Solution to 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)
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

v    s rˆ  s tˆ
(A3)
Q
s+s
s
P


r 
s+s
s
Figure A1
 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
  
v  v   vQ  (  s rˆ  s tˆ )  R tˆ   s rˆ
(A4)

(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
 tˆ
v 
Q

(A5)
v


v
P
v

v
O
12


v  v
Figure A2
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
h
R
s

s sin
R cos
O
Figure A3
P
(g) At the lowest point of its trajectory, the particle’s gravitational potential energy U must
assume its minimum value Um. By differentiating Eq. (A6) with respect to θ and using
Eq. (A1), the angle  m corresponding to the minimum gravitational energy can be
obtained.
dU
ds


 mg  R sin  
sin   s cos 
d
d


 mgR sin   (  R ) sin   s cos 
 mgs cos
At    m ,

dU
 0 . We have  m  . The lowest point of the particle’s trajectory is
2
d  m
shown in Fig. A4 where the length of the string segment of QP is s = L−  R /2.
x
A

Q
2
R
O
s
Figure A4
P
From Fig. A4 or Eq. (A6), the minimum potential energy is then
U m  U  / 2  mg[ R  L  (R / 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
13
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)
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
Q
s

 R
A
O
T
P
mg sin
 mg
Figure B1
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

[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.
14
(B4)
y
Figure B2
30
20
y1  tan
10

y1  tan
2
s

2
0

-10
y2 
-20
-30
0
/2

3
L
(  )
2
R
3/2

5/2
3
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
sin 
( y1  y2 )
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
15
smin  L  R s  R(
9 2
 9
2R

 cot  ) 
cot  3.352R
8 3
16 8
3
16
(B8)
When    s , we have T = 0 and Eqs. (B2) and (B3) then leads to v 2s   gsmin sin  s .
Hence the speed v s is
2 gR


cot sin 
3
16
8
v s   gs min sin  s 
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,
4 gR


cos sin  0.4334 gR
3
16
8
v H  v s sin(  s   ) 
(B10)
The horizontal distance H traveled by the particle from point P to the point of maximum
height is
v s2 sin 2( s   ) v s2
9
H

sin
 0.4535 R
2g
2g
4
x
s  
L
m
vs

R
y
(B11)
vH
P  ( xs , y s )
O

s
H
smin
s
Q
Figure B3
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
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.
16
 smin cos

Part C
(j) Assume the weight is initially lower than O by h as shown in Fig. C1.
x
L
A

m
R
O
h
M
Figure C1
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
Noting that D =  L and introducing ℓ = L−D, we may write
  L  D  (1   ) L
(C3)
(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
2 MgD
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 )
From the last two equations, it follows
17
(C6)
(C7)
(C8)
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  D )  R , we may write
3
f ( )  1 
sin   cos  1  A sin(    )
2R
where we have introduced
 2R 
3 2
A  1 (
) ,   tan 1 

2R
 3 
From Eq. (C11), the minimum value of f() is seen to be given by
(C9)
(C10)
(C11)
(C12)
2
3  
f min  1  A  1  1  
(C13)

2 R
Since the tension T remains nonnegative as the particle swings around the rod, we have
from Eq. (C9) the inequality
MD
M ( L  )
 3 
 f min 
1 1 
 0
mR
mR
 2R 
2
(C14)
or
2
 ML 
 M 
 3 
 M   3 

 1  
  1 
 


 mR 
 mR 
 2R 
 mR   2 R 
(C15)
From Eq. (C4), Eq. (C15) may be written as
 ML 
 ML 3L 


 1  
(1   )
 mR 
 mR 2 R 
Neglecting terms of the order (R/L) or higher, the last inequality leads to
 ML 
2R
3L

 1
1
1
1
mR

3L 
  1 
 2R

 ML 3L  ML  3L 2 M  1 1  2 M



3m
3m
 mR 2 R  mR 2 R
The critical value for the ratio D/L is therefore
1
c 
2M
1
3m
18
(C16)
(C17)
(C18)
Marking Scheme
Theoretical Question 1
A Swing with a Falling Weight
Total
Scores
Part A
Sub
Scores
(a)
4.3 pts.
0.5
(b)
0.5
(c)
0.7
(d)
0.7
(e)
0.7
(f)
0.5
(g)
0.7
Marking Scheme for Answers to the Problem
Relation between  and s . ( s   R )
 0.2 for   s .
 0.3 for proportionality constant (-R).

Velocity of Q relative to O. ( vQ  R tˆ )
 0.2 for magnitude R  .
 0.3 for direction tˆ .

Particle’s velocity at P relative to Q. ( v    s rˆ  s tˆ )
 0.2+0.1 for magnitude and direction of r̂ -component.
 0.3+0.1 for magnitude and direction of tˆ -component.
  
Particle’s velocity at P relative to O. ( v  v   vQ   s rˆ )


 0.3 for vector addition of v  and vQ .

 0.2+0.2 for magnitude and direction of v .
tˆ -component of particle’s acceleration at P.


 0.3 for relating a or a  tˆ to the velocity in a way that implies

| a  tˆ |  v 2 / s .

 0.4 for a  tˆ   s 2 (0.1 for minus sign.)
Potential energy U.
 0.2 for formula U   mgh .
 0.3 for h  R(1  cos  )  s sin  or U as a function of  s, and R.
Speed at lowest point vm.
 0.2 for lowest point at    / 2 or U equals minimum Um.
2
 0.2 for total mechanical energy E  mvm
/ 2 Um  0 .
 0.3 for vm   2U m / m  2 g[ R  ( L  R / 2)] .
Part B
(h)
Particle’s speed vs when QP is shortest.
4.3 pts.
2.4
 0.4 for tension T becomes zero when QP is shortest.
 0.3 for equation of motion  T  mg sin   m(s 2 ) .
 0.3 for E  0  m(s) 2 / 2  mg[ R(1  cos )  s sin  ] .

3
L
 0.4 for ( s  )  tan s .
2
R
2
 0.5 for  s  9 / 8 .
 0.3+0.2 for vs  4gR / 3 cos  / 16  1.133 gR
19
(i)
1.9
Part C
(j)
3.4 pts
3.4
The speed vH of the particle at its highest point.
 0.4 for particle undergoes projectile motion when    s .
 0.3 for angle of projection   (3 / 2   s ) .
 0.3 for v H is the y-component of its velocity at    s .
 0.4 for noting particle does not strike the surface of the rod.
 0.3+0.2 for
vH  4 gR / 3 cos( / 16) sin(  / 8)  0.4334 gR .
The critical value  c of the ratio D/L.
 0.4 for particle’s energy E   MgD when the weight has stopped.
 0.3 for s  L  D  R .
 0.3 for E   MgD  mv 2 / 2  mg[ R(1  cos )  s sin  ] .
 0.3 for  T  mg sin   m(s 2 ) .




0.3 for concluding T must not be negative.
0.6 for an inequality leading to the determination of the range of D/L.
0.6 for solving the inequality to give the range of  = D/L.
0.6 for  c  (1  2M / 3m) .
20