EdExcel Mechanics 2
Centres of mass
Chapter Assessment
Take g = 9.8 ms-2 throughout this test.
1.
y
A
4m
D
2m
4
2
-4
-2
B 3m
C m
0
2
4
x
6
In this question, coordinates are referred to the axes shown above.
Four particles at A, B, C and D with masses 4m, 3m, m and 2m lie in a plane with
positions (-3, 4), (0, 0), (2, 0) and (5, 4) respectively.
(i)
Calculate the coordinates of the centre of mass of the four particles.
[5]
A thin, uniform, rigid wire of mass 12m connects A to B, B to C and C to D, as
shown below.
D
A
B
C
(ii) Calculate the coordinates of the centre of mass of the wire.
[4]
(iii) Calculate the coordinates of the combined centre of mass of the wire and the
particles at A, B, C and D.
[3]
A is connected to D by means of a further straight rigid wire of negligible mass.
The combined system of wires and the particles at A, B, C and D is suspended
freely from the midpoint of AD.
(iv) What extra mass must be added at C if the system is to hang in equilibrium
with AD horizontal?
[3]
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2. In this question, the units on the axes are metres. Answers should be given in
exact numbers or correct to three significant figures. Coordinates are referred to
the axes shown in the diagram below.
The diagram shows a shape ABCDEF with the dimensions given.
y
F
0.2
E
0.4
D
C
0.2
A
(i)
B
0.4
x
Regarding ABCDEF as a uniform lamina, show that the coordinates of its
centre of mass are  16 , 16  .
[5]
(ii) Now regarding ABCDEF as a framework of uniform rods, show that the
coordinates of its centre of mass are (0.175, 0.175) .
[5]
y
3.
6 cm
L
F
E
P
4 cm
C
D
2 cm
O
M
B
A
x
2 cm
In this question, coordinates refer to the axes shown in the diagram above and the
units are centimetres. Answers should be given correct to three significant figures.
A thin, uniform sheet of metal is cut to form the shape shown above. OAEF is a
rectangle and ABCD is a square with the dimensions shown. A square of side 2.5
cm and centre (4, 2) has been removed leaving the shaded part P.
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(i)
Show that the x-coordinate of the centre of mass of P is 3.45, correct to three
significant figures, and calculate the y-coordinate.
[5]
L and M are the mid-points of EF and OA respectively. The shape P is freely
suspended from L.
(ii) Calculate the angle that LM makes with the vertical.
[3]
The mass of the square removed is 0.05 kg.
(iii) What vertical force must be applied at O so that when P is freely suspended
from L, the line LM is vertical?
[4]
The lamina is now placed on a plane inclined at an angle  to the horizontal. The
edge OF is in contact with the plane, O being lower than F.
(iv) Assuming that the lamina does not slide down the plane, find the maximum
value of  for which the lamina remains in equilibrium.
[3]
Total 40 marks
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Solutions to Chapter Assessment
1. (i)
Particle
Mass
x-coordinates
y-coordinates
For x-coordinates:
A
4m
-3
4
B
3m
0
0
C
D
2m
5
4
m
2
0
Total
10m
x
y
(4m  3)  (3m  0)  (m  2)  (2m  5 )  10mx
0  10mx
x 0
For y-coordinates: (4m  4)  (3m  0)  (m  0)  (2m  4)  10my
24m  10my
y  2.4
The centre of mass is at (0, 2.4)
(ii) Length AB = CD = 5, length BC = 2.
Mass of AB = 5m, mass of BC = 2m, mass of CD = 5m
AB, BC and CD can be considered as particle masses located at their
centres.
Wire
Mass
x-coordinate
y-coordinate
AB
5m
-1.5
2
BC
2m
1
0
CD
5m
3.5
2
Total
12m
x
y
By symmetry, the x-coordinate of the centre of mass = 1.
For y-coordinates: (5 m  2)  (2m  0)  (5 m  2)  12my
20m  12my
y
5
3
Centre of mass is  1, 53 
(iii)
Wire
12m
1
Mass
x-coordinate
y-coordinate
For x-coordinates:
5
3
Particles
10m
0
2.4
Total
22m
x
y
(12m  1)  (10m  0)  22mx
12m  22mx
x
6
11
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For y-coordinates: (12m  53 )  (10m  2.4)  22my
44m  22my
y 2
The centre of mass is at ( 116 , 2)
(iv)
X
5
11
G
1
M
C
Mg  1  22mg  115  0
M  10m
Taking moments about X:
The extra mass to be added at C is 10m.
2. (i)
y
F
0.4
0.2
E
D
P
C
Q
A
Lamina
Area
x-coordinate
y-coordinate
0.2
P
0.08
0.1
0.2
x
B
0.4
Q
0.04
0.3
0.1
Total
0.12
x
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y
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EdExcel Mechanics 2
For x-coordinates:
(0.08  0.1)  (0.04  0.3)  0.12 x
0.02  0.12 x
x  61
By symmetry y  x  61
Centre of mass is  61 , 61 
(ii) Each rod can be considered to be a point mass located at the rod’s centre.
y
F
0.2
E
0.4
D
C
0.2
A
Rod
Length
x-coordinate
y-coordinate
AB
0.4
0.2
0
x
B
0.4
BC
0.2
0.4
0.1
CD
0.2
0.3
0.2
DE
0.2
0.2
0.3
EF
0.2
0.1
0.4
FA
0.4
0
0.2
Total
1.6
x
y
For x-coordinates:
(0.4  0.2)  (0.2  0.4)  (0.2  0.3) (0.2  0.2) (0.2  0.1)  (0.4  0)  1.6 x
0.28  1.6 x
x  0.175
By symmetry y  x  0.175
Centre of mass is at (0.175, 0.175).
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EdExcel Mechanics 2
y
3. (i)
6 cm
F
L
E
P
4 cm
C
D
2 cm
O
A
M
Lamina
Area
x-coordinate
y-coordinate
For x-coordinates:
OAEF
24
3
2
ABCD
4
7
1
2 cm
Square
6.25
4
2
x
B
P
21.75
x
y
(24  3)  (4  7)  (6.25  4)  21.75 x
75  21.75 x
x  3.45 (3 s.f.)
For y-coordinates: (24  2)  (4  1)  (6.25  2)  21.75 y
39.5  21.75 y
y  1.82 (3 s.f.)
(ii)
L
2.18 
0.45
G
M
tan  
0.4483
2.1839
  11.6 (3 s.f.)
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(iii)Area of square = 6.25
Mass of square = 0.05
0.05
Mass per unit area 
 0.008
6.25
Mass of P  21.75  0.008  0.174
L
0.4483
3
O
0.174g
X
Taking moments about L: 0.174  9.8  0.4483  X  3  0
X  0.255 (3 s.f.)
The vertical force required is 0.255 N.
(iv)
G
y
x


tan  
y 39.5

x
75
  27.8 (3 s.f.)
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