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Mechanics 2
Centre of Mass
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Centre of Mass
We use moments to find the centre of mass of uniform plane figures and
discrete mass distributions. In M1 we used moments to calculate the centre
of mass for non uniform rods but in M2 we need to consider two dimensions.
M1 RECAP .......................................................................................................................... 3
CENTRE OF MASS OF A SYSTEM OF PARTICLES DISTRIBUTED IN TWO
DIMENSIONS ................................................................................................................ 4
QUESTIONS 1 ................................................................................................................ 7
CENTRE OF MASS OF A UNIFORM PLANE LAMINA. ....................................... 9
APPLICATION TO COMPOSITE FIGURES. ......................................................... 12
FRAMEWORKS CONSTRUCTED FROM UNIFORM RODS AND WIRES. ..... 16
QUESTIONS 2 ............................................................................................................. 18
EQUILIBRIUM OF A UNIFORM LAMINA ON AN INCLINED PLANE. ...... 23
EQUILIBRIUM OF A PLANE LAMINA ................................................................. 25
QUESTIONS 3 ............................................................................................................ 28
Past Examination Questions………………………………………………………………..31
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M1 Recap
Example 1
Three particles of mass 6kg, 3kg and 2.5kg are attached to a light rod PQ of
length 3m at the points P, Q and R, where PR = 0.9m. Find the position of the
centre of mass of the system.
Start by adding the centre of mass to the diagram and let the distance PG
be x.
Taking moments about P gives:
11.5g × x = 3g × 0.9 + 3 × 2.5g
x = 0.89m
6g 3g 2.5g
P Q R
0.9m
G
11.5g
6g 3g 2.5g
P Q R
0.9m
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Centre of mass of a system of particles distributed in two dimensions
The principle applied above is simply applied firstly in the horizontal
direction and then vertically.
Example 2
Particles of mass 2kg, 4kg, 5kg and 6kg are attached to the corners of a
light rectangular plate PQRS. Given that PQ = 5cm and QR = 12cm calculate
the distance of the centre of mass of the system from
a) PQ
b) PS
It is very easy to make simple numerical mistakes with these questions and
therefore you are advised to set the question out in a table. It is assumed
that the centre of mass horizontally is at X and vertically at Y .
Separate Masses Total Mass
Mass 2 4 5 6 17
x co-ord 0 0 12 12 X
Y co-ord 0 5 5 0 Y
To find the distance of the centre of mass from PQ we use the formula:
Q
P S
R
4kg
2kg 6kg
5kg
5cm
12cm
i i imx mX
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To find the distance from PS we use the formula:
i i imy mY
20 25 17Y
Y 2.65
Example 3
The diagram below shows a series of particles that make up a system. The
centre of mass of the system is at the point (x,y). Find the coordinates of
the centre of mass of the system.
With an example such as this it is once again easy to make a simple numerical
mistake as students may miss the minus signs.
Y
X
3kg
4Kg
2Kg
2.5kg
(x,y)
60 72 17X
X 7.76
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Using the same tabular approach:
Separate Masses Total Mass
Mass 2 2.5 4 3 11.5
x co-ord -2 1 3 5 X
Y co-ord 2 -1 -2 2 Y
Using the formula: i i imx mX
4 2.5 12 15 11.5X
X 2.22
And similarly for the y direction:
i i imy mY
4 2.5 8 6 11.5Y
Y 0.04
Therefore the centre of mass of the system is at (2.22,0.04)
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Questions 1
1 A light rod PR of length 3.5m has particles of mass 1.5kg, 3kg and 2.5kg
attached to it at P Q and R respectively, where PQ = 1.5m. Determine the
distance of the centre of mass from R.
2 Particles of mass 4kg, 2.5kg, 6kg and 3kg are attached to the corners of a
light rectangular plate PQRS. Given that PQ = 8cm and QR = 16cm calculate
the distance of the centre of mass of the system from
a) PQ
b) PS
3 Four particles of mass 1kg, 2kg, 5kg and 2.5kg lie in the (x,y) plane at the
points with coordinates (1,2), (-2,3), (4,2) and (-3,2) respectively. Calculate
the coordinates of the centre of mass of the system.
Q
P S
R
2.5kg
4kg 3kg
6kg
8cm
16cm
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4 The system below is made up of three light rods. Three masses of value
2.5kg, 4kg and 1kg are placed at the vertices A, B and C respectively.
Calculate the distance of the centre of mass from
a) AB
b) AC.
A
B
C
18cm
12cm
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Centre of mass of a uniform plane lamina.
Obviously there are some standard results to be taken for granted.
Uniform rectangular lamina:- at centre of the shape
Uniform circular disc:- at centre of disc
Uniform triangular lamina:-
o Equilateral:- at centre
o Isosceles:- at the intersection of the
medians
A median is a line that joins a vertex of a triangle to the centre of the side
opposite to the vertex. The centre of mass of a scalene triangle is at a point
one third of the way along the median (from the edge).
The centre of mass of the triangle ABC is at the point G, where EG = ⅓EC.
