didn't need to worry about air resistance and the bearings on your car were perfect, then, on a flat road you would be able to corner (without using any petrol) without losing any speed. But: try it on an ice skating rink: the car will keep going in a straight line unless you apply a force to it (normally provided by the friction between the tyres and the road).

As it turns out any object moving with a constant speed in a circle is actually accelerating towards the centre of the circle:

a= u²/r (where a is the acceleration u the speed and r the radius of curvature).

Acceleration is a vector quantity as is force (although people normally don't specify the direction of either). Now acceleration is proportional to force, and, in the direction of the force. If you think about it acceleration has to be towards the centre of a circle because that's the way the forces act in simple examples like planets orbiting stars (in a circle), and, swinging a ball round on the end of a string (ignore gravity to keep this one simple).

The question is very specific about ignoring the effect of air resistance on the ball, so that it follows the normal parabolic flight, but, allowing the fly to, well, fly so that it follows the same flight path, but maintains its original scalar speed.

As Fyz explained brilliantly (as always) the velocity of the ball can be divided into two parts at the beginning of its flight.

Since no forces are acting on it horizontally the horizontal component remains the same (V/√2). At the top of its flight gravity is acting perpendicular to its direction of flight so the problem is exactly like the circular motion case. The acceleration of the ball was clearly g at this point (notice how you can use that term g to mean either force of acceleration), and since the acceleration is proportional to the speed squared over the radius of curvature (the same for the ball and the fly) the acceleration of the fly is ((V/(V/√2))2)g = 2g.

EDIT NOTE: started this before lunch and finished after lunch so didn't see Tkot similar explanation.

I too didn't find ddk's contribution too illuminating, as the example was overcomplicated by gravity.

One way to start is to look at the difference between a bicycle going at a steady speed along a straight line, and another bicycle going at the same speed in a circle.
For the straight-line case, I think we all agree that the acceleration is zero.
For the case of a circle, consider what happens to the velocity when the cycle goes 90^O round the circle. It's velocity in the original direction has reduced to zero, but it is now moving in a direction at right-angles to the original path. Effectively, the cycle has stopped going in the original direction and restarted in a new direction at right angles. This change of velocity can only be due to acceleration. Of course, the cycle didn't simply start and stop - it went at a constant speed, but continuously changed direction.
The change of direction is provided by acceleration perpendicular to the direction of motion; that dose not change the speed along the direction of motion, but adds a very small component at right angles to it - effectively changing the direction, but not the value.

[A small practical note: the reason you need to lean a cycle or motor-cycle into a corner is to allow the tyres to provide the force that allows the direction of motion to change. You may also have noticed spectacle cases, papers etc. sliding across a car seat when you go around a corner]

N.B. Equivalent in many ways to TKOT's, but I hope the differences might just swing it for some

The speed is constant but the direction is changing constantly. take ball with a string attached and swing it around with the ball at constant speed but as direction is changing there is acceleration. Don't you feel the force on your hand to keep it in circular motion

"A body remains at rest or uniform motion in a straight line unless acted upon by an external force"

I would be happy with " approximately -18.4m/s² ". But you write: "acceleration due to gravity (g)" which I take to be a good definition, and that has the same direction - downwards.

The fly's acceleration is zero. Constant velocity equals zero acceleration. The given problem directs us to neglect air resistance. However, it doesn't tell us to neglect gravity. But i believe the poser of the question also wanted us to neglect gravity. On the other hand, the fly, in order to maintain level flight, even for an instant, is also developing constant lift that equals gravity and opposes it. Therefore, level flight could be said to be acceleration upward against gravity at 32 ft/s/s.

GRG

I will give my vote to all people that answered -2g (at this time it's jim35848 and BobD)
I just would like to add my explanations:

The equations for the ball are:

Vx = V0 * cos(A)
Vy = V0 * sin(A) - g * t

Sx = V0 * t * cos(A) i)
Sy = V0 * t * sin(A) - 1/2 * g * t^2

The maximim heigh is reached when Vy = 0, that is
t1 = V0 * sin(A) / g
x1 = V0^2 * sin(A) * cos(A) / g

The slope of the ball (and fly) trajectory is:

dy/dx = dy/dt * dt/dx = (dy/dt) / (dx/dt)= Vy / Vx
= sin(A)/cos(A) - g*t / (V0 * cos(A))

from i) we know that t = x / (V0*cos(A)), therefore
dy/dx = sin(A)/cos(A) - g*x / (V0*cos(A))^2

Subtituting with
A=45 deg
cos(A)=sin(A)=sqrt(2)/2, we get finally

dy/dx = 1 - 2*g*x/V0^2 (slope of the trajectory)

The velocity vector of the fly follows exactly this slope. The magnitude is V0=constant, as postulated initially.

