Chapter 3
A Divide the surface of the cone up into triangular Use symmetry to obtain this result.
strips with vertices at O and with bases on the
circumference of the circular base of the cone.
One is shown in the diagram.
Each of these triangles has centre of mass _23 of
the distance from O to the base of the cone.
So the centre of mass of the hollow cone is
also _ 32 of the distance from O to the base of
the cone, but is on the axis of symmetry.
The distance of the centre of mass from the
base is h − _ 23_ h = _ 31_ h
■ The centre of mass of a conical shell lies on the axis of symmetry and is at a distance _31 h from
the base.
Example 12
Find the position of the centre of mass of the frustum of a right circular uniform cone, of end
radii 1 cm and 4 cm, and of height 7 c m.
1 cm h A frustum is a portion of the cone lying
7 cm H between two parallel planes. It may be
considered as a large cone with a small
4 cm cone removed from the top.
Let the large cone have height H and the small cone Problem-solving
have height h. The masses are in the ratio 64 : 1 : 63
From similar triangles _H h_ = _ 41_ or H = 4h and you can use these ratios in your
But H = 7 + h, so h = _ 73 , and H = _ 23_8 moments equation to simplify the
working.
Shape Mass Centre of mass So you would have 64 × _ 37 − 1 × _ 1921_ = 63_x_
Large cone = _31 ρ_ 6π3_4 4 ρ2πHh ( _41 H, 0 ) = ( _73 , 0 )
_31 ρπ 12h = _ 31 ρπh
Small cone _ 63_3 ρπh = 21ρπh ( 7 + _41 h, 0 ) = ( _912_1 , 0 )
Frustum 0)
(x_ ,
Taking moments about the base of the frustum:
_63_4 ρπh × _ 37 − _ 31 ρπh × _ 912_1 = 21ρπhx_
So x_ = ( _4_94 8_ − _39_61 ) ÷ 21 = _ 94_
The centre of mass is _94 _ c m above the base on the axis
of symmetry.
94
Further centres of mass
Exercise 3B
A 1 The finite region bounded by the curve Hi nt In questions 1–4 use symmetry to find the
y = x2 − 4x and the x-axis is rotated coordinates of the centre of mass of the solid.
through 360° about the x-axis to form
a solid of revolution. Find the coordinates of its centre of mass.
2 The finite region bounded by the curve (x − 1) 2 + y 2 = 1 is rotated through 180° about the
x-axis to form a solid of revolution. Find the coordinates of its centre of mass.
3 The finite region bounded by the curve y = cos x, _π2_ < x < _ 32_π_ , and the x-axis, is rotated
through 360° about the x-axis to form a solid of revolution. Find the coordinates of its centre
of mass.
4 The finite region bounded by the curve y 2 + 6y = x and the y-axis is rotated through 360° about
the y-axis to form a solid of revolution. Find the coordinates of its centre of mass.
5 Find, by integration, the coordinates of the Hi nt In questions 5–10 use integration to find
centre of mass of the solid formed when the the position of the centre of mass of the solid.
finite region bounded by the curve y = 3x2,
the line x = 1 and the x-axis is rotated through 360° about the x-axis.
6 rFeignido,nbbyoiunntedgerdatbioynt,htehceucrovoerydi=na√t _xe_ ,s of the centre of mass of the solid formed when the finite
the line x = 4 and the x-axis is rotated through 360° about
the x-axis.
7 Find, by integration, the coordinates of the centre of mass of the solid formed when the finite
region bounded by the curve y = 3x2 + 1, the lines x = 0, x = 1 and the x-axis is rotated through
360° about the x-axis.
8 Find, by integration, the coordinates of the centre of mass of the solid formed when the
finite region bounded by the curve y = _ x3_ , the lines x = 1, x = 3 and the x-axis is rotated
through 360° about the x-axis.
E 9 Find, by integration, the coordinates of the centre of mass of the solid formed when the finite
region bounded by the curve y = 2e x, the lines x = 0, x = 1 and the x-axis is rotated through
360° about the x-axis. (10 marks)
E/P 10 Find, by integration, the coordinates of the centre of mass formed when the region bounded
by the curve xe y = 3, the lines y = 1, y = 2 and the y-axis is rotated through 360° about the
y-axis. (10 marks)
95
Chapter 3
A 11 Find, by integration, the coordinates of the centre of mass formed when the region bounded
E by the curve y = _1_+2__ x_ , the lines x = 0, x = 4 and the x-axis is rotated through 360° about
the x-axis. (10 marks)
E/P 12 Find the position of the centre of mass of the frustum of a 2 cm
right circular uniform solid cone, where the frustum has end 4 cm
radii 2 cm and 5 cm, and has height 4 cm. (8 marks)
5 cm
E/P 13 The diagram shows a frustum of a right circular uniform cone. 4 cm
The frustum has end diameters of 4 cm and 8 cm, and 2 cm
height 8 cm. A cylindical hole of diameter 2 cm is drilled
through the frustum along its axis. Find the distance of 8 cm 8 cm
the centre of mass of the resulting solid from the larger
face of the frustum. (12 marks)
E/P 14 A thin uniform hemispherical shell has a circular base of
the same material. Find the position of the centre of mass
above the base in terms of its radius r. (6 marks)
r
E/P 15 A thin uniform hollow cone has a circular base of the same
material. Find the position of the centre of mass above the
base, given that the radius of the cone is 3 cm and its
height is 4 cm. (6 marks) 4 cm
3 cm
Hi nt The curved surface area of a cone of
slant height l and base radius r has area πrl. h
a
E/P 16 A cap of a sphere is formed by making a plane cut across the
sphere. Show that the centre of mass of a cap of height h cut
from a uniform solid sphere of radius a lies a perpendicular
distance _ h4_((43_aa__−−__hh_)) from the flat surface of the cap. (12 marks)
96
Further centres of mass
A 17 The finite region R, shown shaded on the diagram, is bounded y
E/P by the curve with equation y = 8 − x3 and the coordinate axes. y = 8 – x3
The region is rotated through 2π radians about the y-axis to R
form a uniform solid of revolution, S. Ox
a Find, using algebraic integration, the y coordinate of the S
centre of mass of S. (8 marks)
X
A solid chocolate egg is formed by attaching a uniform solid
hemisphere to the base of the solid S, as shown in the diagram.
The units of measurement are cm.
b Find the distance of the centre of mass of the chocolate
egg from its base, X, giving your answer correct to
3 significant figures. (5 marks)
E/P 18 A uniform solid is formed from two right circular cones. The cones each kx
have base radius r and heights x and kx respectively, where k is a constant.
The point O lies at the centre of the common plane face of both cones. Or
x
Given that the centre of mass of the solid is a distance _1 1_0 x above O,
find the value of k. (5 marks)
E/P 19 A uniform solid cylinder has height h and radius 2r. r r
A hole is drilled in the cylinder in the shape of an O h
inverted cone of base radius r and height h.
The vertex of the cone lies on the base of the
cylinder, and the axes of the cone and the
cylinder are both vertical. The centre of the
top plane face of the cylinder, O, lies on the
circumference of the base of the cone.
a Find the distance of the centre of mass of the
solid from its top plane face, giving your answer in the form kh where k is a rational constant
to be found. (7 marks)
b Fully describe the position of the centre of mass of the solid. (5 marks)
E/P 20 A game piece is modelled as a solid cone of base radius 5 cm 12 cm
and height 12 cm, sitting on top of a solid cylinder of radius
5 cm and height 3 cm. The cone and cylinder are made of the 5 cm
same uniform material. 3 cm
97
Chapter 3
A a Find the distance of the centre of mass of the game piece from the base of the cylinder.
(6 marks)
In order to save money, the manufacturer decides to make the game piece hollow. The game
piece is now modelled as the curved surfaces of a cone and cylinder, and a single circular face
on its base. All the surfaces are made from the same uniform material.
b Find the distance of the centre of mass of the hollow game piece from the base of the
cylinder. (6 marks)
Challenge y Hi nt Divide the shell
R into small elemental
Using calculus, find the
centre of mass of the δθ cylindrical rings, centred
uniform hemispherical on the x-axis, with radius
shell with radius R. R sin θ, and height R δθ,
x where θ is the angle
R sin θ between the radius R and
the x-axis.
θ
O
Rδθ
3.3 Non-uniform bodies
You can find the centres of mass of composite bodies made from materials of different densities.
Example 13
A uniform solid right circular cone of height 2R and base radius R is joined at its base to the base
of a uniform solid hemisphere. The centres of their bases coincide at O and their axes are collinear.
The radius of the hemisphere is 2R.
Find the position of the centre of mass of the composite body if the cone has four times the
density of the hemisphere.
98
Further centres of mass
Ay
2R x Problem-solving
O
Although each separate
R 2R solid has uniform
density, the composite
Let ρ be the mass per unit volume for the hemisphere and 4ρ be the body does not. Use ρ to
mass per unit volume for the cone. represent the density
(mass per unit volume)
Shape Mass Ratio of Distance from O of the hemisphere,
masses to centre of mass so that the cone has
Cone _ 31 4ρπR2(2R) 1 _41 (2R) = R_2_ density 4ρ.
_23 ρπ(2R)3 − _38_ (2R) = − 3_4_R_
Hemisphere _31 4ρπR2(2R) + _ 32 ρπ(2R)3 2 The masses are in the
Composite x_ ratio 1 : 2 : 3 and you
body 3 can use these ratios
in your moments
The centre of mass lies on the common axis of symmetry. equation to simplify
Take moments about an axis through O: the working.