A
B
C
G E
F
D
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Uniform circular arc:- the centre of mass is along the axis of
symmetry at a distance rsin
from the centre where α is measured
in radians.
Uniform Sector:- centre of mass is on the axis of symmetry at a
distance 2rsin
3
from the centre, where α is measured in radians.
(note the use of α in the formula and 2α in the diagram.)
r
2α
r
2α
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Example 4
Calculate the coordinates of the centre of mass of the uniform triangular
lamina ABC if the point A is placed at the origin.
From the definitions above the centre of mass is at the point that is one
third of the way along DC.
Note that D has coordinates (0,6).
Therefore C of M is at (6,4).
D
A
B
C
18m
12m
A
B
C
18m
12m
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Application to composite figures.
Example 5
The centre of mass for the shape below can be found by using the same
ideas as those set out above. All you need to do is treat each part
separately.
Assume that the shape ABCDE is a uniform lamina made up of a rectangle
and an isosceles triangle.
One only needs to consider the distance of the centre of mass from the
edge AB as the shape is symmetrical and therefore the centre of M will lie
along the mirror line.
Since the rectangle ABCE is uniform its centre of mass will be 6 cm from
AB. The centre of mass of the isosceles triangle will be one third of the way
along the line FD which is a distance of 14.5cm from the line AB. By applying
the tabular method we get:-
Separate Masses Total Mass
Mass 108m 33.75m 141.75m
x co-ord 6 14.5 X
Since the separate parts are uniform I have used the areas to represent the
masses.
7.5cm F
A
D
E
C B
12cm
9cm
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Using the formula: i i imx mX
(6 108) (14.5 33.75) 141.75X
1137.375 141.75X
X 8.02cm
Example 6
The object below is formed by removing a uniform semi circular disc of
radius 3m from a second uniform semi circular disc of radius 6m. Calculate
the center of mass of the object.
By definition the centre of mass will lie on the mirror line.
The formula for the centre of mass for a uniform sector is 2rsin
3
where
the angle at the centre is 2α. Obviously in the above example 2α = ∏.
Using the tabular approach again:
Separate Masses Total Mass
Mass 9
2
18 27
2
y co-ord 2 3 sin 42 sin3 3 22
2 6 sin 82 sin3 22
X
3m 6m
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Obviously the disc we are interested in can be found by subtracting the
small semi circular disc from the larger one.
Using the same formula again:
i i imx mX
8 9 4 2718 sin sin X
2 2 2 2
27144 18 X
2
X 2.97
So the centre of mass is 2.97cm from the centre along the axis of
symmetry.
The same principle of adding and subtracting parts can be applied to the
example below.
Example 7
The object below is formed by removing two uniform circular discs from a
uniform rectangular lamina. Calculate the centre of mass of the plate.
A
10cm
16cm
D
C B
1.5cm
6cm
12cm
3cm
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Separate Masses Total Mass
Circle 1 Circle 2 Rectangular
Lamina
Plate
Mass 2.25∏ 2.25∏ 120 120-4.5∏
x co-ord 3 12 6 X
Y co-ord 6 6 5 Y
Plate = Rectangular Lamina - Circle 1 - Circle 2
i i imx mX
(120-4.5∏)X = 720 – (7.75∏ + 27∏)
X = 5.77cm
Looking at it vertically:
i i imy mY
(120-4.5∏)Y = 600 – (13.5∏ + 13.5∏)
Y = 4.87cm
Therefore the centre of mass is 5.77cm from AB and 4.87cm from AD.
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Frameworks constructed from uniform rods and wires.
By definition the centre of mass of a uniform rod is at its centre therefore
a framework can be treated as a series of particles.
Example 8
The framework ABC is made up of three uniform rods and a semi circular
arc. Find the centre of mass of the framework assuming that A is at the
origin.
Remembering that the formula for the centre of mass of a uniform circular
arc is rsin
and that
2
. Also note that the centre of mass of the arc
will have a negative x coordinate.
Separate Masses Total Mass
Rod AB Rod AC Rod BC Arc AB Framework
Mass 12m 16m 20m 6∏m (48 + 6∏)m
x co-ord 0 8 8 6sin 122
2
X
Y co-ord 6 0 6 6 Y
20cm
A
B
C
16cm
12cm
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Looking at the problem horizontally:
i i imx mX
12(16 8) (20 8) (6 ) (48 6 )X
X 3.23cm
And now vertically:
i i imy mY
(12 8) (20 6) (6 6) (48 6 )Y
Y 4.56cm
Therefore the centre of mass is at the point with coordinates (3.23,4.56).
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Questions 2
1 For the uniform laminae below determine the coordinates of the
centre of mass.
a)
b)
c)
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2 Determine the coordinates of the centre of mass of the uniform plate
below assuming that A is at the origin.
3 The diagram shows a uniform lamina PQRS. Find the distance of
centre of mass from
a) PQ
b) PS
R Q
12cm
12cm
18cm S P
F
E
D
B
27cm
14cm
6cm
19cm
A
C
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4 A uniform circular sector has a radius of 25cm and the angle at the
centre is 100º. Find the distance of the centre of mass of the sector from
the centre of circle.