The x- and y- components of the fly velocity are therefore:

Vxf = V0 / sqrt (1 + (dy/dx)^2)
Vyf = V0 * (dy/dx) / sqrt (1 + (dy/dx)^2)

The acceleration in each direction can be calculated as:

a = dV/dT = dV/dX * dX/dT = dV/dX * Vx

Therefore

axf = ... = 2*g*(dy/dx) / (1+(dy/dx)^2)
ayf = ... = vxf^2 * dy/d^2x

At x = x1, that is at the highest point of the path:

axf = 0, and
ayf = -2*g

The acceleration of the flight at the peak is therefore
-2*g

See Fyz's reply #60

http://cr4.globalspec.com/comment/238582/Re-Of-Balls-and-Flies-CR4-Challenge-06-10-08

To BobD's post #41

http://cr4.globalspec.com/blogentry/5981?frmtrk=cr4sd#comment238198

It's all a bit nit picking or academic but to me: g is acceleration towards the centre of the earth; exactly what the ball is doing, and, the fly is accelerating twice as much in the same direction and sense.

It really depends on what your coordinate system is. If your coordinate system is fixed to the earch and has the x direction horizontal, the y direction vertical, and the z direction out of the imaginary paper we have all been using, then the acceleration vector would be a = (0,-2g,0). That is what I implied (not very well) with my answer of -2g. However, the fly would be experiencing a lateral acceleration (perpendicular to its direction of travel) of magnitude 1g but would be observed by an observer fixed to the earth to have an acceleration of 2g toward the earth.

Also, I've tried but I've not been able to do the math to get a general description of the fly's acceleration for the full path. Does anyone know how to do that?

Thanks,

Jim

Thanks Randall - yes, it's nit-picking and apparently academic - but unclear thought and expression often has practical results; so thank you for your (so far unique) support - you have encouraged me to try another explanation (not that anyone else will thank you).

Assuming you choose a coordinate system in which the Earth is stationary, it will not matter what specific axes you choose - so long as you use the same system for your measurement of g as you do for the acceleration of the fly.

To take the conventional system as an example - where height is measured in the usual direction. In that system, you will measure g as ~ -9.81m/s². Measure the fly's acceleration in the same coordinate system, and you get a value of ~ -19.62m/s², which is +2.g.

And, yes, the fly's acceleration for the full path is reasonably straightforward. The following method breaks the task into manageable chunks.
You know the gradient of the parabola at any position, which will allow you to calculate the component of the gravitational force at right angles to the motion (call this perp). You can also calculate the speed of the ball (call this vinst). The force on the fly at that position is then perp.(v/vinst)²

Hi Jim,

Alfabeta gave a general mathematical formulation in post 64. The basic idea is to look at the acceleration in the x and y directions separately. They can then be combined in a vector way to get the combined acceleration of the fly.

The way to do this is to break the velocity vector (V) of the fly into an x component (V_x) and a y component (V_y) such that V² = V_x²+V_y² ….1. Now the slope of the parabola at any point (dy/dx) = V_y/V_x so V_y = (dy/dx)V_x ….2. So substituting 2 into 1 gives

V_x = V(1+(dy/dx)²)^-1/2. Now the formula for the parabola is y=-(g/v²).x²+x (noting that g is the scalar value of the gravity vector) so (dy/dx) = 1-(2g/v²).x.