1 × R_2_ − 2 × 3_4_R_ = 3 × x_
So x_ = − _31 R If you take moments
about a point inside
The centre of mass of the composite body is a distance − _31_ R below O, the body, be careful
on the common axis of symmetry. with your positive and
negative signs. The
centre of mass of the
whole body is below O.
In some cases the density of a body may vary continuously along the length of the body. You can use
integration to find centres of mass of non-uniform rods, where the mass per unit length of the rod is
given as a function of distance.
This cylinder is modelled as a non-uniform rod. At a distance Problem-solving
x along the rod, the mass per unit length of the rod is given
by ρ = f(x). Because the cylinder is
modelled as a rod and density
x δx is given as mass per unit length,
you do not need to consider
The mass of a section of length δx at a distance x from the end the volume of the cylinder.
of the rod is ρδx = f(x)δx. So the mass of the whole rod is ∑f(x)δx. You could use a similar model
for any prism with a constant
cross-section, where density
varies as a function of height.
99
Chapter 3
A If the centre of mass of the rod is a distance x¯ from the end, then
x_ ∑f(x)δx = ∑xf(x)δx
The moment of one section of the cylinder is xρδx = xf( x )δx
As δx → 0 the summation can be replaced by an integral.
■ For a non-uniform rod with a variable mass per unit length, ρ = f(x), and length l, the
distance of the centre of mass from the end of the rod is given by:
x ¯ = _ ∫∫0_ l _ xl ρ_ρ_ d _dx_x_ = _ ∫∫0_ l _ xl f _f(_(x_x)_) d_ d_xx_
00
Example 14
A non-uniform telegraph pole is 10 m long. At any distance x metres from its base, the mass per
unit length of the telegraph pole is given by m = 20 − _ 15 x kg m−1. By modelling the telegraph pole as
a rod, find:
a the mass of the telegraph pole
b the distance of the centre of mass of the telegraph pole from its top.
a Total mass = ∫0 1 0 (20 − _ 51_ x) dx Problem-solving
The mass per unit length of the telegraph pole
= [20x − _ 110_ x 2] 100 varies along the length of the pole. To find the
= [20(10) − _ 110_ (10) 2] total mass you need to integrate the mass per
unit length across the length of the whole pole.
= (200 − 10)
Integrate then substitute in the limits.
= 190 kg
Where x_ is the distance of the centre of mass
b Taking moments about the base of the from the base of the telegraph pole.
telegraph pole:
Integrate then substitute in the limits.
190x ¯ = ∫0 1 0 x(20 − _ 51_ x) dx
= [ 10 x 2 − _ 115_ x 3] 100
= [10 (10) 2 − _ 115_ (10) 3]
= 1000 − _ 2_30 0_
= _ 2_83_0 0_
x¯ = _25_870_ m
So the centre of mass is _2 5_870_ m from the
base of the telegraph pole and therefore
_ 25_970_ m from the top of the telegraph pole.
100
Further centres of mass
Exercise 3C
A 1 A uniform solid right circular cone of height 10 cm and base radius 5 cm is joined at its base to
P the base of a uniform solid hemisphere. The centres of their bases coincide and their axes are
collinear. The radius of the hemisphere is also 5 cm. The density of the hemisphere is twice the
density of the cone. Find the position of the centre of mass of the composite body.
P 2 A solid is composed of a uniform solid right circular cylinder of height 10 cm and base radius
6 cm joined at its top plane face to the base of a uniform cone of the same radius and of
height 5 cm. The centres of their adjoining circular faces coincide at point O and their axes are
collinear. The mass per unit volume of the cylinder is three times that of the cone. Find the
position of the centre of mass of the composite body.
E/P 3 a Prove that the centre of mass of a uniform right square-based pyramid of height h lies a
distance _ 4h_ from its base.
(6 marks)
A solid is composed of a cube of side 8 cm joined at its top plane face to the base of a uniform
square-based pyramid of side 8 cm and perpendicular height 6 cm. The centres of their adjoining
square faces coincide at the point O and their axes are collinear.
b Given that the cube has twice the density of the pyramid, find the position of the centre of
mass of the composite body. (4 marks)
E/P 4 a Find the distance of the centre of mass of a solid tetrahedron from its base, giving your
answer in terms of the height of the tetrahedron, h. (6 marks)
A solid is composed of two solid tetrahedrons of side 9 cm joined together. The centres of their
adjoining faces coincide at the point O and their axes are collinear.
b Given that the bottom tetrahedron has three times the density of the top tetrahedron, find the
position of the centre of mass of the composite body. (4 marks)
E/P 5 A truncated square-based pyramid has a height of 5 cm
5 cm and a base length 10 cm. The top face of the
truncated square-based pyramid has an area of 25 cm2.
a Find the height of the centre of mass of the 5 cm
truncated square-based pyramid above its base.
(6 marks) 10 cm
The truncated square-based pyramid and a cube of side length 5 cm are joined at the smaller
square face of the truncated pyramid. The centres of their adjoining square faces coincide at the
point O and their axes are collinear.
b Given that the cube has twice the density of the truncated square-based pyramid, find the
distance of the centre of mass of the composite body from O. (4 marks)
101
Chapter 3
A 6 A non-uniform fence post is 1.2 m long. At any distance d metres from its base, the mass per
E/P unit length of the fence post is given by m = 10( 1 − _ 1d_2_ ) kg m−1. By modelling the fence post as a
rod, find:
a the mass of the fence post (3 marks)
b the distance of the centre of mass of the fence post from its base. (4 marks)
In reality, the post has a roughly circular cross-section. The model is refined so that the fence
post is modelled as a cylinder of radius 0.1 m with density1 00(1 − _ 1d_2_ ) kg m−3 at a distance d m
from its base.
c State, with reasons, whether your answers to parts a and b will change under the new model.
(2 marks)
E/P 7 A solid wooden bowl is modelled as a uniform solid 2r 2r
right circular cylinder with height 2r and radius 4r. OP
The centre of one face is at O. A hemisphere of
diameter 2r is removed from the bowl. The centre X 4r
of the hemisphere lies on a diameter of the circular
face of the cylinder, and the point O lies on the (6 marks)
circumference of the circular face of the hemisphere.
a Show that the centre of mass of the whole
bowl is at a vertical distance _3 3_87_16_r from the
plane face containing O.
The hemisphere is filled with water. The density of the water is twice the density of the wood
from which the bowl is made.
b Find:
i the vertical distance of the centre of mass of the filled bowl from the plane face containing O
ii the horizontal distance from the axis of the cylinder, OX, to the centre of mass of the filled
bowl.
In each case, give your answer in the form kr where k is a rational number to be found. (6 marks)
P 8 A non-uniform rod of length l m is such that, at a distance x m from the end of the rod, the mass
per unit length of the rod is given by m kg m−1. In each of the following cases, find the distance
of the centre of mass from the end of the rod, giving your answers to 3 significant figures where
appropriate.
a l = 6 and m = (x + 1) 2
b l = 10 and m = 10 − _ 41 x
c l = 2 and m = _ 1_+_1_ x__2
d l = 5 and m = e 0.5x
102
Further centres of mass
A 9 The mast of a ship is 18 m high. The mast is modelled as a non-uniform rod, such that, at a
E/P height h m above its base, the mass per unit length of the mast, m kg m−1, is given by the
formula m = 50e −0.01h
Find, correct to 3 significant figures:
a the mass of the mast (3 marks)
b the height of the centre of mass of the mast above its base. (6 marks)
E/P 10 A non-uniform rod AB of length 2 m is such that, at a distance x m from A, the mass per unit
length of the rod is given by (5 − px) kg m−1.
Given that the mass of the rod is 7 kg, find:
a the value of p (3 marks)
b the distance of the centre of mass of the rod from A. (4 marks)
Challenge
A non-uniform post is 2 m long and has mass 10 kg. At any distance
x metres from its base, the mass per unit length of the post is given by
m = a(1 − bx) kg m−1. Given that the centre of mass of the post is 1.5 m
above its base, work out the possible values of a and b.
3.4 Rigid bodies in equilibrium
You can solve problems about rigid bodies Li nks You can combine resultant forces and moments
which are suspended in equilibrium. to solve problems involving ladders leaning against
If a body is resting in equilibrium then there walls. ← Statistics and Mechanics Year 2, Chapter 7
is zero resultant force in any direction.
This means that the sum of the components of all the forces in any direction is zero, and the sum of
the moments of the forces about any point is zero.
■ When a lamina is suspended freely from a fixed point or pivots freely about a horizontal axis
it will rest in equilibrium in a vertical plane with its centre of mass vertically below the point
of suspension or pivot.
This result is also true for a rigid body.
Let the body be suspended from a point A. The body rests in equilibrium and P
the only forces acting on the body are its weight, W, and the force at point A, P. A
This implies that the forces must be equal and opposite and act in the same
vertical line.
■ When a rigid body is suspended freely from a fixed point and rests in G
W
equilibrium then its centre of mass is vertically below the point
of suspension.
103
Chapter 3
Example 15
A A uniform solid hemisphere has radius r. It is suspended by a string attached to a point A on the
rim of its base. Find the angle between the axis of the hemisphere and the downward vertical when
the hemisphere is in equilibrium.
A
r Draw a diagram showing the centre of mass, G, of
the hemisphere below the point of suspension A.
X θ
G Mark the angle between GA and the axis of the
3r hemisphere and mark the radius and length XG.
8
The distance from the centre of mass to the Use trigonometry to solve the problem.
base is 3_ 8_r_ so XG = _38_r_
Let ∠XGA be θ
Then tan θ = _ _38r__ r_ = _ 83_
So the required angle is 69.4° (3 s.f.)
