1. APhO II 2001 Theoretical Question (spare)
Theoretical Question
The Effect of Spin on the Rebound of a Ping-Pong Ball
In the game of table tennis, also known as ping-pong, players often serve the
ball by striking it with a twisting or slicing motion, thereby giving the ball an initial
spin or angular velocity. In this problem, we consider the effect of various kinds of
initial spin on the final translational and rotational velocities of a ping-pong ball as it
rebounds from a level surface.
A ping-pong ball with mass m and radius R , moving from left to right, strikes
a horizontal table surface obliquely. If we take the +x axis to point in the forward
horizontal direction, i.e. to the right, then the x-component of the incident velocity is
ov with 0>ov . By taking the +y axis to be vertically downward, the +z axis is
horizontal and points perpendicularly into the page in Fig.1a. The only non-vanishing
component of the initial angular velocity oω

is its z-component zω . The magnitude
|| oω

of the vector oω

is denoted as )0( ≥oo ωω .
A ping-pong ball has no spin if its angular velocity component zω is zero. If
zω is of the sign as the x-component of the translational velocity of the ball, then the
ball is said to have a topspin or forward spin. Otherwise, it is said to have a underspin,
backspin or backward spin. For the case depicted in Fig.1a, the initial and final spin
of the ball are both topspin. If the direction of the initial angular velocity were
reversed in Fig.1a, then the initial spin of the ball would be a underspin. Note that if
the horizontal component fv of the rebound velocity has the same sign as ov , then
it is bouncing forward to the right. If the rebound angular velocity fω is positive,
then it is turning clockwise when we look straight into the page.
When considering this problem, the following assumptions may be made:
(a) At the instant of rebound, the horizontal component of the velocity of the bottom
of the ball is zero relative to the table.
1
fv
ov
fω0ω
Fig. 1a

2. APhO II 2001 Theoretical Question (spare)
(b) Air resistance and buoyancy can be neglected.
Assume the moment of inertia of the ball for rotations about an axis passing
through its center is
2
3
2
mR . Let kˆ be the unit vector of the +z direction. Analyze the
horizontal component fv of the velocity and the angular velocity fω of the ping-
pong ball at the instant of rebound for the following three initial states of rotation, and
complete the table in the answer sheet.
(1) no spin: oω

= 0.
(2) topspin: oω

= 0ω kˆ .
(3) underspin: oω

= kˆ
0ω− .
2

3. APhO II 2001 Theoretical Question (spare)
[Answer Sheet] Theoretical Question
The Effect of Spin on the Rebound of a Ping-Pong Ball
oω fv fω condition
oω

= 0
(no spin)
magnitude:
sign:
magnitude:
sign:
koo
ˆωω =

(top spin)
magnitude:
sign:
magnitude:
sign:
magnitude:
sign:
magnitude:
sign:
koo
ˆωω −=

(under spin)
magnitude:
sign:
magnitude:
sign:
magnitude:
sign:
magnitude:
sign:
magnitude:
sign:
magnitude
sign:
Note: (i) If the sign of fv or fω is positive, write down “＋”, and if the sign is
negative, write down “－”.
(ii) If the answers to fv and fω require a condition involving 0v , 0ω and R,
write it down under “condition” in the table. If it does not require a
condition, write down “no”.
(iii) If either fν or fω happens to be zero, write down “no” for sign.
3

4. APhO II 2001 Theoretical Question (spare)
Solution:
(1) No spin
The frictional force between the ball and the tabletop imparts an impulse to the
ball in a direction opposite to `0v , as shown in figure 1a.
0mvmvtF f −=∆− (1)
where F is the average frictional force during a collision time t∆ .
The frictional force also imparts an angular impulse to the ball during collision,
fItRF ω=∆ (2)
where 2
3
2
mRI = is the moment of inertia of the ball for rotations about an axis
passing through the center. The assumption that, at the moment of rebound of the
ball from the tabletop, the ball does not slide, is equivalent to the condition that
the velocity of the bottom of the ball is zero relative to the tabletop, namely,
Rv ff ω= (3)
Eliminating tF∆ from equations (1) and(2), one obtain
)(20 R
mR
I
vv ff ω=− (4)
Substituting ff vR =ω in the above equation, one obtains






+
=
2
0
1
mR
I
v
vf
= 0
5
3
v (5)
fω
R
v
R
v f
5
3 0
== (6)
After rebound, the ball acquires a top spin.
4
fvov
F
Fig. 1a
fω