5 An arc of a circle has a radius 18cm and an angle the centre of 5
8
radians.
Find the distance of the centre of mass of the arc from the centre of
circle.
18cm
5
8
25c
m
100º
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6 A uniform framework is made up of a uniform circular arc and two uniform
rods as displayed in the diagram below. Calculate the distance of the centre
of mass from the centre of the circle.
7 A thin isosceles triangular plate has a corner folded over as shown in
the diagram below. Find the distance of the centre of mass from AB.
14cm
9cm B
A
C
27cm
25c
m
100º
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8 The earring below is formed from a uniform circular disc of radius
2.5cm. As part of the construction process a circular disc of radius 0.5cm is
removed from the original disc where the centre is 3cm from A along the
line AB. The disc is then added back onto the original disc where the centre
of the disc is 0.5cm from A along the line AB. Calculate the distance of the
centre of mass of the earring from A.
A
B
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Equilibrium of a uniform lamina on an inclined plane.
For an inclined lamina to remain in equilibrium on an inclined plane the line of
action of the weight must fall within the side of the lamina that is in contact
with the plane (as shown in the diagram below).
In the second diagram the object will topple over.
G
θ
G
θ
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Example 9
A uniform rectangular lamina is placed on a plane inclined at an angle θ. Given
that the lamina is in limiting equilibrium find the angle θ.
By simple trigonometry:
3Tan
7.5
21.8
In exam questions students need to be careful in choosing the angle, a
diagram and the application of alternate angles is usually sufficient.
6cm
15cm G
θ
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Equilibrium of a plane lamina
A suspended lamina will be in equilibrium when its centre of mass is directly
below the point of suspension.
Exam question
The diagram shows a uniform sheet of metal. Find the distance, in cm, of the
centre of mass of the plate.
a) (i) From AB
(ii) From AF.
The shape hangs in equilibrium from the point C.
b) Find, to the nearest degree the angle made by CD with the downward
vertical.
a) Splitting the shape into two rectangles and applying the tabular approach
gives:
F
E
D
B
15cm
8cm
3cm
9cm
A
C
1
2
F
E
D
B
15cm
8cm
3cm
9cm
A
C
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Using the tabular approach:
Separate Masses Total Mass
Rectangle 1 Rectangle 2 Plate
Mass 45m 30m 75m
x co-ord 1.5 5.5 X
Y co-ord 7.5 12 Y
Looking at the problem horizontally:
i i imx mX
(45 1.5) (30 5.5) 75X
X 3.1cm
Therefore the distance from AB is 3.1cm
And now vertically:
i i imy mY
(45 7.5) (30 12) 75Y
Y 9.3cm
Therefore the distance from AF is 9.3cm
b) Once the object is suspended from C, by definition CG will be vertical
and hence θ needs to be found.
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4.9Tan
5.7
40.7
Therefore the angle between the downward vertical and CD is 40.7º.
θ
G
F
E
D
B
15cm
8cm
3cm
9cm
A
C
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Questions 3
1 The diagram shows a uniform sheet of metal. Find the distance, in cm,
of the centre of mass of the plate.
a) (i) From AB
(ii) From AF
The shape hangs in equilibrium from the midpoint of the edge AB.
b) Find, to the nearest degree the angle made by ED with the downward
vertical.
2 A uniform semicircular lamina is suspended from a point on the
diameter half way between the centre and the outer edge. Calculate in
radians the angle that the diameter makes with the downward vertical.
F
E
D
B
22cm
14cm
5cm
16cm
A
C
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3 The framework ABC is made up of three uniform rods and a semi
circular arc. Find the centre of mass of the framework assuming that A is at
the origin.
If the framework is suspended from B, find the angle that AB makes with
the downward vertical.
4 The metallic lamina below is non-uniform. Areas CDEF and IFGH are
twice as dense as the rest of the plate. Calculate the distance of the centre
of mass from:
a) HG
b) JL
If the object is freely suspended from B and is hanging in equilibrium, find
the angle that BC makes with the downward vertical.
5 If the object in example 7 is freely suspended from D, find the angle
that AD makes with the horizontal.
10cm
2cm
6cm
A B
J
H
E
D C M
I F
G
L
15cm
A
B
C
12cm
9cm
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6 The earring below is formed from three quarters of a circular disc of
radius 3cm. The distance OB is 2.5cm. Find the distance of the centre of
mass, along the line of symmetry, from O.
If the earring is freely suspended from A, find the angle that AB makes
with the mirror line.
7 The object below is formed by removing a uniform semi circular disc
of radius 2.5m from a second uniform semi circular disc of radius 8m.
Calculate the center of mass of the object.
If the object above is placed on an inclined plane, at what angle would the
object topple?
8m 2.5cm
A B O
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Past Examination Questions
1. Figure 1
C A
10 cm
D
20 cm
B
Figure 1 shows a decoration which is made by cutting 2 circular discs from a sheet of
uniform card. The discs are joined so that they touch at a point D on the circumference of
both discs. The discs are coplanar and have centres A and B with radii 10 cm and 20 cm
respectively.
