Pendulum problem
Bertil
above ground.
The ball is made to swing like a pendulum and the string is cut to
allow the ball to fly as far as posible.
Can a formula or equatiion be written to calculate the max distance,D,
the ball can go ?
This problem is the same as asking how far can a kid "jump" from a
swing.
Can the answer be expressed in terms of staring and final angle from
vertical?
Has this problem been solved in any textbook?
Bertil
William Elliot
> A steel ball is suspended with a string L meters long at H meters
> above ground.
> The ball is made to swing like a pendulum and the string is cut to
> allow the ball to fly as far as posible.
>
> Can a formula or equatiion be written to calculate the max distance,D,
> the ball can go ?
>
> Can the answer be expressed in terms of staring and final angle from
> vertical?
>
No. Starting from what? The first nudge of the pendulum?
You need the point of departure, Earth's gravity, assuming
the pendulum is on the ground, and the energy in the pendulum.
mike
@n8g2000prh.googlegroups.com>, stigf...@hotmail.com says...
applied maths courses than you have had hot dinners. You could solve it
yourself, ignoring such complications as wind resistance etc) by
considering:
a) for the pendulum bob the sum of its potential and kinetic energy is
constant.
b) for the pendulum bob, its vertical and horizontal components of its
velocity are equal to its velocity multiplied by simple trignometric
functions of the angle the string makes to the vertical .
c) from the point in time that the string is cut, the horizontal
velocity of the ball remains constant and its vertical velocity, height
etc are simple functions of initial vertical position, initial vertical
velocity, the constant g=9.8m/s^2, and time, which can be found in any
elementary text on projectile motion. Thus you will be able to determine
the time of flight of the ball (and thus the horizontal distance it
moves) before hitting the floor.
d) from b) and c) you will find a simple equation that provides the
distance travelled as a function of angle-when-string-cut that you can
differentiate to find the maximum D.
Note: as well as specifying L and H, you will ned to specify the maximum
angle or height above/below the support point of the pendulum from which
you will release the pendulum bob.
Enjoy
--
Mike
Frederick Williams
>
> On Tue, 20 Jul 2010, Bertil wrote:
>
> > A steel ball is suspended with a string L meters long at H meters
> > above ground.
> > The ball is made to swing like a pendulum and the string is cut to
> > allow the ball to fly as far as posible.
> >
> > Can a formula or equatiion be written to calculate the max distance,D,
> > the ball can go ?
> >
> > Can the answer be expressed in terms of staring and final angle from
> > vertical?
> >
> No. Starting from what?
The point at which the ball will fly as far as possible. (Which, to be
sure, is ambiguous: horizontal distance or length of arc? Length of arc
is non-trivial.)
--
I can't go on, I'll go on.
Frederick Williams
>
> In article <fd656dca-2639-4f30-8f68-96fc1299e7b2
> @n8g2000prh.googlegroups.com>, stigf...@hotmail.com says...
> > A steel ball is suspended with a string L meters long at H meters
> > above ground.
> > The ball is made to swing like a pendulum and the string is cut to
> > allow the ball to fly as far as posible.
>
> Note: as well as specifying L and H, you will ned to specify the maximum
> angle or height above/below the support point of the pendulum from which
> you will release the pendulum bob.
That which maximizes the distance the ball will fly... horizontal
distance I guess.
The Qurqirish Dragon
In addition to what others have stated, the maximum height that the
bob obtains while swinging is needed. This can be given directly, via
a maximum angle, by total energy in the pendulum (which, in turn means
the mass of the pendulum would be needed), or the speed at any given
point in the path the pendulum swings through
Bertil
wrote:
> mike wrote:
>
> > In article <fd656dca-2639-4f30-8f68-96fc1299e7b2
> > > A steel ball is suspended with a string L meters long at H meters
> > > above ground.
> > > The ball is made to swing like a pendulum and the string is cut to
> > > allow the ball to fly as far as posible.
>
> > Note: as well as specifying L and H, you will ned to specify the maximum
> > angle or height above/below the support point of the pendulum from which
> > you will release the pendulum bob.
>
> That which maximizes the distance the ball will fly... horizontal
> distance I guess.
>
> --
> I can't go on, I'll go on.