5. APhO II 2001 Theoretical Question (spare)
(2) top spin
Relative to the tabletop, the incident horizontal component of the velocity of the
bottom of the ball is Rv 00 ω− .
(i) if Rv 00 ω≥ , then the direction of the frictional force is opposite to that of 0v
as shown in figure 1b.
0mvmvtF f −=∆− (7)
)( 0ωω IItRF f −=∆ (8)
Rv ff ω= (9)
From equations (7) ~(9), one obtain,






+=






+
+
= Rv
mR
I
mR
I
v
vf 00
2
0
0
3
2
5
3
1
ω
ω
(10)






+== 0
0
3
2
5
3
ωω
R
v
R
vf
f (11)
(ii)If Roo ων ≤ , then the direction of the frictional force is the same as 0v , as
shown in figure 1c.
5
ωf
ωo vf
v0
F
Fig. 1b
vf
ωf
Figure 1c
ωo
ov
F

6. APhO II 2001 Theoretical Question (spare)
0mvmvtF f −=∆ (12)
0ωω IItRF f −=∆− (13)
ff Rv ω= (14)
the solutions are,






+=






+
+
= Rv
R
I
mR
I
v
vf 00
2
0
0
3
2
5
3
1
ω
ω
(15)






+= 0
0
3
2
5
3
ωω
R
v
f (16)
(3) under spin
0mvmvtF f −=∆− (17)
)( 0ωω IItFR f −−=∆ (18)
ff Rv ω= (19)
The solutions are






−=






+
−
= Rv
mR
I
mR
I
v
vf 00
2
0
0
3
2
5
3
1
ω
ω
(20)






−= 0
0
3
2
5
3
ωω
R
v
f (21)
(i) if 0v > R0
3
2
ω , then fv >0, and 0>fω
(ii) if 0v < R0
3
2
ω , then fv <0, the ball rebounds to the incoming direction, and
0<fω
(iii) if ,
3
2
Rv oo ω= then ,0=fv the ball rebounds vertically, and
,0=fω the ball does not spin.
6
ω o ω f
f
ov
F
Fig.1d

7. APhO II 2001 Theoretical Question (spare)
oω

fv fω condition
oω = 0 magnitude: 5/3 ov
sign: +
magnitude: 5/3 ov R
sign: +
No
koo
ˆωω =

(top spin)
magnitude: R
v
o
o
ω
5
2
5
3
+
sign: +
magnitude:
5
2
5
3 oo
R
v ω
+
sign: +
Roo ων ≥
magnitude: R
v
o
o
ω
5
2
5
3
+
sign: +
magnitude: o
o
R
v
ω
5
2
5
3
+
sign: +
Rv oo ω≤
koo
ˆωω −=

(under spin)
magnitude: R
v
o
o
ω
5
2
5
3
−
sign: +
magnitude: o
o
R
v
ω
5
2
5
3
−
sign:+
Rv oo ω
3
2
>
magnitude:
5
3
5
2 o
o
v
R −ω
sign: -
magnitude:
R
vo
o
5
3
5
2
−ω
sign:-
Rv oo ω
3
2
<
magnitude: 0
sign: No
magnitude: 0
sign: No
Rv oo ω
3
2
=
7

8. APhO II 2001 Theoretical Question (spare)
Marking Scheme for theoretical question 1
1. Understand the change in linear momentum equals impulse ( 0.5 point )
(momentum-impulse theorem)
2. Understand the change in angular momentum equals angular impulse
(angular momentum –angular impulse theorem) (1.0 point)
3. An equation which correctly expresses the condition of no sliding at the instant of
rebound, namely, Rf ων = ( 0.5 point)
4. The remaining 8.0 points are to be graded according to the following table
oω

fv fω condition
oω = 0
magnitude:0.6
sign: 0.1
magnitude:0.6
sign: 0.1
0.1
koo
ˆωω =

(top spin)
magnitude:0.6
sign: 0.1
magnitude:0.6
sign: 0.1
magnitude:0.6
sign: 0.1
magnitude:0.6
sign: 0.1
0.1
0.1
koo
ˆωω −=

(under spin)
magnitude:0.6
sign: 0.1
magnitude:0.6
sign:0.1
magnitude: 0.1
sign: 0.1
magnitude:0.6
sign:0.1
magnitude:0.6
sign:0.1
magnitude:0.1
sign:0.1
0.1
0.1
0.1
8