(a) Find the distance of the centre of mass of the decoration from B.
(5)
The point C lies on the circumference of the smaller disc and CAB is a right angle. The
decoration is freely suspended from C and hangs in equilibrium.
(b) Find, in degrees to one decimal place, the angle between AB and the vertical.
(4)
Q2, June 2001
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2. Figure 1
A
W Z
2a
12a N O C
X Y
B
8a
Figure 1 shows a template made by removing a square WXYZ from a uniform triangular
lamina ABC. The lamina is isosceles with CA = CB and AB = 12a. The mid-point of AB is
N and NC = 8a. The centre O of the square lies on NC and ON = 2a. The sides WX and
ZY are parallel to AB and WZ = 2a. The centre of mass of the template is at G.
(a) Show that NG = a11
30 .
(7)
The template has mass M. A small metal stud of mass kM is attached to the template at C.
The centre of mass of the combined template and stud lies on YZ. By modelling the stud
as a particle,
(b) calculate the value of k.
(4)
Q4, Jan 2002
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3. Figure 2
A B
10 cm
D 10 cm C
A uniform lamina L is formed by taking a uniform square sheet of material ABCD, of
side 10 cm, and removing the semi-circle with diameter AB from the square, as shown in
Fig. 2.
(a) Find, in cm to 2 decimal places, the distance of the centre of mass of the lamina L
from the mid-point of AB.
(7)
[The centre of mass of a uniform semi-circular lamina, radius a, is at a distance 3
4a
from the centre of the bounding diameter.]
The lamina is freely suspended from D and hangs at rest.
(b) Find, in degrees to one decimal place, the angle between CD and the vertical.
(4)
Q4, June 2002
4. Three particles of mass 3m, 5m and m are placed at points with coordinates (4, 0), (0,
3) and (4, 2) respectively. The centre of mass of the system of three particles is at (2, k).
(a) Show that = 2.
(4)
(b) Calculate the value of k.
(3)
Q1, Jan 2003
:
Figure 1
L
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4a
5. Figure 2
A 8a B
6a G X C
E D
Figure 2 shows a uniform lamina ABCDE such that ABDE is a rectangle, BC = CD, AB =
8a and AE = 6a. The point X is the mid-point of BD and XC = 4a. The centre of mass of
the lamina is at G.
(a) Show that GX = a1544 .
(6)
The mass of the lamina is M. A particle of mass M is attached to the lamina at C. The
lamina is suspended from B and hangs freely under gravity with AB horizontal.
(b) Find the value of .
(3)
Q1, Jan 2003
6. Figure 1
a
A B
C ED
2a
3a
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A uniform lamina ABCD is made by taking a uniform sheet of metal in the form of a
rectangle ABED, with 3AB a and 2AD a , and removing the triangle BCE, where C
lies on DE and CE = a, as shown in Fig. 1.
(a) Find the distance of the centre of mass of the lamina from AD.
(5)
The lamina has mass M. A particle of mass m is attached to the lamina at B. When the
loaded lamina is freely suspended from the mid-point of AB, it hangs in equilibrium with
AB horizontal.
(b) Find m in terms of M.
(4)
Q3, June 2003
7. Figure 3
A O B
4m m
2a
2m
D 5a C
A loaded plate L is modelled as a uniform rectangular lamina ABCD and three particles.
The sides CD and AD of the lamina have lengths 5a and 2a respectively and the mass of
the lamina is 3m. The three particles have mass 4m, m and 2m and are attached at the
points A, B and C respectively, as shown in Fig. 3.
(a) Show that the distance of the centre of mass of L from AD is 2.25a.
(3)
(b) Find the distance of the centre of mass of L from AB.
(2)
The point O is the mid-point of AB. The loaded plate L is freely suspended from O and
hangs at rest under gravity.
(c) Find, to the nearest degree, the size of the angle that AB makes with the horizontal.
(3)
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A horizontal force of magnitude P is applied at C in the direction CD. The loaded plate L
remains suspended from O and rests in equilibrium with AB horizontal and C vertically
below B.
(d) Show that P = 4
5 mg.
(4)
(e) Find the magnitude of the force on L at O.
(4)
Q7, Jan 2004
8. Figure 1
Figure 1 shows a decoration which is made by cutting the shape of a simple tree from a
sheet of uniform card. The decoration consists of a triangle ABC and a rectangle PQRS.
The points
P and S lie on BC and M is the mid-point of both BC and PS. The triangle ABC is
isosceles with
AB = AC, BC = 4 cm, AM = 6 cm, PS = 2 cm and PQ = 3 cm.
(a) Find the distance of the centre of mass of the decoration from BC.
(5)
The decoration is suspended from Q and hangs freely.
(b) Find, in degrees to one decimal place, the angle between PQ and the vertical.
(4)
Q3, June 2004
6 cm
C
R
B
M P
Q
S
A
3 cm
2 cm
4 cm
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9. Figure 1
A uniform rod AB, of length 8a and weight W, is free to rotate in a vertical plane about a
smooth pivot at A. One end of a light inextensible string is attached to B. The other end is
attached to point C which is vertically above A, with AC = 6a. The rod is in equilibrium
with AB horizontal, as shown in Figure 1.