Example 16
A non-uniform cylinder has height 20 cm and radius 2 cm. At any height h cm above its base,
the density of the cylinder is given by ρ = 3( 2 − _ 5h_0_ ) g cm–3. The cylinder is freely suspended from
a point P that lies on the circumference of its top face. Find the angle between the downward
vertical and the top face of the cylinder.
δh 20 cm Problem-solving
h
You need to work out the mass of a disc of
4 cm height δx, then integrate to find the sum
of these masses. Because the cylinder has
Mass of elemental disc = ρ π r 2h = 3π( 2 − _ 5h_0_ ) 2 2δh a constant, uniform cross-section, and you
know the centre of mass lies on its axis
of symmetry, you could also tackle this
problem by modelling the cylinder as a
non-uniform rod with mass per unit
length of 3 ( 2 − _ 5h_0_ ) g cm−1.
104
A Total mass = 12π ∫ 0 2 0 (2 − _ 5h_0_ ) dh Further centres of mass
= 12π [2h − _ 1h0__ 20_ ] 02 0 Use M y ¯ = ∫ρπr2h dh to find the position
= 12π(2(20) − _ ( 21_00_0)_ 2 ) of y ¯ .
Take moments from the base of the
= 12π(40 − 4) cylinder.
= 432π
Draw a clear diagram to represent the
432π y ¯ = 12π ∫ 0 2 0 h( 2 − _ 5h_0_ ) dh situation. You need to find the angle
between the top face and the vertical
where y ¯ is the distance of the centre of mass above which is marked as α on this diagram.
the base of the cylinder.
36y ¯ = ∫0 2 0 ( 2h − _ 5h_0 2_ ) dh
= [ h 2 − _ 1h5__ 30_ ] 02 0
= 202 − _ 21_50_03_
= 400 − _ 163_0_
= _ 10_34_ 0_
y¯ = _110_04_80_ = _ 22_670_
P
Oα
C
y
W Problem-solving
tan α = _ OO__CP_ The density of the cylinder reduces as the
height increases, so it is bottom-heavy.
where O is the centre of the top face of the cylinder The centre of mass will be less than half-
and C is the centre of mass. way up the cylinder, and the angle will be
tan α = _ 2_0__−2__ _ 2_2_6_70__ = _124_70_ greater than arctan _2 4_0 , the angle made
between the top face and the diagonal.
α = 79.1° (3 s.f.)
105
Chapter 3
Example 17 T2
A The diagram shows a uniform solid right circular cone, T1 r Q
of mass M with radius r and height 3r. P and Q are points XG
at opposite ends of a diameter on the circular plane face P
of the cone. The cone is suspended from a horizontal beam
by two vertical inelastic strings fastened at P and Q. Mg
Given that the cone is in equilibrium with its axis at an angle
of 45° to the horizontal, find the values of the tensions in the
strings, giving your answers in terms of M and g.
X
3r
45° 4
rG
Y
PW
Z
Mg
Let the tensions in the strings be T1 and T2.
Resolve vertically:
T1 + T2 = Mg (1) Moments could be taken about a number
of points, but choosing point P eliminates
Take moments about point P: (2) T1 and simplifies the algebra.
( )T2 × 2r cos 45° = Mg r cos 45° + _34_r_ sin 45° The distance PW is PZ + ZW.
Also ZW = YG.
Divide equation (2) through by 2r cos 45°, So from the diagram:
then T2 = _ 87_ Mg PW = r cos 45° + _ 34_r sin 45°
Substitute into equation (1) to give T1 = _ 81_ Mg
So the tensions are _81 _ Mg and _87 _ Mg
106
Further centres of mass
Example 18
A Two smooth uniform spheres of radius 4 cm and mass 5 kg are suspended from the same point A
by identical light inextensible strings of length 8 cm attached to their surfaces. The spheres hang in
equilibrium, touching each other. What is the reaction between them?
A
4 cm θ T
8 cmT
4 cm R From symmetry, the tensions in the two strings
R are equal.
5g 5g Divide equation (2) by equation (1) to eliminate T.
Resolve vertically: As sin θ = _ 13 the triangle XYZ can be X
T cos θ = 5g (1) used to find XY using Pythagoras’ 3θ
Resolve horizontally: theorem, and then to find ta_n_ θ: __ Z 1Y
T sin θ = R (2) XY 2 = 32 − 12 ∴ XY = √ 8 = 2√ 2
Eliminating T gives R = 5g tan θ
From the geometry of the diagram
sin θ = _ 142_ = _ 31 and so tan θ = _ 2_√1_ _2 __
So R = _25_√g_2_ __ = 17 (2 s.f.)
Exercise 3D
1 A uniform solid right circular cone is suspended by a string attached to a point on the rim of
its base. Given that the radius of the base is 5 cm and the height of the cone is 8 cm, find the
angle between the vertical and the axis of the cone when it is in equilibrium.
2 A uniform solid cylinder is suspended by a string attached to a point on the rim of its base.
Given that the radius of the base is 6 cm and the height of the cylinder is 10 cm, find the angle
between the vertical and the circular base of the cylinder when it is in equilibrium.
3 A uniform hemispherical shell is suspended by a string attached to a point on the rim of
its base. Find the angle between the vertical and the axis of the hemisphere when it is in
equilibrium. (8 marks)
107
Chapter 3
A 4 a Find the position of the centre of mass of the frustum of 4 cm
a uniform solid right circular cone, of end radii 4 cm and 6 cm
E 5 cm and of height 6 cm. (Give your answer to 3 s.f.) (8 marks)
5 cm
This frustum is now suspended by a string attached to a
point on the rim of its smaller circular face, and hangs (3 marks)
in equilibrium.
T2
b Find the angle between the vertical and the axis of the frustum. B
E 5 The diagram shows a uniform solid cylinder of T1
mass 2 kg, height 2 m and radius 0.5 m. The cylinder A
is suspended from two light, inelastic, vertical
strings attached to the upper rim of the cylinder
at points A and B. The line AB forms a diameter
of the top face of the cylinder and is inclined at
an angle of 10° to the horizontal.
Given that the cylinder is in equilibrium,
work out the tension in each string. (5 marks)
E 6 A non-uniform rod PQ of length 1 m is freely T1 T2
suspended from its ends by two light, inelastic P
strings which hang vertically. The tensions in 45º
the strings are T1 and T2 respectively, and PQ Q
makes an angle of 45° with the vertical,
as shown in the diagram. (8 marks)
(4 marks)
At a distance h along the rod from Q, the
mass per unit length of the rod is given by:
ρ = _√ __11_+0_ h_ kg m–1
a Find the distance of the centre of mass of the rod from Q.
b Find T1 and T2
E/P 7 The diagram shows a non-uniform rod AB of length 10 m, which rests on two pivots P and Q
that are positioned 1 m from each end of the rod.
AP QB
At a distance x along the rod from A, the mass per unit length of the rod is (1 + 3x) kg m−1.
a Find the reaction on the rod of each pivot. (9 marks)
A mass of m kg is attached to the rod at B so that the rod is on the point of turning about Q.
b Find the value of m. (4 marks)
108
Further centres of mass
A 8 A non-uniform solid cylinder has height 30 cm and base radius 10 cm. At a height h cm above
E/P the base of the cylinder, its density is given by e 0.1h g cm−3.
a Show that the centre of mass of the cylinder lies a distance _1 _0_(e2_3_e−_3 _+1_ _1_) cm above the base of
the cylinder. (8 marks)
The cylinder is suspended from a point on its upper rim.
b Find, to the nearest degree, the angle that the upper plane face of the cylinder makes with
the vertical. (4 marks)
E 9 A solid is formed by removing a solid hemisphere of diameter T1 T2
6 cm and centre O from a solid, uniform right circular A B
cylinder of height 10 cm and diameter 10 cm. The point O
lies at the centre of one circular face of the cylinder.
The cylinder is suspended by two light, inextensible 6 cm O 10 cm
strings, which hang vertically. The strings are attached
to points A and B on the rims of the solid, such that
AB is parallel to the axis of the cylinder.
The solid has a density of _π1 _ kg m3 , and hangs in equilibrium
with AB horizontal.
a Find the tensions in the strings. (10 marks) 10 cm
The string at B breaks, and the solid hangs in equilibrium
from point A.
b Find the angle between the circular plane face of the solid and the vertical. (3 marks)
E/P 10 A uniform solid right circular cone has base radius r, height 4r θ
and mass m kg. One end of a light inextensible string is attached 4r
to the vertex of the cone and the other end is attached to a
point on the rim of the base. The string passes over a smooth θ r
peg and the cone rests in equilibrium with the axis horizontal,
and with the strings equally inclined to the horizontal at an
angle θ, as shown in the diagram.
a Show that angle θ satisfies the equation tan θ = _21
(4 marks)
b Find the tension in the string, giving your answer as an exact multiple of mg. (4 marks)
E/P 11 A plastic mould is formed by removing a solid hemisphere of 60 cm
radius 40 cm and centre O from a solid hemisphere of metal 40 cm
of radius 60 cm and centre O, as shown in the diagram. O
The mould is filled to the brim with molten plastic,
which is allowed to solidify. The metal used to form (10 marks)
the mould has 10 times the density of the plastic.
After the mould is filled and set, it is suspended
from a point on the outer rim of its plane face,
and hangs in equilibrium.
Find the angle that the plane face makes with the vertical.
109
Chapter 3
3.5 Toppling and sliding
A You can determine whether a body will remain in equilibrium or if equilibrium will be broken by
sliding or toppling.
