Imagine steel ball radius R and mass M held under a swing seat by an electro magnet with Xo being the point on the level ground directly under the ball if system had zero Kinetic Energy and Xm is point directly under ball when all the KE has been converted into potential energy. Assume whatever else is needed: For example swing ropes are L long and center of ball is H above ground at Xo, mass of swing /electromagnet is also M. Etc. as needed or convenient. At what X should the electro-magnet current be turned off to have ball hit the ground with the max value of X?
As no one has tried to solve, I decided to. First let's change L to be the length from the top of the rope to the center of the ball. Then the equation for the circular arc of the ball's (center, always implied) pre- release trajectory is: L^2 = (L -y)^2 + x^2 and when ball is at (X,Y), the assumed release point, L^2 = (L -Y)^2 + X^2 . Step 1: Differentiate this to get the slope of the circular arc path at point (X,Y) with result: Y' = X / (L -Y) as a check, note at x=0 the slope is 0, and if x is max (= L ) when ball has enough energy to swing up thru 90 degrees, slope is infinite as then Y = L. Step2: Ball velocity V at point (X,Y) is function of Y only, V(Y) as energy is conserved. At x = 0, all the energy is kinetic, E = (M/2)[V(0)]^2 = (M/2)[V(Y)]^2 + MGY or Eo = 2E/M = [V(Y)]^2 + 2GY where the just defined Eo is twice the energy per unit mass. Or V(Y) = sq root{Eo - 2GY}, but we are more interested in the vertical component of V(Y), Vy(Y), to find what is the max altitude the ball released from the circular path can achieve above the release altitude Y. I'll call that H. I.e. the peak of the ball's free falling parabolic trajectory is Y+H above the x-axis. [V(Y)]^2 = [Vy(Y)]^2 + [Vx]^2 Note Vx is a constant after release until ball hits the ground, so: [Vy(Y)]^2 = [V(Y)]^2 - [Vx]^2 = 2GH (the "2" as our units of energy are twice the normal ones) but we must get Vx in terms of Vy using the known slope at point (X,Y). Step 3: Vy/Vx = slope = Y' = X / (L -Y) So Vx = Vy (L -Y) / X With this and prior expression for [V(Y)]^2 = Eo - 2GY this gives Y + H, the peak altitude of the free ball as: Y + H = Y + {Eo - 2GY - [Vy (L -Y) / X]^2}/ 2G = Yp where subscript refers to the peak point. Thus we now know Ball is on a parabolic trajectory which passes thru (X,Y) and (Xp, Yp) points but we do not yet know Xp. To find Xp we need to know how far past X the ball travels in x direction at constant speed Vx, while climbing to the peak. The average vertical speed to the peak is Vy /2 and the peak is H above the release point (X,Y) so the duration of the climb, T, is: T = 2H / Vy(Y) Then Xp = X + T[Vx(Y)] = X + [2H /Vy(Y)] [Vy (L -Y) / X] Now , I think*, we know the assumed release point coordinates, (X, Y) and the peak coordinates (Xp, Yp) and that the axis of the parabola is line x = Xp and it opens downward. Thus we can write equation for the post release trajectory of the ball. The ground is the line y = -g where g is an assumed distance below the x-axis, a parameter of the problem which will change the answer. For example, if g >> L then the release for max X ground impact will be relatively close to x = 0 as we want Vx to be large. Climbing up a little more to the peak is of little help and reduces the speed Vx. Conversely, if g<< L we need to get some upward speed to avoid hitting the ground too soon. Step 4: solve equation of Parabola and line x= -g for the intercept or ground impact point, of zero radius ball or use "false ground line" at x = -g + R for real ball. Note we do not need to find when ball hits the ground. This intercept is at Xg. Step 5: We have an expression for Xg in terms of the energy Eo, g, & L and the assumed release point (X,Y) but due to the constraint of being on the circular arc, the release point can be taken as a function of X only. Thus, Xg is a now known but complex function of Eo, g , L and X or Xg(Eo, g, L, X) which we differentiate wrt x and set equal to zero to get equation Xg' (Eo,g,L,X) = 0 Step 6: solve Xg' (Eo,g,L,X) = 0 and that gives the maximum X or Xm in terms of the Eo , g , & L. I can understand why no one, me included, has actually done this but that is how it is done. I think it interesting that such a simply stated classical problem is so hard to do. --------------- *We need to use the first equation for the circular arc, prior to release, to express the release point's y coordinate, Y in terms of X and L The after Y is eliminated only the parameters, Eo, g, L and variable X remain in Xg' (Eo, g, L, X) = 0 equation.