(a) By taking moments about A, or otherwise, show that the tension in the string is 65 W.
(4)
(b) Calculate the magnitude of the horizontal component of the force exerted by the
pivot on the rod.
(3)
Q1, Jan 2005
10. Figure 2
6a
A B
C
8a
A
5 cm
D
6 cm
20 cm
C
B
10 cm
O
3 cm
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Figure 2 shows a metal plate that is made by removing a circle of centre O and radius 3
cm from a uniform rectangular lamina ABCD, where AB = 20 cm and BC = 10 cm. The
point O is 5 cm from both AB and CD and is 6 cm from AD.
(a) Calculate, to 3 significant figures, the distance of the centre of mass of the plate from
AD.
(5)
The plate is freely suspended from A and hangs in equilibrium.
(b) Calculate, to the nearest degree, the angle between AB and the vertical.
(3)
Q2, Jan 2005
11. Figure 1
A thin uniform wire, of total length 20 cm, is bent to form a frame. The frame is in the
shape of a trapezium ABCD, where AB = AD = 4 cm, CD = 5 cm and AB is perpendicular
to BC and AD, as shown in Figure 1.
(a) Find the distance of the centre of mass of the frame from AB.
(5)
The frame has mass M. A particle of mass kM is attached to the frame at C. When the
frame is freely suspended from the mid-point of BC, the frame hangs in equilibrium with
BC horizontal.
(b) Find the value of k.
(3)
Q2, June 2005
4 cm
5 cm 4 cm
D
C
A
B
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12. Figure 1
Figure 1 shows a triangular lamina ABC. The coordinates of A, B and C are (0, 4), (9, 0)
and (0, –4) respectively. Particles of mass 4m, 6m and 2m are attached at A, B and C
respectively.
(a) Calculate the coordinates of the centre of mass of the three particles, without the
lamina.
(4)
The lamina ABC is uniform and of mass km. The centre of mass of the combined system
consisting of the three particles and the lamina has coordinates (4, ).
(b) Show that k = 6.
(3)
(c) Calculate the value of .
(2)
The combined system is freely suspended from O and hangs at rest.
(c) Calculate, in degrees to one decimal place, the angle between AC and the vertical.
(3)
Q5, Jan 2006
4m
4
4
2m
6m O 9
C
A
B
y
x
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13. Figure 1
Figure 1 shows four uniform rods joined to form a rigid rectangular framework ABCD,
where AB = CD = 2a, and BC = AD = 3a. Each rod has mass m. Particles, of mass 6m and
2m, are attached to the framework at points C and D respectively.
(a) Find the distance of the centre of mass of the loaded framework from
(i) AB,
(ii) AD.
(7)
The loaded framework is freely suspended from B and hangs in equilibrium.
(b) Find the angle which BC makes with the vertical.
(3)
Q4, June 2006
14. Figure 1
B
3a
D (2m)
2a
A
C (6m)
X
8cm 24 cmBA
O
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Figure 1 shows a template T made by removing a circular disc, of centre X and radius
8 cm, from a uniform circular lamina, of centre O and radius 24 cm. The point X lies on
the diameter AOB of the lamina and AX = 16 cm. The centre of mass of T is at the point
G.
(a) Find AG.
(6)
The template T is free to rotate about a smooth fixed horizontal axis, perpendicular to the
plane of T, which passes through the mid-point of OB. A small stud of mass 41 m is fixed
at B, and T and the stud are in equilibrium with AB horizontal. Modelling the stud as a
particle,
(b) find the mass of T in terms of m.
(4)
Q3, Jan 2007
15. Figure 1
A uniform lamina ABCDEF is formed by taking a uniform sheet of card in the form of a
square AXEF, of side 2a, and removing the square BXDC of side a, where B and D are
the mid-points of AX and XE respectively, as shown in Figure 1.
(a) Find the distance of the centre of mass of the lamina from AF.
(4)
The lamina is freely suspended from A and hangs in equilibrium.
(b) Find, in degrees to one decimal place, the angle which AF makes with the vertical.
(4)
Q3, June 2007
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16.
Figure 1
A set square S is made by removing a circle of centre O and radius 3 cm from a triangular
piece of wood. The piece of wood is modelled as a uniform triangular lamina ABC, with
ABC = 90°, AB = 12 cm and BC = 21 cm. The point O is 5 cm from AB and 5 cm from
BC, as shown in Figure 1.
(a) Find the distance of the centre of mass of S from
(i) AB,
(ii) BC.
(9)
The set square is freely suspended from C and hangs in equilibrium.
(b) Find, to the nearest degree, the angle between CB and the vertical.
(3)
Q4, Jan 2008
43 | P a g e
17.
Figure 3
Figure 3 shows a rectangular lamina OABC. The coordinates of O, A, B and C are (0, 0),
(8, 0), (8, 5) and (0, 5) respectively. Particles of mass km, 5m and 3m are attached to the
lamina at A, B and C respectively.