So V_x=V.(1+(1-(2gx/V²)²)^-1/2

Now acceleration of fly in x direction a_x= (d(V_x)/dt) = (dV_x/dx).(dx/dt) = (dV_x/dx).V_x Doing the mathematics is tedious and results in

a_x = 2.g.(1-(2gx/V²).(1+(1-(2gx/V²)²)^-2

Now acceleration of fly in y direction a_y= (d²y/dt²) = (d²y/dx²).(dx/dt)² = (d²y/dx²).V_x. Doing the maths gives;

ay=-2.g.(1+(1-(2gx/V²))²)^-1

Noting that g is the SCALAR value of the gravity vector.

You can put in the various values for x and see what the acceleration is at any point on the path of the fly. At the apex of the parabola when x = V²/2g, a_y=-2g, a_x=0.

I think you mean MODULUS. (SCALAR can have sign - as in the vertical component)

I made a mistake in my previous post. Here is the corrected version.

A way to get a general formulation for the acceleration at any point is to break the velocity vector (V) of the fly into an x component (V_x) and a y component (V_y) such that V² = V_x²+V_y² ….1. Now the slope of the parabola at any point (dy/dx) = V_y/V_x so V_y = (dy/dx)V_x ….2. So substituting 2 into 1 gives

V_x = V.(1+(dy/dx)²)^-1/2. Now the formula for the parabola is y=-(g/V²).x²+x (noting that g is the numerical value of the gravity vector without direction information) so (dy/dx) = 1-(2gx/V²)

So V_x=V.(1+(1-(2gx/V²))²)^-1/2

Now acceleration of fly in x direction a_x= (d(V_x)/dt) = (dV_x/dx).(dx/dt) = (dV_x/dx).V_x Doing the mathematics results in

a_x = 2.g.(1-(2gx/V²)).(1+(1-(2gx/V²))²)^-2

Now acceleration of fly in y direction a_y= (dV_y/dt) = (dV_y/dx).(dx/dt) = (dV_y/dx).V_x. Using Vy = Vx.(dy/dx) and doing the maths gives;

a_y=-2.g.(1+(1-(2gx/V²))²)^-2

Noting that g is the numerical value of the gravity vector without direction information.

You can put in the various values for x and see what the acceleration is at any point on the path of the fly. At the apex of the parabola when x = V²/2g, a_y=-2g & a_x=0 so resultant acceleration is -2g. When x=0 a_x=g/2 & a_y=-g/2 so the resultant acceleration has value of g.2^-1/2 perpendicular to the tangent.

The answer is 2g as given by FYZ in posting#4 where it was pointed out that the acceleration at the apex is a ratio of the squares of the speeds.

In order to justify this solution, we should verify that the acceleration at the apex is axial and centripetal only. A general acceleration of a point at the end of an R,θ vector in a plane, has accelerations which are:

Axial [R'']; Centripetal [R*(θ')²]; Angular [R*θ''] and Coriolis [2*R'*θ']. note ' denotes differentiation w.r.t. time. It turns out that at the apex, Angular and Coriolis = 0

A further point is that several of the solutions used the "radius of curvature" to determine the Centripetal acceleration. As post #4 noted, the ratio of the accelerations was V1²/Rad / V2²/Rad and the value of Rad cancels out, so it doesn't matter if you get it wrong (as at least one early posting did, and there was further talk about fitting circles to the parabola). In fact, the "curvature" and "radius of curvature" of a curve are precisely definedand the radius of curvature at the apex is (in parametric t form)

Rad = {[(x')²+(y')²]^(3/2)}/{(x'*y'')-(x''*y')}

and since y'=0 and y''=g this reduces to Rad=(x')²/g= v²/2g in terms of the original question. It is interesting to note that this is 2*hmax.

All of this may seem to be seem to be over-kill when the correct answe was explained in a simple sentence, but I hope it will give some insight in case the fly asks. "I'm half way to the apex. What's my acceleration doing?"

Regarding the "final" solution:

The fly has a restricted motion to a trajectory of a ball lounched close to Earth. Although fly's trajectory parameters include terms refering to Earth's gravity (g) there is no basis on equating the centripetal force at the peak with fly's weight because the centripetal force is caused by restrictions only. Using the following operators:

1) d/dt = v(x) d/ dx, and

2) d2/ dt2 = a(x)d/dx + v(x)**2(d2/dx2) (restricted motion)

and the fact that

|v|**2 = constant, implying that the scalar product <v, dv/dt> = 0 (speed and acceleration due to restrictions are allways perpendicular), use

a(x)v(x) + a(y)v(y) = a(x)v(x)+ v(x)a(y)dy/dx=0 to find that

3) a(x) = - a(y)dy/dt.