■ If a body rests in equilibrium on a rough inclined plane, G
B
then the line of action of the weight of the body must pass
through the face of the body which is in contact with A
the plane.
In the diagram, the weight of the body produces a clockwise αW
moment about A which keeps it in contact with the plane.
If the angle of the plane is increased so that the line of action of GB
A
the weight passes outside the face AB then the weight produces
an anticlockwise moment about A. If the plane is sufficiently
rough to preventy sliding, then the body will topple over.
On line Explore toppling and sliding using GeoGebra. βW
The only forces acting on the body are its weight and the total reaction G
between the plane and the body. The total reaction between the plane R
and the body consists of a normal reaction force and a friction force.
As the body is in equilibrium, these forces must be equal and opposite W
and act in the same vertical line.
Example 19
A solid uniform cylinder of radius 3 cm and height 5 cm has a solid uniform hemisphere made from
the same material, of radius 3 cm, joined to it so that the base of the hemisphere coincides with one
circular end of the cylinder.
a Find the position of the centre of mass of the composite body.
The composite body is placed with the circular face of the cylinder on a rough inclined plane,
which is inclined at an angle α to the horizontal. Given that the plane is sufficiently rough to
prevent sliding,
b show that equilibrium is maintained provided that tan α , _3238_
a Shape Mass Ratio Height of centre
of mass above
of base in cm xG
5 cm
masses 2.5 3 cm
5 + _83_ × 3 = 6.125
Cylinder ρπ 32 × 5 = 45ρπ 5 Draw a diagram and use a
_ 32 ρπ 33 = 18ρπ 2 x_ table to show masses and
Hemisphere positions of centres of mass.
63 ρπ 7
Composite
body
110
Further centres of mass
A So 5 × 2.5 + 2 × 6.125 = 7 x_ 99 Take moments to find the
and x_ = _ 2_47_.7_ 5_ = _ 92_89_ 28 position of the centre of
mass of the composite body.
b From the diagram, tan α = _ _ 293__89 = 3 × _ 29_89 = _ 32_83 G
If α is smaller than this value, G is above α The limiting case is shown
a point of contact and equilibrium is 3 in the diagram, where the
maintained. Equilibrium is maintained point vertically below the
provided that tan α , _ 32_83 α centre of mass, G, is on the
edge of the area of contact.
W
If tan α . _ 2383_ then the body
will topple over because
there will be a turning effect
about the point A.
Example 20
A uniform solid cube of mass M and side 2a rests on a 2a P
rough horizontal plane. The coefficient of friction is μ.
A horizontal force of magnitude P is applied at the 2a
midpoint of an upper edge, perpendicular to that R
edge, as shown in the diagram. The reaction between
the plane and the cube comprises a normal Mg
reaction force R and a friction force F and acts at A
a distance h from a lower edge of the cube as shown
in the diagram. Fh
Find whether the cube remains in equilibrium, or whether the equilibrium is broken by sliding or
toppling, in each of the following cases. Also determine the value of F and the value of h in each
case, giving your answers in terms of M, g and a as appropriate:
a P = 0 b P = _ 21 M g and μ = _ 43 c P = _ 41 Mg and μ = _ 15
For equilibrium: First use the conditions for
equilibrium in the general case,
Resolve horizontally → i.e. resolve horizontally, resolve
P − F = 0, so F = P (1) vertically and take moments.
Resolve vertically ↑ A is the point on the bottom edge of
R − Mg = 0 so R = Mg (2) the cube shown in the diagram.
Take moments about point A: For part a substitute
P × 2a + R × h = Mg × a (3) P = 0 in equations (1), (2) and (3).
a When P = 0 In part a the normal reaction acts at
Substituting result from equation (2) into equation (3): the centre of the base.
0 + Mgh = Mga
h=a
Substituting P = 0 into equation (1) gives F = 0.
The cube remains in equilibrium. It does not slide and does
not topple.
111
Chapter 3
A b When P = _ 21 M g For part b substitute P = _ 21 mg in
Substituting result from equation (2) into equation (3): equations (1), (2) and (3).
_ 21 M g × 2a + Mgh = Mga
In part b the normal reaction acts at
h = 0, and the cube is about to topple. A and so toppling is about to occur
Substituting P = _ 21 M g into equation (1) gives F = _ 21 M g around the edge through A.
But μR = _ 34 M g so F , μR and the body does not slide.
This implies that the body does not
c When P = _ 41 M g topple as the reaction is within the
Substituting result from equation (2) into equation (3): area of the base.
_ 41 Mg × 2a + Mgh = Mga
h = _ 21 a, and the cube does not topple. Assuming equilibrium leads to a
Substituting P = _ 41 M g into equation (1) would give F = _ 41 M g contradiction. This cube will slide as
the force P exceeds the maximum
But this is impossible as the maximum value that F can friction force.
take is μR and μR = _ 51 M g so F = _ 51 M g
The cube will slide if the force P is maintained.
■ A body is on the point of sliding when F = μR
■ A body is on the point of toppling when the reaction acts at the point about which turning
can take place.
Example 21
A uniform solid cube of mass M and side 2a, is placed on a rough inclined plane which is at an
angle α to the horizontal, where tan α = _12 . The coefficient of friction is μ.
Show that if μ , _ 21 the cube will slide down the slope.
R(↖) R − Mg cos α = 0 Draw a diagram showing G the centre of mass
of the cube, and points A and B on the edges
R = Mg cos α of the cube in the same vertical plane as G.
R G B R(↗) F − Mg sin α = 0
CF F = Mg sin α First use the conditions for equilibrium, i.e.
Z resolve along and perpendicular to the plane
A x and use F < μR, where F is the force of
α Mg friction and R is the normal reaction.
For equilibrium F < μR, i.e. μ > _FR_ As tan α = _ 12 the limiting case is when μ = _ 12
So μ > _MM_g_g _cs_ion_s α_ α_
i.e. μ > tan α and so μ > _ 21 for equilibrium.
So if μ < _21 the cube will slide.
112
Further centres of mass
A Let the point Z, vertically below the centre of Draw an enlarged G
mass G, be at a distance x from A up the plane. triangle if it helps. αa
Let C be the midpoint of AB. C
a–x
From triangle GCZ, a__−a__ x_ = tan α = _ 21_
Z
So 2a − 2x = a
x = _21 a The normal reaction would act at point Z,
which could also be established by taking
So Z is between A and B and the cube does not
topple. moments and using the equilibrium conditions.
So the weight of the cube acts through a point
within the area of contact.
Example 22
The cube in the previous example is now replaced by a cylinder of mass M with base radius a and
height 6a. The cylinder is placed on the rough inclined plane, which is inclined at an angle α to
the horizontal, where tan α = _21 . The coefficient of friction is μ. Show that if μ . _ 12 the cylinder will
topple about the lower edge.
As μ . _ 21 G This condition was shown in Example 21, and the
the cylinder working here would be identical.
will not slip. i.e. Resolve along and perpendicular to the plane
and use F < μR.
Z AC B The points A, B, C, G and Z are defined as in
Example 21.
α Mg
Draw an enlarged triangle G
Let the point Z, vertically below the centre of GZC.
mass G, be at a distance x from A down the α
plane. 3a
Let C be the midpoint of AB. C
From triangle GCZ, _ a_3+_a_ x_ = tan α = _ 21_ a+x
Z
So 2a + 2x = 3a
x = _ 21 a Another method you could use is to let the
reaction act at point A and show that even in this
So Z is not between A and B and the position there will be a turning moment about A
cylinder will topple. and therefore no equilibrium.
113
Chapter 3
Exercise 3E
A
1 A uniform solid cylinder with base diameter 4 cm stands on a rough plane inclined at 40° to the
horizontal. Given that the cylinder does not topple, find the maximum possible height of the
cylinder.
2 A uniform solid right circular cylinder with base radius 3 cm and height 10 cm is placed with
its circular plane base on a rough plane. The plane is gradually tilted until the cylinder topples
over. Given that the cylinder does not slide down the plane before it topples,
a find the angle that the plane makes with the horizontal at the point where the cylinder
topples
b find the minimum possible value of the coefficient of friction between the cylinder and the
plane.
P 3 A uniform solid right circular cone with base radius 5 cm and height _√h 3_3_ c_ _ .mTihsepplalacnede with
its circular plane base on a rough plane. The coefficient of friction is is
gradually tilted.
When the plane makes an angle of θ with the horizontal, the cone is about to slide and topple
at the same time. Find:
a θ
b h
P 4 A uniform solid right circular cone of mass M with base P
radius r and height 2r is placed with its circular plane V 60°
base on a rough horizontal plane. A force P is applied
to the vertex V of the cone at an angle of 60° above the 2r
horizontal as shown in the diagram. r
The cone begins to topple and to slide at the same time.
a Find the magnitude of the force P in terms of M.
b Calculate the value of the coefficient of friction.
P 5 A frustum of a right circular solid cone has two plane circular end faces with radii r and 2r
respectively. The distance between the end faces is 2r.
a Show that the centre of mass of the frustum is at a distance _1 1_14_r from the larger circular face.
b Find whether this solid can rest without toppling on a rough plane, inclined to the horizontal
at an angle of 40°, if the face in contact with the inclined plane is:
i the large circular end ii the small circular end.
c In order to answer part b you assumed that slipping did not occur. What does this imply
about the coefficient of friction μ?
114
Further centres of mass
A 6 A uniform cube with edges of length 6a and weight W stands on a rough horizontal plane.