The x-coordinate of the centre of mass of the three particles without the lamina is 6.4.
(a) Show that k = 7.
(4)
The lamina OABC is uniform and has mass 12m.
(b) Find the coordinates of the centre of mass of the combined system consisting of the
three particles and the lamina.
(6)
The combined system is freely suspended from O and hangs at rest.
(c) Find the angle between OC and the horizontal.
(3)
Q6, May 2008
44 | P a g e
18.
Figure 2
A uniform lamina ABCD is made by joining a uniform triangular lamina ABD to a
uniform semi-circular lamina DBC, of the same material, along the edge BD, as shown in
Figure 2. Triangle ABD is right-angled at D and AD = 18 cm. The semi-circle has
diameter BD and BD = 12 cm.
(a) Show that, to 3 significant figures, the distance of the centre of mass of the lamina
ABCD from AD is 4.69 cm.
(4)
Given that the centre of mass of a uniform semicircular lamina, radius r, is at a distance
3
4r from the centre of the bounding diameter,
(b) find, in cm to 3 significant figures, the distance of the centre of mass of the lamina
ABCD from BD.
(4)
The lamina is freely suspended from B and hangs in equilibrium.
(c) Find, to the nearest degree, the angle which BD makes with the vertical.
(4)
Q5, Jan 2009
45 | P a g e
19.
Figure 2
A shop sign ABCDEFG is modelled as a uniform lamina, as illustrated in Figure 2.
ABCD is a rectangle with BC = 120 cm and DC = 90 cm. The shape EFG is an isosceles
triangle with EG = 60 cm and height 60 cm. The mid-point of AD and the mid-point of
EG coincide.
(a) Find the distance of the centre of mass of the sign from the side AD.
(5)
The sign is freely suspended from A and hangs at rest.
(b) Find the size of the angle between AB and the vertical.
(4)
Q5, May 2009
46 | P a g e
20. [The centre of mass of a semi-circular lamina of radius r is 3
4r from the centre.]
Figure 3
A template T consists of a uniform plane lamina PQROS, as shown in Figure 3. The
lamina is bounded by two semicircles, with diameters SO and OR, and by the sides SP,
PQ and QR of the rectangle PQRS. The point O is the mid-point of SR, PQ = 12 cm and
QR = 2x cm.
(a) Show that the centre of mass of T is a distance 38
324 2
x
x cm from SR.
(7)
The template T is freely suspended from the point P and hangs in equilibrium.
Given that x = 2 and that θ is the angle that PQ makes with the horizontal,
(b) show that tan θ =
622
948
.
(4)
Q7, Jan 2010
47 | P a g e
21.
Figure 1
A triangular frame is formed by cutting a uniform rod into 3 pieces which are then joined
to form a triangle ABC, where AB = AC = 10 cm and BC = 12 cm, as shown in Figure 1.
(a) Find the distance of the centre of mass of the frame from BC.
(5)
The frame has total mass M. A particle of mass M is attached to the frame at the mid-
point of BC. The frame is then freely suspended from B and hangs in equilibrium.
(b) Find the size of the angle between BC and the vertical.
(4)
Q3, June 2010
48 | P a g e
22.
Figure 2
The uniform L-shaped lamina ABCDEF, shown in Figure 2, has sides AB and FE
parallel, and sides BC and ED parallel. The pairs of parallel sides are 9 cm apart. The
points A, F, D and C lie on a straight line.
AB = BC = 36 cm, FE = ED = 18 cm.
ABC = FED = 90°, and BCD = EDF = EFD = BAC = 45°.
(a) Find the distance of the centre of mass of the lamina from
(i) side AB,
(ii) side BC.
(7)
The lamina is freely suspended from A and hangs in equilibrium.
(b) Find, to the nearest degree, the size of the angle between AB and the vertical.
(3)
Q5, Jan 2011
49 | P a g e
23.
Figure 1
The trapezium ABCD is a uniform lamina with AB = 4 m and BC = CD = DA = 2 m, as
shown in Figure 1.
(a) Show that the centre of mass of the lamina is 9
34 m from AB.
(5)
The lamina is freely suspended from D and hangs in equilibrium.
(b) Find the angle between DC and the vertical through D.
(5)
Q4, Jan 2012
50 | P a g e
24.
Figure 2
A uniform circular disc has centre O and radius 4a. The lines PQ and ST are
perpendicular diameters of the disc. A circular hole of radius 2a is made in the disc, with
the centre of the hole at the point R on OP where OR = 2a, to form the lamina L, shown
shaded in Figure 2.
(a) Show that the distance of the centre of mass of L from P is 3
14a.
(4)
The mass of L is m and a particle of mass km is now fixed to L at the point P. The system
is now suspended from the point S and hangs freely in equilibrium. The diameter ST
makes an angle with the downward vertical through S, where tan = 65 .
(b) Find the value of k.
(5)
Q4, May 2012
25. Two uniform rods AB and BC are rigidly joined at B so that ABC = 90°. Rod AB has
length 0.5 m and mass 2 kg. Rod BC has length 2 m and mass 3 kg. The centre of mass of
the framework of the two rods is at G.
