Re: question on how to kick yourself to center of merry go round.

Dear Sircular,

I will explain how to determine the initial radial speed that the
child must have in order to reach the center of the merry-go-round in
half of a rotation, but it is necessary to know one other piece of
information to calculate the average force that the child must exert
on the railing to give herself this initial radial speed.  The missing
quantity is the time over which the child exerts the force.

The physics involved in this problem can be analyzed most simply from
the rider's point of view as long as one recognizes that the rider is
accelerating (anything moving in a circle accelerates, even if the
speed is constant) and is therefore a "non-inertial" observer.
"Non-inertial" means that Newton's First Law does not hold, and the
rider will be subject to so-called "fictitious" forces.  Of course,
from the rider's point of view, the forces will feel quite real.  They
are only called fictitious because these forces have no obvious cause,
and they result from an accelerating observer attempting to apply
Newton's Laws as if the observer were not accelerating.

One fictitious force which we are going to negate is the Coriolis
force.  Let's assume that the merry-go-round is rotating
counterclockwise as seen from above.  If you actually try the
experiment that you describe, you will feel a force to the right as
you move radially inward.  If you are not careful to brace yourself,
the Coriolis force will push you off the skateboard sideways.  Let's
assume that the wheels of the skateboard are only frictionless in the
radial direction, and that the friction between the wheels and the
merry-go-round floor is large enough to cancel the sideways Coriolis
force.  This is the usual way in which a skateboard works.

The only other fictitious force acting on the rider is the centrifugal
force.  That's right, not centripetal (toward the center) but
centrifugal (away from the center).  The magnitude of the centrifugal
force is:

F_c = m * (v_t)^2 / r = m * w^2 * r

where m is the mass of the rider (25 kg), v_t is the tangential speed
of the rider, r is the distance of the rider from the center of the
merry-go-round, and w (omega) is the angular speed of the
merry-go-round.  The second form of the centrifugal force is more
convenient since the angular speed of the merry-go-round is constant
in time and does not change as the rider changes distance from the
center.

The time it takes for the merry-go-round to complete one rotation is
the period

T = 2 * pi * R / V

where R is the radius of the merry-go-round (1 meter) and V is the
tangential speed of a point on the rim of the merry-go-round (10 m/s).
With your numbers, the time to go all the way around once is T=0.628
seconds.  The rider takes half of this time to reach the center 
t_f = 0.314 seconds.  The angular speed that we introduced above is

w = 2 * pi / T = 10 radians/second

Now Newton's Second Law can be applied to the rider (even though the
First Law can not):

F_total = m a

and the only force is the centrifugal force F_c, while "a" is the
radial acceleration of the rider.

F_c = m * w^2 * r = m a

Cancel the mass from both sides of the last equality, and the
acceleration is seen to be

a = w^2 * r (outward)

At this point, the problem moves out of physics into the realm of
mathematics, differential equations to be specific.  The acceleration
"a" is the second derivative of position with respect to time, so the
equation above becomes

d^2 r / dt^2 = w^2 * r

and you either know how to solve this because you've done it several
times before, or you look it up in a book.  The solution is the radial
distance as a function of time

r(t) = A sinh(wt) + B cosh (wt)

where sinh is the hyperbolic sine and cosh is the hyperbolic cosine.
"A" and "B" are two constants which we will now determine from the
initial conditions.  First, the radial distance at time zero (just as
the rider pushes off the railing) is R (= 1 meter, but we won't use
numbers until the very end).

r(0) = R

A sinh(0) + B cosh(0) = R,    but sinh(0)=0 and cosh(0)=1

B=R ,   this fixes one of the two constants.

The second condition that must be satisfied is that the radial
distance after one half of a rotation (t = T/2 = pi/w = 0.314 seconds)
must be zero, the middle of the merry-go-round.

r(pi/w) = 0

A sinh(pi) + R cosh(pi) = 0,    where we used B=R

A = - R coth(pi),    where coth is the hyperbolic cotangent

Now the formula for radial distance for any time t is

r(t) = - R coth(pi) sinh(wt) + R cosh(wt)

The first derivative of this equation gives the radial velocity at any
time t

v(t) = dr / dt = -w*R coth(pi) cosh(wt) + w*R sinh(wt)

and if you want to know the initial velocity of the rider, that's the
velocity at time zero

v(0) = -w*R coth(pi) cosh(0) + w*R sinh(0) = -w*R coth(pi)

The negative sign indicates that the initial velocity is inward.
Now we can introduce numerical values

v(0) = - (10 radians/second)*(1 meter) coth(pi) = - 10.0374 m/s

________________________________________________________________________

Next, let's call the amount of time that the rider pushes on the
railing of the merry-go-round to give herself this initial speed
t_push.  This will enable us to derive a formula for the average force
that she exerts on the railing.

The railing must supply the centripetal (inward) force, m V^2 / R,
plus the force required to give the rider an initial radial speed
|v(0)| (with absolute value brackets to make it positive).

F_on_railing = m * V^2 / R   +   m * |v(0)| / t_push

         = (25 kg)*(10 m/s)^2 / (1 meter) + (25 kg)*(10.0374 m/s) / t_push

         = 2500 newtons + 250.935/t_push

I suppose that a typical push-off time might be one tenth of a second,
t_push = 0.1 second, in which case

F_on_railing = 5009 newtons = 1124 pounds!

________________________________________________________________________

Some comments:

That's a lot of force -- over half a ton!  Not just for a child rider,
but even for an adult.  You may know about schemes for creating
artificial gravity on spacecraft by spinning them about some axis.
(If not, watch the movie "2001: A Space Odyssey".)  In these schemes,
the radial acceleration w^2 * R should be close to Earth gravity, 
9.8 m/s^2 to maintain a comfortable environment.  The maximum radial
acceleration experienced by the rider in your problem is 
(10 radians/second)^2 * (1 meter) = 100 m/s^2 or about 10 times Earth
gravity!  I'm not sure that a person could survive this acceleration
for very long.

Also, the speed of the child when she reaches the center of the
merry-go-round is not zero it is

v(t_f) = v(pi/w) = - w*R csch(pi) = 0.866 m/s

where csch is the hyperbolic cosecant.  If there is nothing at the
center (like a post) to stop the rider, she will pass over the center
and pick up speed as she heads back toward the railing.  It will
require the same force to stop the child as it did to get her moving.
The trip from the center back out to the railing will also take one
half of a rotation, so she will end her trip exactly where she began!


--Randall J. Scalise    http://www.phys.psu.edu/~scalise/