The Golfer’s Nemesis

Kirk T. McDonald
Joseph Henry Laboratories, Princeton University, Princeton, NJ 08544
(November 23, 1980; updated August 9, 2021)

1 Problem
Can a golf ball roll into the cup, roll around on its vertical wall and pop back out?1
Consider a sphere of radius a that rolls without slipping inside a vertical cylinder of
radius b > a.

If Ω = φ̇ = angular velocity of the point of contact about the vertical, 1̂ points from
the center of the sphere to the point of contact, ẑ is vertical, and 2̂ = ẑ × 1̂, show that the
component equations of motion are,

ẑ : Ω̇ = 0, (1)
1̂ : a ω̇ 1 = Ωz, (2)
 
2̂ : I + ma2 z̈ = −ma2g − Ia ω 1 Ω. (3)

Show that z of the center of mass executes simple harmonic motion, and if at t = 0,
z = 0, ż = ż0, and ω 1 = ω 10, then,

ma2g + Ia Ω ω 10 ż0 I
z= 2
(cos αt − 1) + sin αt, where α=Ω . (4)
IΩ α I + ma2
With what velocity and angular velocity must the ball arrive at the rim of the cup to fall
in and execute the above oscillatory motion, and possibly pop back out?

2 Solution
This problem is discussed in §421, p. 357 of E.A. Milne, Vectorial Mechanics (Metheun;
Interscience Publishers, 1948),
http://kirkmcd.princeton.edu/examples/mechanics/milne_mechanics.pdf
1
This behavior is distinct from the possibility that the ball bounces off the flagpole in the hole, or the
plastic insert therein, as occurs from time to time.

1
 We consider a sphere, of mass m and radius a with moment of inertia I about its center,
that rolls without slipping on a fixed, vertical cylinder of radius b > a. We use a set of
principal axes (but not body axes) about the center of the sphere of radius a, where 1̂ points
outward along the horizontal line from the center of the spheres to the point of contact with
the cylinder. Axis 3̂ is vertical (parallel to ẑ), and axis 2̂ = ẑ × 1̂ is also horizontal).
The center of the sphere of radius a is at position r = (b − a) 1̂ + z ẑ with respect to the
origin at the bottom center of the cylindere. Then, the velocity of the center of the sphere
of radius a is,
dr d1̂
v= = (b − a) + ż ẑ . (5)
dt dt
The (nonholonomic) constraint of rolling without slipping is that the point of contact of
sphere with the cylinder is instantaneously at rest in the lab frame,
d1̂
vcontact = 0 = v + ω × a = (b − a) + ż ẑ + aω × 1̂, (6)
dt
where ω is the total angular velocity of the sphere in the lab frame, and a = a 1̂ is the vector
from the center of the sphere of radius a to the point of contact.
The force and torque equations of motion for (the center of) the sphere of radius a are,
dv d2 1̂ d21̂
m = m(b − a) 2 + mz̈ ẑ = F − mg ẑ, F = m(b − a) 2 + m(g + z̈) ẑ, (7)
dt dt dt
dL dω d2 1̂
=I = τ = a × F = ma(b − a) 1̂ × 2 − m(g + z̈)a 2̂, (8)
dt dt dt
where I is the moment of inertia of the sphere about its center.
We define Ω = Ω ẑ as the angular velocity of the center of the sphere (and also of the
point of contact, as well as of the triad 1̂-2̂-3̂) about the vertical axis, such that,
d1̂ d2 1̂ d2 1̂
= Ω × 1 = Ω 2̂, = Ω̇ 2̂ + Ω Ω × 2̂ = −Ω2 1̂ + Ω̇2̂, 1̂ × = Ω̇ ẑ. (9)
dt dt dt
The velocity (5) of the center of the sphere can now be written as,
v = −Ω(b − a) 2̂ + ż ẑ , (10)
so the 2̂-component of the total angular velocity ω of the sphere about its center (and also
that about the point of contact) is vz /a = ż/a, and the ẑ-component is v2 /a = −(b − a)/a.
Thus,
ż b−a dω z̈ Ωż b−a
ω = ω 1 1̂ + 2̂ − Ω ẑ, = ω̇ 1 1̂ + Ω ω 1 2̂ + 2̂ − 1̂ − Ω̇ ẑ, (11)
a a dt a a a

