IPHO 2003
Solution − Theoretical Question 1
(2003/07/12)
p.1/7
Solution- Theoretical Question 1
A Swing with a Falling Weight
Part A
(a) Since the length of the string L  s  R is constant, its rate of change must be
zero. Hence we have
(A1)*1
s  R  0
(b) Relative to O, Q moves on a circle of radius R with angular velocity  , so

(A2)*
vQ  R tˆ   s tˆ
(c) Refer to Fig. A1. Relative to Q, the displacement of P in a time interval t

is r   (s )(  rˆ)  (s)tˆ  [( s)( rˆ)  s tˆ]t . It follows

(A3)*
v   s rˆ  s tˆ
Q
s+s
Figure A1
s


P
r 
s+s
s
 r̂
tˆ
(d) The velocity of the particle relative to O is the sum of the two relative velocities
given in Eqs. (A2) and (A3) so that
  
(A4)*
v  v   vQ  (s rˆ  s tˆ )  R tˆ  s rˆ

(e) Refer to Fig. A2. The (  tˆ )-component of the velocity change v is given by

(tˆ)  v  v  v t . Therefore, the tˆ -component of the acceleration


a  v / t is given by tˆ  aˆ  v . Since the speed v of the particle is s
according to Eq. (A4), we see that the tˆ -component of the particle’s
acceleration at P is given by

a  tˆ  v  (s)  s 2
 r̂
(A5)*
 tˆ
v 
Q
Figure A2

v


v
P
1
v

v
O
An equation marked with an asterisk contains answer to the problem.


v  v
IPHO 2003
Solution − Theoretical Question 1
(2003/07/12)
p.2/7
Note that, from Fig. A2, the radial component of the acceleration may also be

obtained as a  rˆ  dv / dt  d ( s) / dt .
(f) Refer to Fig. A3. The gravitational potential energy of the particle is given by
U  mgh . It may be expressed in terms of s and  as
U ( )  mg[ R(1  cos  )  s sin  ]
(A6)*
x
A
R −R cos
Q
Figure A3
h
R
s

s sin
R cos
O
P
(g) At the lowest point of its trajectory, the particle’s gravitational potential energy
U must assume its minimum value Um. If the particle’s mechanical energy E
were equal to Um, its kinetic energy would be zero. The particle would then
remain stationary and be in the static equilibrium state shown in Fig. A4. Thus,
the potential energy reaches its minimum value when  =  /2 or s = L−  R /2.
x
A

Figure A4
Q
2
R
O
s
P (at rest)
From Fig. A4 or Eq. (A6), the minimum potential energy is then

U m  U ( )  mg[ R  L  (R / 2)] .
2
(A7)
Initially, the total mechanical energy E is 0. Since E is conserved, the speed vm
of the particle at the lowest point of its trajectory must satisfy
E0
1 2
mv  U m .
2 m
(A8)
From Eqs. (A7) and (A8), we obtain
vm   2U m / m  2 g[ R  ( L  R / 2)] .
(A9)*
Solution − Theoretical Question 1
IPHO 2003
(2003/07/12)
p.3/7
Part B
(h) From Eq. (A6), the total mechanical energy of the particle may be written as
1 2
1
mv  U ( )  mv 2  mg[ R(1  cos )  s sin  ]
2
2
From Eq. (A4), the speed v is equal to s . Therefore, Eq. (B1) implies
E0
v 2  (s) 2  2 g[ R(1  cos )  s sin  ]
(B1)
(B2)
Let T be the tension in the string. Then, as Fig. B1 shows, the tˆ -component of
the net force on the particle is –T + mg sin . From Eq. (A5), the tangential
acceleration of the particle is (s 2 ) . Thus, by Newton’s second law, we have
(B3)
m(s 2 )  T  mg sin 
x
A
Q
s
Figure B1
O
T
P
mg sin

 R
 mg
According to the last two equations, the tension may be expressed as
mg
T  m( s 2  g sin  ) 
[2 R(1  cos )  3s sin  ]
s
2mgR
 3
L
(B4)

[tan  (  )]( sin  )
s
2 2
R
2mgR

( y1  y 2 )(sin  )
s
The functions y1  tan( / 2) and y2  3(  L / R) / 2 are plotted in Fig B2.
Figure B2
y
30
20
y1  tan
10

y1  tan
2
s
y2 
-20
-30
0
/2

2

0
-10

3
L
(  )
2
R
3/2

5/2
3
IPHO 2003
Solution − Theoretical Question 1
(2003/07/12)
p.4/7
From Eq. (B4) and Fig. B2, we obtain the result shown in Table B1. The angle at
which .y2 = y1 is called  s (    s  2 ) and is given by