(a) Find the distance of G from BC.
(2)
The distance of G from AB is 0.6 m.
The framework is suspended from A and hangs freely in equilibrium.
(b) Find the angle between AB and the downward vertical at A.
(3)
Q1, Jan 2013
51 | P a g e
26.
Figure 1
The uniform lamina ABCDEF is a regular hexagon with centre O and sides of length 2 m,
as shown in Figure 1.
Figure 2
The triangles OAF and OEF are removed to form the uniform lamina OABCDE, shown
in Figure 2.
(a) Find the distance of the centre of mass of OABCDE from O.
(5)
The lamina OABCDE is freely suspended from E and hangs in equilibrium.
(b) Find the size of the angle between EO and the downward vertical.
(6)
Q4, June 2013
52 | P a g e
27.
Figure 2
A uniform triangular lamina ABC of mass M is such that AB = AC, BC = 2a and the
distance of A from BC is h. A line, parallel to BC and at a distance 2
3
h from A, cuts AB at
D and cuts AC at E, as shown in Figure 2.
It is given that the mass of the trapezium BCED is 5
9
M.
(a) Show that the centre of mass of the trapezium BCED is 7
45
h from BC.
(5)
Figure 3
The portion ADE of the lamina is folded through 180° about DE to form the folded
lamina shown in Figure 3.
(b) Find the distance of the centre of mass of the folded lamina from BC.
(4)
The folded lamina is freely suspended from D and hangs in equilibrium. The angle
between DE and the downward vertical is α.
(c) Find tan α in terms of a and h.
(4
Q6, June 2013_R
53 | P a g e
28.
Figure 2
Figure 2 shows a lamina L. It is formed by removing a square PQRS from a uniform
triangle ABC. The triangle ABC is isosceles with AC = BC and AB = 12 cm. The midpoint
of AB is D and DC = 8 cm. The vertices P and Q of the square lie on AB and PQ = 4 cm.
The centre of the square is O. The centre of mass of L is at G.
(a) Find the distance of G from AB.
(4)
When L is freely suspended from A and hangs in equilibrium, the line AB is inclined at
25° to the vertical.
(b) Find the distance of O from DC.
(6)
Q4, June 2014_R
54 | P a g e
29.
Figure 1
The uniform lamina ABCDEF, shown shaded in Figure 1, is symmetrical about the line
through B and E. It is formed by removing the isosceles triangle FED, of height 6a and
base 8a, from the isosceles triangle ABC of height 9a and base 12a.
(a) Find, in terms of a, the distance of the centre of mass of the lamina from AC.
(5)
The lamina is freely suspended from A and hangs in equilibrium.
(b) Find, to the nearest degree, the size of the angle between AB and the downward
vertical.
(4)
Q3, June 2014
55 | P a g e
30.
Figure 1
The uniform lamina OABCD, shown in Figure 1, is formed by removing the triangle
OAD from the square ABCD with centre O. The square has sides of length 2a.
(a) Show that the centre of mass of OABCD is 9
2a from O.
(4)
The mass of the lamina is M. A particle of mass kM is attached to the lamina at D to form
the system S. The system S is freely suspended from A and hangs in equilibrium with AO
vertical.
(b) Find the value of k.
(4)
Q2, June 2015
31.
56 | P a g e
The uniform lamina OBC is one quarter of a circular disc with centre O and radius 4 m.
The points A and D, on OB and OC respectively, are 3m from O. The uniform lamina
ABCD, shown shaded in Figure 1, is formed by removing the triangle OAD from OBC.
Given that the centre of mass of one quarter of a uniform circular disc of radius r is at a
distance 4 2
3r
from the centre of the disc,
(a) find the distance of the centre of mass of the lamina ABCD from AD.
(5)
The lamina is freely suspended from D and hangs in equilibrium.
(b) Find, to the nearest degree, the angle between DC and the downward vertical.
(4)
Q4, June 2016
32.
Figure 1
A uniform lamina ABCD is formed by removing the isosceles triangle ADC of height
h metres, where h < 2√3, from a uniform lamina ABC in the shape of an equilateral
triangle of side 4 m, as shown in Figure 1. The centre of mass of ABCD is at D.
(a) Show that h = √3.
(7)
The weight of the lamina ABCD is W newtons. The lamina is freely suspended from A. A
horizontal force of magnitude F newtons is applied at B so that the lamina is in
57 | P a g e
equilibrium with AB vertical. The horizontal force acts in the vertical plane containing the
lamina.
(b) Find F in terms of W.
(4)
Q4, Jan 2014, IAL
33.
Figure 2
The uniform square lamina ABCD shown in Figure 2 has sides of length 4a. The points E
and F, on DA and DC respectively, are both at a distance 3a from D.
The portion DEF of the lamina is folded through 180° about EF to form the folded
lamina ABCFE shown in Figure 3 below.
Figure 3
58 | P a g e
(a) Show that the distance from AB of the centre of mass of the folded lamina is 55
32a .