2
With these, the equation of motion (8) becomes,
    
Ωż z̈ b−a
I ω̇ 1 − 1̂ + Ω ω 1 + 2̂ − Ω̇ ẑ = ma(b − a) Ω̇ ẑ − m(g + z̈)a 2̂, (12)
a a a

The components of the equation of motion imply,

ẑ : Ω̇ = 0, Ω = constant, (13)
Ωż Ωz
1̂ : ω̇ 1 = , ω1 = + ω 10 , (14)
 a  a
2̂ : I + ma2 z̈ + I Ω2 z = −ma2g − I Ω ω 10 . (15)

The center of the sphere executes simple harmonic motion in z,2 and if at time t = 0, z = 0,
ż = ż0 , ω 1 = ω 10 , then,

ma2g + Ia Ω ω 10 ż0 I
z= 2
(cos αt − 1) + sin αt, where α=Ω . (16)
IΩ α I + ma2
We now consider under what conditions a golf ball could roll into a cup/vertical cylinder
such that at time t = 0 the motion is described by eq. (16).
According to eqs. (10) and (11), the velocity v0 and the angular velocity ω 0 at this time
must be,
ż0 b−a
v0 = −Ω(b − a) 2̂ + ż0 ẑ , ω 0 = ω 10 1̂ + 2̂ − Ω ẑ. (17)
a a

The figure above shows side and top views of the ball as it enters the cup, after rolling
into it from the left while
√ on the horizontal surface. At time t = 0, the ball has fallen through
height a, so ż0 = − 2ag. If the ball arrived at the top of the cup with horizontal velocity
v0 (in the −2̂ direction), then this is also the horizontal velocity when the center of the ball
has fallen to z = 0, and so Ω = v0/(b − a). The angular velocity of the ball did not change
while it fell into the cup, so the angular velocity at the time of arrival was,

2g v0
ωarrival = ω0 = ω 10 1̂ − 2̂ − ẑ, varrival = −v0 2̂ = −Ω(b − a) 2̂. (18)
a a
If the ball had been simply rolling without slipping prior to arrival at the cup, then ω 10 = v0/a
and the 2̂- and ẑ-components of ωarrival would be zero. Hence, only under special conditions
2
This motion can be regarded as a nutation about steady motion with angular velocity Ω in a horizontal
circle at z = −(ma2 g + Ia Ω ω 10 )/I Ω2 .

3
of rolling with slipping at the moment of arrival at the cup could the ball roll into it and
pop back out after following motion of the form (16).
For a golf ball of uniform mass density, I = 2ma2/5, and α = 2/7 Ω = Ω/1.87. If the
golf ball does pop out of the hole, it does so in somewhat less than one period of the vertical
oscillation, i.e., in less the 1.87 revolutions of the ball around the vertical axis of the cup.

An early discussion of the problem is on p. 354 of Besant, Treatise on Dynamics (1914).

http://kirkmcd.princeton.edu/examples/mechanics/besant_14.pdf
It was briefly mentioned on p. 26 of Littlewood’s Miscellany (1953),
http://kirkmcd.princeton.edu/examples/mechanics/littlewood_miscellany.pdf
It is solved via Lagrange’s method on p. 95 of J.I. Neı̌mark and N.A. Fufaev, Dynamics of
Nonholonomic Systems (Am. Math. Soc., 1972),
http://kirkmcd.princeton.edu/examples/mechanics/neimark_72.pdf
Other discussions of it include, M. Gaultieri et al., Golfer’s dilemma, Am. J. Phys. 74, 497
(2006), http://kirkmcd.princeton.edu/examples/mechanics/gaultieri_ajp_74_497_06.pdf
O. Pujol and J.P. Pérez, On a simple formulation of the golf ball paradox, Eur. J. Phys. 28,
379 (2007), http://kirkmcd.princeton.edu/examples/mechanics/pujol_ejp_28_379_07.pdf
O. Pujol and J.-P. Pérez, The golfer’s curse revisited with motion constants, Am. J. Phys.
90, 657 (2022), http://kirkmcd.princeton.edu/examples/mechanics/pujol_ajp_90_657_22.pdf