Then, from 2) and 3)

4) a(y) = [v(x)**2 d2y/dx2] /[1+(dy/dx)**2], using parameters calculated by BobD and the apex abscisa it results that

a(y) = -2g due to restrictions. Gravity may jus add another g to the fly.

Are you really writing what you mean?

Naturally, the combined force provided to the fly's body by its wings and the air is proportional to the mass of its body (m). To clarify this together with the effective values of acceleration (which I believe is equivalent to your posting mean):

With the fly hovering stationary, that force would be m.g, where g is the modulus of the acceleration due to gravity. If the fly were to follow the ball's trajectory at the speed of the ball, the wings would need to provide zero force. And with the fly going continuously at the starting speed of the ball the force at the peak of the trajectory would be -m.g.

Note that the total forces on the fly's body for the three cases (including gravity) are 0, -m.g, ang -2.m.g respectively, corresponding to the Earth-referred acceleration as given by your good self (and others).

(all "forces" being given as the vertical component of force, with upwards forces taking a positive sign; this is acceptable because for all these cases the horizontal components of the force are zero).

See also Jorrie's postings for comments on "Proper Acceleration"

Ball's motion is ballistic while the fly's motion is restricted to a trajectory described by a quadratic function of parameters discussed previously.

The acceleration on such trajectory depends on trajectory's parameters. In our case the first condition is y (x) = ax² + bx +c and the second is |v|² = constant.

The condition for constant modulus of v shows that the restriction (force) is allways normal to the trajectory. In this circumstances, at the apex

a_y = d²y/dt² = 2av²/[1+(2a(-b/2a)+b)²] = 2av² and

a_x = -a_y dy/dx non zero everywhere else but the apex, where:

a_x = -a_y(-2ab/2a +b) = 0

Centripetal acceleration at the apex is v²/(b²/4a) = (2/b²) (2av²) = (2/b²)a_y

since b² = 1 , then a_cp = 2a_y = - 2 g v²/ v²sin(alpha)cos(alpha) = -4 g

only if b² = 2 (contradicting the initial condition for the ball) a_cp = a_y = -2g

Therefore I am skeptical invoking the centripetal acceleration as fly's weight only because a_y is simply the effect of forced trajectory; it stands even if we neglect fly's weight.

Your solution applies to any fly flying parallel with the ground, at apex height with constant velocity v. At each point and moment, -g could be seen as "cetripetal" acceleration maintaining the fly on trajectory.

BobD,

thanks for you comment on curvature which highlighted the mistake in my previous post: centripetal acceleration at the apex is v²(2a)/b², since the radius is (2a)^-1 . Then , b² =1 gives

a_y = a_{cp =} 2av² = -2g

and the comment where b² = 2 can now be ignored. However, I am still skeptical to use etc., etc.

Regards

All I am saying is that part of the accelerating force on the fly is its weight (=m.g), the other part (also = m.g) is provided by the air (via the fly's wings). Total force of 2.g to provide the 'centripetal' acceleration of 2.g
(No sign needed as g is a downwards-acting vector).

We could long discuss about the sign, if it should be
2g or -2g
I think that both solution should be considered right.
It is just a matter on how you interpret the term "g" and how you define the reference system. Anyway, it has been a nice "challange questions". Many thanks to the issuer !

Either can be correct - if you take care to define your terms. However, contrary to your apparent implication, the difference in the various answers is not the reference system (which everyone seems to have taken with positive values of position being upwards) but the use of g as the modulus (= undirected and unsigned value) of the acceleration versus g (the vector value of the acceleration). The difference is that you don't need to specify the orientation of the reference system if you use g; but if you use g the orientation of the reference system is critical. On that basis, if the orientation of the reference system is not given, the assumption should be that the acceleration is taken as a vector - in which case the answer is +2.g

P.S. I'm wondering why I'm bothering - but I seem to have started on this pedantry when I was 'taken to task' for giving a positive answer without specifying the axes (and being a bit sloppy using emboldened scalars to make the distinction between text and equation clearer - I should have known better).