P The coefficient of friction is μ. A gradually increasing force P is applied at right angles to a
vertical face of the cube at a point which is a distance a above the centre of that face.
a Show that equilibrium will be broken by sliding or toppling depending on whether
μ , _43 or μ . _ 34 .
b If μ = _41 , and the cube is about to slip, find the distance from the point where the normal
reaction acts, to the nearest vertical face of the cube.
E/P 7 Two uniform, rough, solid cuboids of dimensions 20 cm 30 cm
20 cm × 20 cm × 30 cm are stacked as shown in the A
diagram shown opposite. Cuboid A has density ρ and 20 cm
cuboid B has density kρ, for some constant k.
B 30 cm
20 cm 20 cm
The cuboids are placed on a rough plane inclined at an A
angle of θ to the horizontal, as shown in the diagram opposite. B
The line of greatest slope of the plane is parallel to the plane
faces shown in the diagram, and the angle of inclination of θ
the plane is gradually increased.
a In the case when k = 5, find the angle of inclination at
which the whole body begins to topple. (8 marks)
b Given that cuboid A topples off cuboid B before the
whole body topples, find the range of possible values
of k. (4 marks)
E/P 8 A non-uniform cylinder is 1.5 m high and has radius 0.25 m. At any height, x m, above its base,
the density of the cylinder is given by ρ = cosh x kg m−3. The cylinder is placed on a rough slope.
Assuming that the cylinder does not slide down the slope, what is the maximum possible angle
of the slope to the horizontal, before the cylinder topples over? (10 marks)
E/P 9 A uniform solid cylinder of base radius r and height h has the same h
density as a uniform solid hemisphere of radius r. The plane face of
the hemisphere is joined to a plane face of the cylinder to form the r
composite solid S shown. The point O is the centre of the plane base of S. O
a Show that the distance from O to the centre of mass of S is _ 6 _h_42_(+_3_h8_h _+r_2_+_r)_3 _r_2
(8 marks) 115
Chapter 3
A The solid is placed on a rough plane which is inclined at an angle α to the
horizontal. The plane base of S is in contact with the inclined plane.
b Given that h = 3r and that S is on the point of toppling, find α to the nearest degree.
(4 marks)
c Given that the solid did not slip before it toppled, find the range of possible values for the
coefficient of friction. (4 marks)
E/P 10 A uniform solid paperweight is in the shape of a frustum of a cone. hθ
It is formed by removing a right circular cone of height h from a 2h
right circular cone of height 2 h and base radius 2 r.
2r
a Show that the centre of mass of the paperweight lies at a
height of _21 8_1 h from its base. (6 marks) (4 marks)
When placed with its curved surface on a horizontal plane, the
paperweight is on the point of toppling.
b Find θ, the semi-vertical angle of the cone, to the nearest degree.
E/P 11 A uniform solid cone of mass M, height 8 cm and radius 3 cm, is placed with its circular base
on a horizontal plane. The coefficient of friction is μ. A horizontal force of magnitude P is
applied at the vertex of the cone.
a Find the value of P which will cause the cone to slide giving your answer in terms of μ, M
and g. (4 marks)
b Find the value of P which will cause the cone to topple, giving your answer in terms of M
and g. (2 marks)
c State whether the cone will slide or topple if: (1 mark)
i μ = _ 41 (1 mark)
ii μ = _ 12 (1 mark)
iii μ = _83
E/P 12 A uniform solid cylinder of mass 200 g, height 4 cm P
and radius 3 cm rests with its circular base on a plane.
The plane makes an angle α with the horizontal where 4 cm
tan α = _34 . The coefficient of friction is _1 6_7 . 3 cm
A horizontal force P is applied to the highest point of
the cylinder as shown in the diagram. α
a Find the value of P which will just cause the cylinder to topple about the highest point of the
base. (4 marks)
b Find the value of P which would cause the cylinder to slide up the plane. (6 marks)
c Show that the cylinder topples before it slides. (1 mark)
116
Further centres of mass
Challenge
A A child’s toy is made from joining a right circular h
r
uniform solid cone, radius r and height h, to a
uniform solid hemisphere of the same material
and radius r. They are joined so that their plane
faces coincide as shown in the diagram.
a Show that the distance of the centre of mass
of the toy from the base of the cone is:
| | _4h_(22_−r__+3_r_h 2_)
The toy is placed with its hemisphere in contact with a horizontal plane
and with its axis vertical. It is slightly displaced and released from rest.
b Given that the plane is sufficiently rough to prevent slipping, explain
clearly, with reasons, what will happen in each of the following cases:
__ __ __
i h . r√3 ii h , r√ 3 iii h = r√ 3
Mixed exercise 3
E/P 1 The curve shows a sketch of the region R bounded
by the curve with equation y2 = 4x and the line
with equation x = 4. The unit of length on both y2 = 4x
the axes is the centimetre. The region R is rotated R
through π radians about the x-axis to form a solid S.
a Show that the volume of the solid S is 32π cm3. O 4
(4 marks)
b Given that the solid is uniform, find the x=4
distance of the centre of mass of S from O.
(6 marks)
E/P 2 The region R is bounded by the curve with equation y = 1
y = _ x1_ , the lines x = 1, x = 2 and the x-axis, as shown x
in the diagram. The unit of length on both the axes
is 1 m. A solid plinth is made by rotating R through R
2π radians about the x-axis.
a Show that the volume of the plinth is _2π _ m3. O x =1 x=2
(4 marks)
(6 marks)
b Find the distance of the centre of mass of the
plinth from its larger plane face, giving your
answer in cm to the nearest cm.
117
Chapter 3
A 3 The diagram shows a cross-section of a uniform solid standing on horizontal (6 marks)
E/P ground. The solid consists of a uniform solid right circular cylinder, of
diameter 80 cm and height 40 cm, joined to a uniform solid hemisphere of
the same density. The circular base of the hemisphere coincides with the
upper circular end of the cylinder and has the same diameter as that
of the cylinder. Find the distance of the centre of mass of the solid
from the ground.
E/P 4 A simple wooden model of a rocket is made by taking a uniform cylinder,
of radius r and height 3r, and carving away part of the top two-thirds to
form a uniform cone of height 2r as shown in cross-section in the diagram.
Find the distance of the centre of mass of the model from its plane
face. (6 marks)
E/P 5 The diagram shows a cross-section containing the axis of symmetry of a C
uniform body consisting of a solid right circular cylinder of base
radius r and height kr surmounted by a solid hemisphere of radius r. (5 marks)
a Given that the centre of mass of the body is at the centre C of the (1 mark)
common face of the cylinder and the hemisphere, find the value of k,
giving your answer to 2 significant figures.
b Explain briefly why the body remains at rest when it is placed with any part of its
hemispherical surface in contact with a horizontal plane.
E/P 6 A uniform lamina occupies the region R bounded by y Figure 1
the x-axis and the curve with equation O y = 41x(4 – x)
y = _14 x(4 − x), 0 < x < 4, as shown in Figure 1.
R
a Show by integration that the y-coordinate of the
centre of mass of the lamina is _52 . (8 marks)
x
A uniform prism P has cross-section R. The prism is placed R
with its rectangular face on a slope inclined at an angle θ θ
to the horizontal. The cross-section R lies in a vertical
plane as shown in Figure 2. The surfaces are sufficiently Figure 2
rough to prevent P from sliding.
b Find the angle θ, for which P is about to topple. (2 marks)
E/P 7 A uniform semicircular lamina has radius 2a and the midpoint C
of the bounding diameter AB is O.
a Using integration, show that the centre of mass of the
lamina is at a distance _38 _πa_ from O.
(8 marks)
AO B
118
Further centres of mass
A The two semicircular laminas, each of radius a and with AO
and OB as diameters, are cut away from the original lamina to
leave the lamina AOBC shown in the diagram, where OC is perpendicular to AB.
b Show that the centre of mass of the lamina AOBC is at a distance _4 π_a_ from O.
(6 marks)
The lamina AOBC is of mass M and a particle of mass M is attached to the lamina at B to
form a composite body.
c State the distance of the centre of mass of the body from OC and from OB. (1 mark)
The body is smoothly hinged at A to a fixed point and rests in equilibrium in a vertical plane.
d Calculate, to the nearest degree, the acute angle between AB and the horizontal. (3 marks)
E/P 8 A uniform wooden ‘mushroom’, used in a game, is made by joining O
a solid cylinder to a solid hemisphere. They are joined symmetrically,
such that the centre O of the plane face of the hemisphere coincides
with the centre of one of the ends of the cylinder. The diagram shows
the cross-section through a plane of symmetry of the mushroom, as it
stands on a horizontal table.
The radius of the cylinder is r, the radius of the hemisphere is 3r, and the centre of mass of the
mushroom is at the point O.
__
a Show that the height of the cylinder is r √ _82_1 (6 marks)
The table top, which is rough enough to prevent the mushroom from sliding, is slowly tilted
until the mushroom is about to topple.
b Find, to the nearest degree, the angle with the horizontal through which the table top has
been tilted. (3 marks)
E/P 9 Figure 1 shows a finite region A which is bounded by the y y2 = 4ax
curve with equation y2 = 4ax, the line x = a and the x-axis.
A uniform solid S1 is formed by rotating A through 2π A
radians about the x-axis.