The gist of the problem is to determine the starting angle and release
angle to obtain max horizontal travel.
Bertil
Chip Eastham
To be fair, Bertil did address the initial and control conditions
when he asked:
> Can the answer be expressed in terms of star[t]ing and final
> angle from vertical?
I've never seen this problem before. Typical treatment of a
pendulum in an ODE class shows (by way of linearization)
that the period is independent of the starting angle. I'd
guess that with a similar linearization a more-or-less explicit
answer can be derived, i.e. given L (pendulum chain length),
H (height above ground at minimum), and starting angle A,
find the optimal "cut" angle C which maximizes the horizontal
distance traveled by the "dropped" pendulum bob.
Of course the linearization is an accurate approximation only
for "small" starting angles A. For the nonlinear pendulum the
exact solution can be expressed in terms of the Jacobi
elliptic function (assuming starting velocity zero):
[Exact solution for the nonlinear pendulum]
(Belendez et al, 2007)
http://www.sbfisica.org.br/rbef/pdf/070707.pdf
The "ballistic" portion of the pendulum's travel (after the cut)
is of course reasonably modelled as a quadratic (neglecting
air resistance as we do with the nonlinear pendulum), giving
an explicit value for the final horizontal position as a function
of the cut angle C.
Obviously the periodic motion of the pendulum guarantees
us the maximum travel corresponds to a critical point. It
seems unlikely that an explicit solution (setting derivative
to zero) can be found, but it's perhaps a good exercise for
computer algebra software.
regards, chip
mike
@d17g2000yqb.googlegroups.com>, hard...@gmail.com says...
if we needed to know the period of the pendulum or the time it takes to
get from one point to anhother.
Conservation of energy is sufficient, assuming the string is massless
and ignoring wind resistance.
If the bob is dropped from rest at some starting angle Theta1, then at
any angle Theta2 the square of the velocity of the ball will be equal to
2g* the distance fallen. i.e :
V^2 = 2gL(cos(Theta2)-cos(Theta1)).
Note that for any Theta2 the velocity will be higher the closer Theta1
is to pi/2 (assuming the bob can only be released with the string
parallel to or angled down from its point of attachment), so for any
release angle Theta2, D will be maximised by setting Theta1 to pi/2 and
the equation can be simplified to:
V^2 = 2gLcos(Theta)
where Theta is the release angle.
Mike.
Tim Little
> A steel ball is suspended with a string L meters long at H meters
> above ground.
> The ball is made to swing like a pendulum and the string is cut to
> allow the ball to fly as far as posible.
>
> Can a formula or equatiion be written to calculate the max distance,D,
> the ball can go ?
Not in terms of just that information alone. You've given the minimum
height above ground of the ball, but you also need the maximum.
If the ball swings higher than L+H, it won't swing "like a pendulum"
as the tension in the string will go to zero near the top of its swing
and it will go slack. Then when the ball falls again it will bounce,
not very pendulum-like. So let's say the ball swings up to L+H.
The speed at angle A will be given by conservation of energy, and you
can then resolve that into vertical and horizontal components as usual
for ballistic motion. Solve the quadratic equation for time it takes
to hit the ground and therefore distance (still as a function of A).
Then apply calculus to find the maximum.
It is not the sort of problem one would want to try to solve by hand.
The derivative of distance with respect to angle is very messy, and
its roots are probably not expressible in elementary functions.
Even the simplest H = 0 case ends up with a fairly nontrivial solution
sin(A) = (sqrt(13) - 1) / 6, if I have not messed up the math.
> Has this problem been solved in any textbook?
I very much doubt it.
- Tim
The Qurqirish Dragon
> On 2010-07-20, Bertil <stigfjor...@hotmail.com> wrote:
>
> > A steel ball is suspended with a string L meters long at H meters
> > above ground.
> > The ball is made to swing like a pendulum and the string is cut to
> > allow the ball to fly as far as posible.
>
> > Can a formula or equatiion be written to calculate the max distance,D,
> > the ball can go ?
>
> Not in terms of just that information alone. You've given the minimum
> height above ground of the ball, but you also need the maximum.
>
As was pointed out to me elsewhere in the thread, the OP wants an
answer based on the starting angle, which implies a starting velocity
of 0, and a max height of that starting point.