3
L
( s  )  tan s
(B5)
2
R
2
or, equivalently, by

L
2
  s  tan s
R
3
2
(B6)
Since the ratio L/R is known to be given by
L 9 2

 2
1


 cot  (  )  tan (  )
R 8 3
16
8 3
2
8
one can readily see from the last two equations that  s  9 / 8 .
(B7)
Table B1
( y1  y2 )
sin 
tension T
0  
positive
positive
positive
 
  s
 s
 s    2
+
0
positive
negative
negative
positive
zero
negative
zero
positive
negative
negative
Table B1 shows that the tension T must be positive (or the string must be
taut and straight) in the angular range 0<  <  s. Once  reaches  s, the tension
T becomes zero and the part of the string not in contact with the rod will not be
straight afterwards. The shortest possible value smin for the length s of the line
segment QP therefore occurs at    s and is given by
9 2
 9
2R

 cot  ) 
cot  3.352R
8 3
16 8
3
16
When    s , we have T = 0 and Eqs. (B2) and (B3) then leads to
smin  L  R s  R(
v 2   gs sin  . Hence the speed v s is
(B8)
IPHO 2003
Solution − Theoretical Question 1
v s   gs min sin  s 
(2003/07/12)
2 gR


cot sin 
3
16
8
p.5/7
4 gR

cos
3
16
(B9)*
 1.133 gR
(i) When    s , the particle moves like a projectile under gravity. As shown in Fig.
B3, it is projected with an initial speed v s from the position P  ( x s , y s ) in a
direction making an angle   (3 / 2   s ) with the y-axis.
The speed v H of the particle at the highest point of its parabolic trajectory is
equal to the y-component of its initial velocity when projected. Thus,
v H  v s sin(  s   ) 
4 gR


cos sin  0.4334 gR
3
16
8
(B10)*
The horizontal distance H traveled by the particle from point P to the point of
maximum height is
H
v s2 sin 2( s   ) v s2
9

sin
 0.4535 R
2g
2g
4
x
s  
L
m
vs

R
vH
P  ( xs , y s )
O
y
Figure B3
(B11)
s

s
H
smin
Q
The coordinates of the particle when    s are given by
xs  R cos s  smin sin  s   R cos
y s  R sin  s  smin cos s   R sin

8

 smin sin
 smin cos

8
 0.358R

(B12)
(B13)
 3.478R
8
8
Evidently, we have | y s |  ( R  H ) . Therefore the particle can indeed reach its
maximum height without striking the surface of the rod.
Part C
(j) Assume the weight is initially lower than O by h as shown in Fig. C1.
x
L
A

m
R
O
Figure C1
h
M
IPHO 2003
Solution − Theoretical Question 1
(2003/07/12)
p.6/7
When the weight has fallen a distance D and stopped, the law of conservation of
total mechanical energy as applied to the particle-weight pair as a system leads
to
 Mgh  E   Mg (h  D)
(C1)
where E is the total mechanical energy of the particle when the weight has
stopped. It follows
E   MgD
(C2)
Let  be the total length of the string. Then, its value at  = 0 must be the same
as at any other angular displacement . Thus we must have
  L


R  h  s  R(  )  (h  D)
2
2
(C3)
Noting that D =  L and introducing ℓ = L−D, we may write
  L  D  (1   ) L
(C4)
From the last two equations, we obtain
(C5)
s  L  D  R    R
After the weight has stopped, the total mechanical energy of the particle
must be conserved. According to Eq. (C2), we now have, instead of Eq. (B1),
the following equation:
E   MgD 
1 2
mv  mg[ R(1  cos )  s sin  ]
2
The square of the particle’s speed is accordingly given by
2MgD
s
v 2  ( s) 2 
 2 gR[(1  cos )  sin  ]
m
R
Since Eq. (B3) stills applies, the tension T of the string is given by
 T  mg sin   m(s 2 )
(C6)
.
(C7)
(C8)
From the last two equations, it follows
T  m( s 2  g sin  )
mg 2 M
[
D  2 R (1  cos  )  3s sin  ]
s m
2mgR MD
3 

[
 (1  cos  )  (   ) sin  ]
s
mR
2 R

where Eq. (C5) has been used to obtain the last equality.
We now introduce the function
3 
f ( )  1  cos  (   ) sin 
2 R
From the fact ℓ = (L−D) >> R, we may write
(C9)
(C10)
IPHO 2003
Solution − Theoretical Question 1
f ( )  1 
(2003/07/12)
3
sin   cos  1  A sin(    )
2R
p.7/7
(C11)
where we have introduced
3
2R
(C12)
  tan 1
3 2
1 ( )
2R
From Eq. (C11), the minimum value of f() is seen to be given by
3 2
(C13)
f min  1  A  1  1  (
)
2R
Since the tension T remains nonnegative as the particle swings around the rod,
we have from Eq. (C9) the inequality
3 2
A  1 (
) ,
2R
M ( L  )
MD
3
 f min 
 1  1  ( )2  0
mR
mR
2R
(C14)
ML
M
3
M
3
) 1  (
)  1  ( )2  (
)( )
mR
mR
2R
mR
2R
(C15)
or
(
From Eq. (C4), Eq. (C15) may be written as
(
ML
ML
3L
)  1  [(
)  ( )](1   )
mR
mR
2R
(C16)
Neglecting terms of the order (R/L) or higher, the last inequality leads to
ML
2R
3L
) 1
1
( ) 1
1
mR
3L 
2
R
  1


ML
3L
ML
3L
2M
2M
(
)( ) (
)( )
1 1
mR
2R
mR
2R
3m
3m
(
The critical value for the ratio D/L is therefore
1
c 
2M
(1 
)
3m
(C17)
(C18)*