(6)
The folded lamina is freely suspended from E and hangs in equilibrium.
(b) Find the size of the angle between ED and the downward vertical.
(4)
Q4, June 2014, IAL
34.
The uniform plane lamina ABCDEF shown in Figure 1 is made from two identical
rhombuses. Each rhombus has sides of length a and angle BAD = angle DAF = 𝜃. The
centre of mass of the lamina is 0.9a from A.
(a) Show that cos 𝜃 = 0.8.
(5)
The weight of the lamina is W. A particle of weight kW is fixed to the lamina at the point
A. The lamina is freely suspended from B and hangs in equilibrium with DA horizontal.
59 | P a g e
(b) Find the value of k.
(4)
Q4, Jan 2015, IAL
35. A thin uniform wire of mass 12m is bent to form a right-angled triangle ABC. The lengths
of the sides AB, BC and AC are 3a, 4a and 5a respectively. A particle of mass 2m is
attached to the triangle at B and a particle of mass 3m is attached to the triangle at C. The
bent wire and the two particles form the system S.
The system S is freely suspended from A and hangs in equilibrium.
Find the size of the angle between AB and the downward vertical.
(10)
Q3, June 2015, IAL
36.
The uniform lamina ABCDEF, shown in Figure 2, consists of two identical rectangles
with sides of length a and 3a. The mass of the lamina is M. A particle of mass kM is
attached to the lamina at E. The lamina, with the attached particle, is freely suspended
from A and hangs in equilibrium with AF at an angle θ to the downward vertical.
Given that tan θ = 4
7find the value of k.
(10)
Q4, Jan 2015, IAL
60 | P a g e
37.
The uniform lamina ABCDE is made by joining a rectangular lamina ABDE to a
triangular lamina BCD along the edge BD. The rectangle has length 6a and width 3a. The
triangle is isosceles, with BC = CD, and the distance from C to BD is 3a, as shown in
Figure 2.
(a) Find the distance of the centre of mass of the lamina, ABCDE, from AE.
(5)
The lamina ABCDE is freely suspended from A. A horizontal force of magnitude F
newtons is applied to the lamina at D. The line of action of the force lies in the vertical
plane containing the lamina. The lamina is in equilibrium with AE vertical. The mass of
the lamina is 4 kg.
(b) Find the magnitude of the force exerted on the lamina at A.
(5)
Q4, June 2016, IAL
61 | P a g e
38.
The uniform lamina L shown shaded in Figure 2 is formed by removing two circular
discs,
C1 and C2, from a circular disc with centre O and radius 8a. Disc C1 has centre A and
radius a. Disc C2 has centre B and radius 2a. The diameters PR and QS are perpendicular.
The midpoint of PO is A and the midpoint of OR is B.
(a) Show that the centre of mass of L is 484
59a from R.
(5)
The mass of L is M. A particle of mass kM is attached to L at S. The lamina with the
attached particle is suspended from R and hangs freely in equilibrium with the diameter
PR at an angle of arctan 1
4
to the downward vertical through R.
(b) Find the value of k.
(5)
Q6, Oct 2016, IAL
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6.
P
Q
R
S
A B2a
a
O
8a
Figure 2
The uniform lamina L shown shaded in Figure 2 is formed by removing two circular discs,
C1 and C ,
from a circular disc with centre O and radius 8a. Disc C
1 has centre A and
radius a. Disc C2 has centre B and radius 2a. The diameters PR and QS are perpendicular.
The midpoint of PO is A and the midpoint of OR is B.
(a) Show that the centre of mass of L is 484
59a from R.
(5)
The mass of L is M. A particle of mass kM is attached to L at S. The lamina with the
attached particle is suspended from R and hangs freely in equilibrium with the diameter
PR at an angle of arctan 1
4
æ
èçö
ø÷to the downward vertical through R.
(b) Find the value of k.
(5)
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62 | P a g e
39.
A uniform lamina is in the shape of a trapezium ABCD with AB = a, DA = DC = 2a
and angle BAD = angle ADC = 90°, as shown in Figure 1.
The centre of mass of the lamina is at the point G.
(a) (i) Show that the distance of G from AB is
10a
9
(ii) Find the distance of G from AD.
(6)
The mass of the lamina is 3M. A particle of mass kM is now attached to the lamina at B.
The lamina is freely suspended from the midpoint of AD and hangs in equilibrium with
AD horizontal.
(b) Find the value of k.
(3)
Q2, Jan 2017, IAL
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2.
A B
C2a
2a
a
D
Figure 1
A uniform lamina is in the shape of a trapezium ABCD with AB = a, DA = DC = 2a
and angle BAD = angle ADC = 90°, as shown in Figure 1.
The centre of mass of the lamina is at the point G.
(a) (i) Show that the distance of G from AB is 10
.9
a
(ii) Find the distance of G from AD.
(6)
The mass of the lamina is 3M. A particle of mass kM is now attached to the lamina at B.
The lamina is freely suspended from the midpoint of AD and hangs in equilibrium with
AD horizontal.
(b) Find the value of k.
(3)
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