Tim Little
> As was pointed out to me elsewhere in the thread, the OP wants an
> answer based on the starting angle, which implies a starting
> velocity of 0, and a max height of that starting point.
Fair enough, that just makes it that little bit messier still. As if
it weren't messy enough already. Most interesting problems don't have
algebraic solutions. This is probably one of them.
- Tim
Chip Eastham
> In article <98cbf83f-a768-4c1d-9e53-a1a9ceb9aa92
I grant you this would maximize the velocity
v at release, but it cannot be the answer that
maximizes the distance D traveled in general.
[Note the OP Bertil confirmed up thread that
this is the objective.]
If you like you can think of the initial condition
as being Theta = pi/2 with nonzero velocity v.
If the height H is understood to be the distance
above ground at the nadir of the pendulum's
swing, then the pendulum will travel only an
arbitrarily small distance (as H -> 0) if it were
released at that point.
On the other hand the pendulum's swing is
not dependent on H, so the horizontal travel
D must be maximized by allowing the ball
to pass beyond the nadir (assuming its
velocity is positive at that point).
regards, chip
Peter Webb
specify only small angles and he did not use this in another response.
The position of the object is easily determined by L and theta where L is
the pendulum length and theta the angle. The displacement is L*cos(theta)
and H-L*sin(theta). Its velocity is also easily calculated by conservation
of energy, 0.5 v^2 = L(1 - cos(theta)). Its velocity is
(Lcos(theta), -sin(theta)).
THerefore we know the start position and start trajectory as a function of
H, L and theta only, where H and L are constants, and this becomes a
standard cannonball problem with some sines and cosines thrown in.
As Mike says, this is all conservation of energy. It does not give us or use
the equations of motion of the pendulum swinging about its origon, as it has
no term in "t" (time). We can say what *angle* the pendulum forms to produce
the maximum distance far, far more simply than we can can work out what
*time* this occurs. The original question does not requiire us to work out
the time, or to assume that the arc the pendulum swings through is small,
its quite clever IMHO.
Macavity
It has been a while since trying such problems, but let me bite. Say
A is the angle that the pendulum is released from rest, and B is the
angle at which the "string" is cut.
If A > pi/2 when released from rest, you may find the bob dropping
straight down etc, so
assuming 0 < A <= pi/2, and 0 <= B <= A. WLOG also let the mass of
the bob be 1 unit (should cancel out everywhere otherwise).
Energy when the bob is released = g(H + L - L cosA), is a constant
throughout the problem
Energy when the string has just been cut
= g(H + L - L cosB) + u^2/2, where u is the speed of the bob at the
time of string being cut.
Equating the two, we get
u^2 =2 g L (cosB - cosA) ....(1)
Now we have a projectile motion with speed u and angle B to the
horizontal. This means
horizontal speed = u cosB, remains same
vertical speed = u sinB - g t
When the bob finally hits the ground, its energy must be
(u cos B)^2/2 + (u sin B - gt)^2/2
Equating this with the energy at start gives
g(H + L - L cosB) + u^2 /2 = (u cosB)^2 /2 + (u sinB - gt)^2 /2
=> (H + L) - gt^2 /2 = L cosB - ut sinB ... (2)
And of course the horizontal distance traveled will be (including the
part before the string is cut)
D = L cosB + ut cosB
or D = cosB (L + ut) ...(3)
(I have measured from the axis of the pendulum, else we need to add L
cosA)
We need to maximise D, by choosing A and B.
It seems intuitively clear that A = pi/2 will give best possible
height and energy to the system, thereby maximising the distance
travelled. Equation (1) suggests the same as well. Setting this gives
u^2 = 2 g L cosB
Substituting above for u and for t from (3) in (2) gives after some
jugglery a quadratic in D:
D^2 - D(2L cosB)(1 - sin2B) + L cosB^2(4L cosB^2 + 2L sin2B - 4H - 3L)
= 0
Now solving for D in closed form is easy,
D = L cosB [Sqrt(sin2B^2 - 4 sin2B + 4sinB^2 + 4H/L) - sin2B + 1]
but finding the maximum of D seems very difficult, unless one does it
numerically for given values of H/L. Unless I have made a mistake or
miss some simplification along the way :(
HTH.