I looked at this earlier on then forgot about it. This is a classic playground problem: How do you get the most distance when you launch yourself off a swing? If you let go too early, you miss out on the height and distance advantage, and your trajectory is too flat. But if you leave it until the top of the swing, you don't get any flight; you just drop straight down. Anyway, I've attacked it again now, using a similar approach to Billy, and ended with a very nasty function for the distance traveled by the ball, measured from the ground under the pivot. I will try to texify and post it later (I was about to do that now, but I've just noticed an error that I'll have to correct first). I haven't attempted to find the equation for the function's maximum over the range of interest. Several simplifying assumptions have been made (in decreasing order of playground practicality): Gravity is uniform Wind resistance is negligible The ground is flat The swing's "rope" is actually a massless rod The payload is an inactive point mass The last two assumptions in particular make the real playground problem much more complex. Solving it is therefore best left to the children. Please Register or Log in to view the hidden image!
I don't understand why this is needed. If payload is ball of radius R, then just imagine a "false ground" R above the real ground for the center of the ball to hit. - that is the same as a ball of radius zero hitting the real ground. I was tactly assuming your other items, but good to replace the "rope" with rigid massless rod. You can buy them cheap now as they are on sale in the theoretical physics store. (Find them hanging on the wall, along side the frictionless pulleys.) Yes; they solve it every day and they don't even know calculus!
You're right, it doesn't have to be pointlike. In that item, I was thinking more about the "inactive" assumption - once you give the payload arms, legs, and muscles, things get much more interesting!
I would be courious to know even one answer. For example if point ball is one unit above ground when with zero energy and rigid rod is ten units long and when all intitial energy is converted to potential energy the point ball has X = 2 units, what is X at the best "release ball" and the corresponding max X?
I should have written "that computer can't handle", but you understood what I meant. The optimal X for releasing the ball is 1.88933 units, achieving max X = 2.1169 units. The value of x when the ball hit the ground, as a function of release location, is \( x=\xi+\xi\eta(\eta-\eta_0)+\sqrt{\xi^2(\eta-\eta_0)^2+2(\eta-\eta_0)(h-\eta)} \) where \(\xi\) is the sine of the angle between the ropes and the vertical direction at the release point, \(\eta\) is the cosine, \(\eta_0\) is the cosine of the maximum vertical angle made by the ropes (i.e., the angle when the kinetik energy is zero), and \(h\) is the height of the fixed point above ground (i.e., the length of the rope plus H). The ball is considered of zero radius and all lengths are measured as multiples of the length of the rope. One interesting thing is that this x does not depend on g, so for instance the answer will be the same for the Earth and the Moon. Now we just have to find where this function x will have maximum.
Can I assume the numerical result is for my suggested case L/H = 10/1 & all potential at X = 2? Also it would seem from your last comment that this numerical result was obtained by systematic trial and error application of your expression for x. Is that the case? Congratuatlions on getting that expression for x. I assume also that your general approach was quite like the one I outlined in post 2 (and Pete was using also.) Is that correct?
The general approach is the same: 1 equation for x, which involve the angle, time, and initial velocity, one equation for the condition that the y coordinate is zero, one equation connecting the initial velocity with the angle (energy conservation). Eliminate time and velocity in terms of angle, and we get the above equation. The numerical result was obtained by differentiating the function x and equating it to zero. You can show that there is only one maximum too, because the differential is monotone.
In the example case you gave, the graph of x as a function of \(\eta\) looks like this: Please Register or Log in to view the hidden image! ps: the differential (derivative is more accurate) is monotone means this graph is concave.