O x=a
a Show that the volume of S1 is 2πa3. (3 marks) Figure 1
b Show that the centre of mass of S1 is a distance _2 3_a_ from x
the origin O. (4 marks)
Figure 2 shows a cross-section of a uniform solid S which has
been obtained by attaching the plane base of solid S1 to the
plane base of a uniform hemisphere S2 of base radius 2a. Figure 2 S1 S2
c Given that the densities of solids S1 and S2 are ρ1 and ρ 2
respectively, find the ratio ρ1 : ρ2 which ensures that the
centre of mass of S lies in the common plane face of
S1 and S2. (6 marks)
d Given that ρ1 : ρ 2 = 6, explain why the solid S may rest in equilibrium with any point of the
curved surface of the hemisphere in contact with a horizontal plane. (2 marks)
119
Chapter 3
A 10 A mould for a right circular cone, base radius r and 2r
E/P height h, is produced by making a conical hole in a A rB
uniform cylindrical block, base radius 2r and height 3r.
The axis of symmetry of the conical hole coincides
with that of the cylinder, and AB is a diameter of the h
top of the cylinder, as shown in the diagram. 3r
a Show that the distance from AB of the centre of
mass of the mould is _42 _(1_36_6r_r2_−−__hh__2)
(8 marks)
The mould is suspended from the point A, and hangs
freely in equilibrium.
b In the case h = 2r, calculate, to the nearest degree, the angle between AB and the downward
vertical. (4 marks)
E/P 11 A compound solid is made from an inverted frustum and a 10 cm
hemisphere as shown opposite. 30 cm
The base of the frustum has a radius of 5 cm and the
hemisphere has a radius of 10 cm. 5 cm
a Given that the density, ρ, of the hemisphere is three
times the density of the frustum, find the height of
the centre of mass of the compound solid from its base.
(8 marks)
The compound solid is placed on a rough plane inclined at 45°
to the horizontal.
b Given that the total mass of the compound solid is m kg, A
work out the minimum horizontal force, in terms of m,
that must be applied at A to stop the solid from toppling
over. The plane is sufficiently rough to prevent
slipping. (4 marks)
45°
E/P 12 A solid is formed by rotating the region enclosed y
by the curve y = _ x_+2__ 3_ , the lines x = 2 and x = 4
and the x-axis, though 360° around the x-axis.
The solid is then placed on its smaller circular end
on rough horizontal ground.
a Find the height of the centre of mass of the
solid above the ground. (10 marks)
2
The solid is then placed on a rough inclined ramp y= x+3
angled at θ° to the horizontal.
4x
b Assuming that the solid does not slip, work out O 2
(4 marks)
the value of θ when the solid is on the point of tipping.
120
Further centres of mass
A 13 A concrete pile is used as a foundation for a bridge. The pile is modelled as a vertical non-
E/P uniform rod. The top of the pile is flush with the ground, and the base of the pile is 9 m deep.
The mass +pe4r0u0n√ _ixt_ length of the pile, m kg m−1, at a depth of x m is given by the formula
m= 1000
a Show that the pile has a mass of 16 200 kg. (3 marks)
b Hence find the depth of the centre of mass of the pile. (4 marks)
E/P 14 A non-uniform solid right circular cylinder has base radius 4 cm and height 90 cm. The cylinder
is modelled as a non-uniform rod, with mass per unit length (20 − _ 19 h) grams cm−1 at a height h
above its base.
a Using this model, calculate the height of the centre of mass of the cylinder above its
base. (7 marks)
The plane base of the cylinder is placed on a rough surface, which is inclined at an angle θ to
the horizontal. The cylinder rests in equilibrium on the point of toppling.
b Comment on the suitability of the model for calculating: (2 marks)
i the centre of mass of the cylinder
ii the value of θ.
c Find the value of θ in degrees to 3 significant figures. (3 marks)
E/P 15 A non-uniform rod AB of length 12 m hangs in horizontal equilibrium, supported by two light
inextensible strings, which hang vertically, attached at A and B. The mass per unit length,
m kg m−1, of the rod at a distance x m from A is given by the formula m = 2 + _x4_2
Find the tension in each string. (9 marks)
E/P 16 The diagram shows a non-uniform rod AB resting on horizontally on two pegs. One peg is
positioned at A and the other peg is positioned three-quarters of the way along AB. At a
distance x m from A the mass per unit length of the rod is given by (35 − _ 12 x) kg m−1.
A 3l lB
Given that the reactions of each peg on the rod are the same, find the length of the rod.
(10 marks)
Challenge
A non-uniform solid right circular cone h
has base radius h m and height h m. h
At a distance x m above its base, the cone
has density (x + 1) kg m−3.
a Show that the mass of the cone is _1 1_2 h3 (h + 4) kg.
b Given that the centre of the mass of the cone is a distance _31 h above
its base, find the value of h.
c Show that as h varies the height of the centre of mass of the cone
above its base cannot exceed _52 h.
121
Chapter 3
Summary of key points
A 1 The centre of mass of a uniform lamina may be found using the formulae:
b b
∫ ∫• x_ = _ _a _ b_21_ y_2_ _d_x
xy dx y_ =
_ a__ b____ and
∫ ∫ y dx
y dx
aa
b b
ρxy dx
a
∫ ∫•
M x_ = and M _y = _12 ρy2 dx
a
∫ b
where M = ρ y dx, and is the total mass of the lamina, and ρ is the mass per unit area
a
of the lamina.
2 Standard results for uniform laminas and for arcs
Lamina Centre of mass along axis of symmetry
Semicircle, radius r
Sector of circle, radius r, angle at centre 2α _34_πr_ from the centre
Circular arc, radius r, angle at centre 2α _ 2_r_ 3s_αin_ _ α_ from the centre
_ r _s_αin_ _ α_ from the centre
3 • For a solid body the centre of mass is the point where the weight acts.
• For a uniform solid body the weight is evenly distributed through the body.
• The centre of mass will lie on any axis of symmetry.
• The centre of mass will lie on any plane of symmetry.
4 From symmetry, the centre of mass of some uniform bodies is at their geometric centre. These
include the cube, the cuboid, the sphere, the right circular cylinder and the right circular prism.
5 For a uniform solid of revolution, where the revolution is about the x-axis, the centre of
mass lies on the x-axis, by symmetry, and its position on the axis is given by the formulae
x_ = ∫_∫π_π_y_y2x_2 _dd_xx_ or M x_ = ∫ρπy2x dx
where M is the known mass of the solid and ρ is its density.
6 For a uniform solid of revolution, where the revolution is about the y-axis, the centre of
mass lies on the y-axis, by symmetry, and its position on the axis is given by the formulae
y_ =_ ∫∫_ππ_ _xx_22y _d _dy_y or M _y = ∫ρπx2y dy
where M is the known mass of the solid and ρ is its density.
122
Further centres of mass
A 7 Standard results for uniform bodies
Body Centre of mass along axis of symmetry
Solid hemisphere, radius R
Hemispherical shell, radius R _83 R from the centre
Solid right circular cone, height h _ 21 R from the centre
Conical shell, height h _ 34 h from the vertex, or _41 h from the circular base
_32 h from the vertex, or _31 h from the circular base
8 For a non-uniform rod with a variable mass per unit length, ρ = f(x), and length l, the distance
of the centre of mass from the end of the rod is given by:
∫ l xρ dx ∫ l xf(x) dx
x ¯ = _ ∫0_ _l ρ _ _d_x_ = _ ∫0_ _l f_(x__) _d_x_
00
9 A rigid body is in equilibrium if:
• there is zero resultant force in any direction, i.e. the sum of the components of all the forces
in any direction is zero, and
• the sum of the moments of the forces about any point is zero.
10 • When a lamina is suspended freely from a fixed point or pivots freely about a horizontal
axis it will rest in equilibrium in a vertical plane with its centre of mass vertically below the
point of suspension or pivot.
• When a rigid body is suspended freely from a fixed point and rests in equilibrium then
its centre of mass is vertically below the point of suspension.
11 If a body rests in equilibrium on a rough inclined plane, then the line of action of the weight
of the body must pass through the face of the body which is in contact with the plane.
12 You can establish whether equilibrium will be broken by sliding or by toppling by considering:
• a body is on the point of sliding when F = μR
• a body is on the point of toppling when the reaction acts at the point about which turning
can take place.
123
Review exercise 1
E 1 A circular flywheel of diameter 7 cm is The length of the string is L.
rotating about the axis through its centre
and perpendicular to its plane with a Show that the tension in the string (2)
constant angular speed 1000 revolutions is 2mg.
per minute.
b Find ω in terms of g and L. (4)
Find, in m s−1 to 3 significant figures, ← Section 1.3
the speed of a point on the rim of the (2) E/P 4 A car moves round a bend which is
flywheel.
banked at a constant angle of 10° to the
horizontal. When the car is travelling at
← Section 1.1
E 2 A particle P of mass 0.5 kg is attached a constant speed of 18 m s−1, there is no
to one end of a light inextensible string sideways frictional force on the car. The
of length 1.5 m. The other end of the car is modelled as a particle moving in
string is attached to a fixed point A. The a horizontal circle of radius r metres.
particle is moving, with the string taut, in Calculate the value of r. (6)
a horizontal circle with centre O vertically
below A. The particle is moving with ← Section 1.3
constant angular speed 2.7 rad s−1. Find:
E 5 A cyclist is travelling around a circular
a the tension in the string (4) track which is banked at 25° to the
horizontal. The coefficient of friction
b the angle, to the nearest degree, that between the cycle’s tyres and the track
AP makes with the downward vertical. is 0.6. The cyclist moves with constant
(4) speed in a horizontal circle of radius
40 m, without the tyres slipping.
← Section 1.2
Find the maximum speed of the cyclist.
