Balls rolling around inside of a gravity well, a straightsided funnel and a bowl speed up or slow down as friction drops them into lower orbits. It is surprising to see a ball speed up due to frictional energy loss.
Material
Assembly
Put one ball in each funnel. cover the top with saran wrap.
To Do and Notice
Swirl the funnels around in a circle to get the balls to orbit near the top of the funnel.
Stop swirling and watch what happens.
The ball in the gravity well goes faster and faster.
The ball in the bowl goes slower and slower.
What does the ball in the funnel appear to do?
Think about whether it is the speed of the ball you are observing or the time it takes to complete one circle.
What's Going On? Math Root.
The balls lose total energy due to friction.
In all cases they drop down into the various funnels.
As they drop they lose potential energy, however, in the case of the
gravity well, and indeed in the case of the orbiting bodies that this
represents, the balls gain kinetic energy as they drop. In the case
of the bowl the balls lose kinetic and potential energy as they
drop.
The kinetic energy, K, of each ball of mass m and velocity v is K=1/2 mv^{2}
The potential energy, U, is U = mgz where z is the height of the ball and g is the acceleration of gravity.
If r represents the distance from the center of the well to the orbital radius of the ball then the shape of the potential wells are:
gravity well z = 1/r
funnel z = r
bowl z = r^{2} (for example)
The total energy E is the sum of the kinetic and the potential E = K + U
Now each radius has a required velocity for the ball in order to allow it to stay in a circular orbit without dropping.
The wall pushes in on the rolling ball to provide the centripetal force to keep the ball moving in a circle F_{c }= mv^{2}/r
The wall pushes on the ball with a contact force. The horizontal component of this contact force is the centripetal force, F_{c },the vertical component must be equal to gravity, F_{g} = mg. The contact force is always perpendicular to the wall. The relationship when the wall makes an angle of q degrees with respect to the horizontal is.
F_{c}/ F_{g} = tan( theta)
so F_{c} = F_{g} tan( theta)
The Funnel
Now in the case of the funnel, the angle between the ground and the funnel is constant and independent of z or r. Let's say the angle is q = 45 °. Then the centripetal force must equal the gravity force. F_{c }= F_{g }since tan theta = 1. So mv^{2}/r =mg and v^{2} = gr
For this funnel r = z
The total energy is E = K + U = 1/2 mv^{2} + mgz = 1/2mgr + mgr = (3/2mg)r
As the total energy decreases, r decreases, v decreases.
The time, T, to complete an orbit or radius r is T = 2pi r/v = 2pi r/(gr)^{0.5}= 2pi r^{0.5}/g^{0.5}
^{And so as the height decreases and the radius decreases the period decreases.}
The Gravity Well
The angle between the wall of the gravity well and the horizontal depends on the radius. The angle goes to zero at large distances.
Recall for the gravity well z = 1/r
U = mgz = mg/r with U = 0 at r = infinity.
The slope of the walls of the well, s, is
s = dz/dr = 1/2 1/r^{2}
s is also the ratio of the rise over the run and so is the tangent of theta.
tan theta = s = 1/2 1/r^{2}
This means that since
F_{c}/ F_{g} = tan theta then
F_{c}/ F_{g} = s = 1/(2r^{2}) and
mv^{2}/r = Fg * s = mg * 1/(2r^{2})
so v^{2} = g/(2r)
and as r decreases velocity increases.
The total energy is E = K + U = 1/2 mv^{2} + mgz = 1/2mg/2r  mg/r = (3/4)mg/r
And as r decreases the energy decreases too.
So friction forces reduce the total energy and also the potential energy while increasing the kinetic energy.
The time to complete an orbit is T = 2pi r/v = 2pi r/(g/2r)^{0.5}= 2pi (g/2r)^{0.5}= pi 2^{1.5}g^{0.5}r^{1.5}
This is Kepler's famous law of the period of planetary orbits, the period is proportional to the three halves power of the radius.
As r goes to zero, the period goes to zero.
The Bowl
The shape of the bowl is z = r^2
The angle between the wall of the bowl and the horizontal depends on the radius.
The angle goes to zero at r = 0. The slope of the walls of the well, s, is
s = dz/dr = 2r
s is also the ratio of the rise over the run and so is the tangent of theta.
tan theta = s = 2r
U =mgz = mgr^{2} with U = 0 at r = 0.
This means that since
F_{c}/ F_{g} = tan theta then
F_{c}/ F_{g} = s = 2r and
mv^{2}/r = F_{g} * s = mg(2r)
so v^{2} = 2gr^{2}
v = (2g)^{0.5}r
and as r decreases velocity decreases.
The total energy is E = K + U = 1/2 mv^{2} + mgz = (1/2)m2gr^{2} + mgr^{2} = 2mgr^2
And as r decreases the energy decreases.
So friction forces reduce the total energy and also the potential energy while decreasing the kinetic energy.
The time to complete an orbit is T = 2pi r/v = 2pi r/(r(2g)^{0.5}= 2^{0.5}pi /g^{0.5}
The period T is independent of the radius r.
Scientific Explorations with Paul Doherty 

9 January 2003 