E/P 3 A (9)
← Section 1.3
60° L E/P 6 A
ωP 5l
3l
A particle P of mass m is attached to one B
end of a light string. The other end of the
string is attached to a fixed point A. The A light inextensible string of length 8l has
particle moves in a horizontal circle with its ends fixed to two points A and B, where
constant angular speed ω and with the A is vertically above B. A small smooth
string inclined at an angle of 60° to the ring of mass m is threaded on the string.
vertical, as shown in the diagram above. The ring is moving with constant speed
124
Review exercise 1
in a horizontal circle with centre B and E/P 9 A
radius 3l, as shown in the diagram. Find: 60°
a the tension in the string (3)
b the speed of the ring. (5)
c State briefly in what way your hP
solutions might no longer be valid if
the ring were firmly attached to the
string. (1)
← Section 1.3 60°
E 7 A B
α° A particle P of mass m is attached to
two light inextensible strings. The other
ends of the string are attached to fixed
points A and B. The point A is a distance
h vertically above B. The system rotates
about the line AB with constant angular
OB speed ω. Both strings are taut and inclined
A metal ball B of mass m is attached to at 60° to AB, as shown in the diagram.
Ta hSehpoawrtitchlaetmr o=v_ e√2_s _3__ i hn. a circle of radius r.
one end of a light inextensible string. (3)
The other end of the string is attached
to a fixed point A. The ball B moves in a
horizontal circle with centre O vertically b Find, in terms of m, g, h and ω, the
below A, as shown in the diagram. The tension in AP and the tension in BP. (3)
string makes a constant angle α° with The time taken for P to complete one
tchoensdtoawntnawnagrudlavrerstpiceaelda√n _2d_g_Bk_ , mwohveerse with
k is a √ circle is T.T,π _ _2g__h__ (2)
constant. The tension in the string is 3mg.
c Show that ← Section 1.3
By modelling B as a particle, find:
a the value of α (4) E 10 A
b the length of the string. (5) θ
h
← Section 1.3
E 8 A particle P of mass m moves on the R
smooth inner surface of a spherical Bω
bowl of internal radius r. The particle
moves with constant angular speed in a One end of a light inextensible string is
horizontal circle, which is at a depth _12 r attached to a fixed point A. The other end
below the centre of the bowl. Find: of the string is attached to a fixed point
B, vertically below A, where AB = h.
a the normal reaction of the bowl (4) A small smooth ring R of mass m is
on P threaded on the string. The ring R moves
in a horizontal circle with centre B, as
b the time for P to complete one (6)
revolution of its circular path.
← Section 1.3
125
Review exercise 1
shown in the diagram. The upper section to a point A which is vertically above the
of the string makes a constant angle θ point O on a smooth horizontal table.
with the downward vertical and R moves The particle P remains in contact with the
with constant angular speed ω. The ring surface of the table and moves in a circle
is modelled as a particle. w_i_t_h centre O and with angular speed
SDheodwuctehathtaωt 2ω=._ hg_ ( _1__ 2h___g+_s__ i ._n s_i _θn_ θ_ ) .
√ a (5) √ _k3_ag_ , where k is a constant. Throughout
(2)
b the motion the string remains taut and
∠ APO = 30°, as shown in the diagram.
√ Given that ω = __ 3h__g__ , a Show that the tension in the string is
_ 2_k3_m_ g_
c find, in terms of m and g, the tension (3)
in the string. (4) b Find, in terms of m, g and k, the
normal reaction between P and the
← Section 1.3 table. (3)
E/P 11 3a c Deduce the range of possible values
of k. (2)
C 4a √ T __ 2ha__ge__ .aTnhgeulpaarrstpiceleedPonfoPwismcohvaensgiendato
V horizontal circle above the table. The
centre of this circle is X.
A hollow cone, of base radius 3a and
d Show that X is the midpoint of OA. (7)
← Section 1.3
height 4a, is fixed with its axis vertical E/P 13 A
and vertex V downwards, as shown in the
diagram. A particle moves in a horizontal
circle with centre C, on the smooth inner l
√ surface_ _98o__agf__ . tFheincdontheewhiethighcot nosftaCntabaonvgeulV(a8.r) 3 l P
2
speed
l
← Section 1.3
E 12 A B
2a A particle P of mass m is attached to the
ends of two light inextensible strings AP
O 30° P and BP, each of length l. The ends A and
B are attached to fixed points, with A
A particle P of mass m is attached to one vertically above B and AB = _ 23 l, as shown
end of a light inextensible string of length in the diagram above. The particle P
2a. The other end of the string is fixed moves in a horizontal circle with constant
angular speed ω. The centre of the circle
is the midpoint of AB and both strings
remain taut.
126
Review exercise 1
a Show that the tension in AP is A a Find:
_ 61 m(3lω 2 + 4g). (7) i the tension in the string
b Find, in terms of m, l, ω and g, an ii the angle θ that the string makes (5)
with the vertical. (4)
expression for the tension in BP. (2) b Show that v = k√ _g_x_ , where k is a
c Deduce that ω 2 > _ 43_gl_ .
(2) rational constant to be found.
← Section 1.3 ← Section 1.3
A 14 A rough disc rotates in a horizontal plane E/P 16 One end of a light inextensible string of
E/P with constant angular velocity ω about a length l is attached to a particle P of mass
m. The other end is attached to a fixed
fixed vertical axis. A particle P of mass m point A. The particle is hanging freely at
lies on the disc at a distance _43 a from the √ rest with the string verticalwhen it__ i5s2__g_ _l
axis. The coefficient of friction between P speed
and the disc is _35 . Given that P remains at projected horizontally with
rest relative to the disc,
a prove that ω 2 < _ 29_0_ga_ . a Find the speed of P when the string (4)
is horizontal.
The particle is now connected to the axis
by a horizontal light elastic string of When the string is horizontal it comes
natural length a and modulus of elasticity into contact with a small smooth fixed
2mg. The disc again rotates with constant peg which is at the point B, where AB is
angular velocity ω about the axis and P horizontal, and AB , l. Given that the
remains at rest relative to the disc at a particle then describes a complete semi-
distance _ 43 a from the axis. circle with centre B,
b Find the greatest and least possible
b find the least possible value of the (9)
values of ω 2. length AB.
← Section 1.3 ← Section 1.4
E/P 15 The diagram shows a particle of mass E/P 17 One end of a light inextensible string of
2 kg attached to a light elastic string length l is attached to a fixed point A.
of natural length x m and modulus of The other end is attached to a particle P
elasticity 10g N. The other end of the of mass m, which is hanging freely at rest
string is attached to a fixed point A, and hatopriozionnttBal.lyTfhreompaBrtiwcliethPsipseperdo√je _3c_gt_le .d
the particle moves with constant speed
v m s−1 in a horizontal circle with centre O, When AP makes an angle θ with the
where O is vertically below A and downward vertical and the string remains
OA = x m. taut, the tension in the string is T.
A a Show that T = mg(1 + 3 cos θ). (6)
θ
b Find the speed of P at the instant
xm when the string becomes slack. (3)
OP c Find the maximum height above the (5)
level of B reached by P.
← Section 1.4
127
Review exercise 1
A 18 O A c Find the value of u. (3)
E/P θ ← Section 1.5
P E 20 A smooth solid sphere, with centre O and
radius a, is fixed to the upper surface of
Au a horizontal table. A particle P is placed
on the surface of the sphere at a point A,
A particle of mass m is attached to one where OA makes an angle α with the
end of a light inextensible string of length
l. The other end of the string is attached upward vertical, and 0 , α , _π2_ . The
to a fixed point O. The particle is hanging particle is released from rest. When
at the point A, which is vertically below OP makes an angle θ with the upward
O. It is projected horizontally with speed vertical, and P is still on the surface of
u. When the particle is at the point P, the sphere, the speed of P is v.
∠ AOP = θ, as shown in the diagram. The a Show that v 2 = 2ga(cos α − cos θ). (4)
string oscillates through an angle α on
either side of OA where cos α = _ 23 Given that cos α = _ 43 , find:
a Find u in terms of g and l. (3) b the value of θ when P loses contact (5)
with the sphere
When ∠ AOP = θ, the tension in the string c the speed of P as it hits the table. (4)
is T.
b Show _m3_g_ (9 ← Section 1.5
that T = cos θ − 4). (4) E/P 21
c Find the range of values of T during (2) C
the oscillations. a
← Section 1.4 θ
OB
E 19 A 0.2 m
u m s–1
B
θ A
The diagram shows a fixed hollow sphere
of internal radius a and centre O.
A smooth solid hemisphere, of radius A particle P of mass m is projected
0.8 m and centre O, is fixed with its plane horizontally from the l√o _ _27w_ a_eg_s . t IptominotvAesoinf
face on a horizontal table. A particle of the sphere with speed
mass 0.5 kg is projected horizontally with
speed u m s−1 from the highest point A a vertical circle, centre O, on the smooth
of the hemisphere. The particle leaves inner surface of the sphere. The particle
the hemisphere at the point B, which is a passes through the point B, which is in
vertical distance of 0.2 m below the level the same horizontal plane as O. It leaves
of A. The speed of the particle at B is the surface of the sphere at the point
v m s−1 and the angle between OA and OB C, where OC makes an angle θ with the
is θ, as shown in the diagram. upward vertical.
a Find the value of cos θ. (1) a Find, in terms of m and g, the normal
reaction between P and the surface of
b Show that v 2 = 5.88. (3) the sphere at B. (4)
128
Review exercise 1
A b Show that θ = 60°. (7) A 23 A
After leaving the surface of the sphere, P E 50 m
meets it again at the point A. 60°
c Find, in terms of a and g, the time P B D 30°
takes to travel from C to A. (4) C
← Section 1.5 The diagram represents the path of a
skier of mass 70 kg moving on a ski-
E/P 22 5m O slope ABCD. The path lies in a vertical
120° plane. From A to B, the path is modelled
A as a straight line inclined at 60° to the
P Q horizontal. From B to D, the path is
modelled as an arc of a vertical circle of
A trapeze artist of mass 60 kg is attached radius 50 m. The lowest point of the arc
to the end A of a light inextensible rope BD is C.
OA of length 5 m. The artist must swing
in an arc of a vertical circle, centre O, At B, the skier is moving downwards with
from a platform P to another platform Q, speed 20 m s−1. At D, the path is inclined
where PQ is horizontal. The other end at 30° to the horizontal and the skier is
of the rope is attached to the fixed moving upwards. By modelling the slope
point O which lies in the vertical plane as smooth and the skier as a particle, find:
containing PQ, with ∠ POQ = 120° and
OP = OQ = 5 m, as shown in the diagram. a the speed of the skier at C (2)
b the normal reaction of the slope on the
skier at C (2)
As part of her awcitt,hthspeeaerdti√s t _1_p5_ r moj es−c1tsin a c the speed of the skier at D (2)
herself from P
direction perpendicular to the rope OA d the change in the normal reaction of
and in the plane POQ. She moves in a the slope on the skier as she passes B.
circular arc towards Q. At the lowest (4)
point of her path she catches a ball of The model is refined to allow for the
mass m kg which is travelling towards influence of friction on the motion of the
her with speed 3 m s–1 and parallel to QP. skier.
After catching the ball, she comes to rest e State briefly, with a reason, how the
answer to part b would be affected
at the point Q. by using such a model. (No further
By modelling the artist and the ball as calculations are expected.) (1)
particles and ignoring her air resistance,
find: ← Section 1.4
a the speed of the artist immediately (4) E 24 A particle P of mass v
before she catches the ball m is attached to P
one end of a light θT
b the value of m (6) inextensible string of O
length a. The other
c the tension in the rope immediately (2) end of the string is A 3ag
after she catches the ball. attached to a point
O. The point A is
← Section 1.4 vertically below O,
and OA = a.
129
Review exercise 1
A The particle issppeerodje√ c _(t3_ea_d_g_)h .oWrizhoenntOallPy A After the string becomes slack, P moves
from A with freely under gravity and is at the same
makes an angle θ with the upward vertical level as O when it is at the point A.
through O and the string is still taut, the
e Explain briefly why the speed of P at A
tension in the string is T and the speed of √ is __ 32_ g_a_ .
P is v, as shown in the diagram. (1)
a Find, in terms of a, g and θ, an (3) The direction of motion of P at A makes
expression for v 2. an angle ϕ with the horizontal.
b Show that T = (1 − 3 cos θ)mg. (3) f Find ϕ. (4)
The string becomes slack when P is at the ← Section 1.5
point B.
E 26 AC
c Find, in terms of a, the vertical height
of B above A. (2)
θ
After the string becomes slack, the O
highest point reached by P is C.
a
d Find, in terms of a, the vertical
height of C above B. (5) B
← Section 1.5 A particle is at the highest point A on
E 25 the outer surface of a fixed smooth
sphere of radius a and centre O. The
lowest point B of the sphere is fixed to a
Oθ T horizontal plane. The particle is projected
uh o,riz√ o _(an_g_ta_) .llTy hfreopmarAticwleitlheasvpeesedthue, where
P sphere
v at the point C, where OC makes an angle
A particle P of mass m is attached to one θ with the upward vertical, as shown in
end of a light inextensible string of length the diagram above.
a. The other end of the string is fixed at a
point O. The particle is held with the string a Find an expression for cos θ in terms (7)
taut and OP horizontal. It is then projected of u, g and a.
vertically downwards with speed u, where
u2 = _ 32 ga. When OP has turned through an The particle strikes the plane with speed
angle θ and the string is still taut, the speed
of P is v and the tension in the string is T, √ __ 9__2a__ g__
as shown in the diagram above. b Find, to the nearest degree, the value
of θ. (7)
a Find an expression for v 2 in terms of a, ← Section 1.5
g and θ. (2) E/P 27 C B
b Find an expression for T in terms of (3) O 120°
m, g and θ.
c Prove that the string becomes slack (2) a
when θ = 210°.
d State, with a reason, whether P would A
complete a vertical circle if the string
were replaced by a light rod. (2) Part of a hollow spherical shell, centre O
and radius a, is removed to form a bowl
130
Review exercise 1
A with a plane circular rim. The bowl is E 30 Three particles of masses 0.1 kg, 0.2 kg
fixed with the circular rim uppermost and and 0.3 kg are placed at the points with
horizontal. The point A is the lowest point position vectors (2i − j) m, (2i + 5j) m and
of the bowl. The point B is on the rim of (4i + 2j) m respectively. Find the position
the bowl and ∠ AOB = 120°, as shown vector of the centre of mass of the
in the diagram above. A smooth small particles. (5)
marble of mass m is placed inside the
bowl at A and given an initial horizontal ← Section 2.2
speed u. The direction of motion of the
marble lies in the vertical plane AOB. The E 31 Three particles of mass 2M, M and
marble stays in contact with the bowl until kM, where k is a constant, are placed at
it reaches B. When the marble reaches B, points with position vectors 6i m, 4j m
its speed is v. and (2i − 2j) m respectively. The centre
of mass of the three particles has
a Find an expression for v 2. (3) position vector (3i + cj) m, where c is
a constant.
b For the case when u2 = 6ga, find the
normal reaction of the bowl on the a Show that k = 3. (4)
marble as the marble reaches B. (3)
b Hence find the value of c. (3)
c Find the least possible value of u for
← Section 2.2
the marble to reach B. (3)
The point C is the other point on the rim E 32 A 20 cm B
of the bowl lying in the vertical plane 10 cm
OAB. 5 cm 3 cm
d Find the value of u which will enable O
the marble to leave the bowl at B
and meet it again at the point C. (7) 6 cm
← Section 1.5 DC
E 28 Three particles of masses 3m, 5m and λm The figure shows a metal plate that is
are placed at the points with coordinates made by removing a circle of centre
(4, 0), (0, −3) and (4, 2) respectively. O and radius 3 cm from a uniform
The centre of mass of the three particles rectangular lamina ABCD, where
is at (2, k). AB = 20 cm and BC = 10 cm. The point
O is 5 cm from both AB and CD, and is
a Show that λ = 2. (4) 6 cm from AD.
b Calculate the value of k. (2) a Calculate, to 3 significant figures, the
distance of the centre of mass of the
← Section 2.2 plate from AD. (6)
E 29 Particles of masses 2M, xM and yM The plate is freely suspended from A and
are placed at points whose coordinates hangs in equilibrium.
are (2, 5), (1, 3) and (3, 1) respectively.
Given that the centre of mass of the three b Calculate, to the nearest degree, the
particles is at the point (2, 4), find the angle between AB and the vertical. (2)
values of x and y. (6)
← Section 2.4, 2.6
← Section 2.2
131
Review exercise 1
E/P 33 A 8a B A uniform lamina ABCD is made by
G X 4a C taking a uniform sheet of metal in the
6a form of a rectangle ABED, with AB = 3a
and AD = 2a, and removing the triangle
BCE, where C lies on DE and CE = a, as
shown in the figure.
a Find the distance of the centre of mass
of the lamina from AD. (5)
ED
The figure shows a uniform lamina The lamina has mass M. A particle
ABCDE such that ABDE is a rectangle, of mass m is attached to the lamina at
BC = CD, AB = 8a and AE = 6a. The point B. When the loaded lamina is freely
X is the midpoint of BD and XC = 4a. The suspended from the midpoint of AB, it
centre of mass of the lamina is at G. hangs in equilibrium with AB horizontal.
a Show that GX = _ 4154_ a.
(6) b Find m in terms of M. (4)
The mass of the lamina is M. A particle ← Sections 2.4, 2.6
of mass λM is attached to the lamina at C.
The lamina is suspended from B and hangs E/P 36
freely under gravity with AB horizontal.
CA
b Find the value of λ. (3) 10 cm
← Sections 2.4, 2.6 D
E/P 34 A uniform square plate ABCD has mass 20 cm
10M and the length of a side of the plate B
is 2l. Particles of masses M, 2M, 3M
and 4M are attached at A, B, C and D The figure shows a decoration which is
respectively. Calculate, in terms of l, the made by cutting two circular discs from
distance of the centre of mass of the a sheet of uniform card. The discs are
loaded plate from: joined so that they touch at a point D on
the circumference of both discs. The discs
a AB (7) are coplanar and have centres A and B
with radii 10 cm and 20 cm respectively.
b BC (3)
The loaded plate is freely suspended from a Find the distance of the centre of mass
the vertex D and hangs in equilibrium. of the decoration from B. (5)
c Calculate, to the nearest degree, the
angle made by DA with the downward
vertical.
← Sections 2.4, 2.6
E/P 35 A 3a B The point C lies on the circumference
of the smaller disc and ∠ CAB is a right
2a angle. The decoration is freely suspended
from C and hangs in equilibrium.
b Find, in degrees to one decimal (4)
place, the angle between AB and
the vertical.
D CaE ← Sections 2.4, 2.6
132
Review exercise 1
E 37 A B and radius 8 cm, from a uniform circular
lamina, of centre O and radius 24 cm.
10 cm The point X lies on the diameter AOB of
L the lamina and AX = 16 cm. The centre
of mass of T lies at the point G.
a Find AG. (4)
D 10 cm C The template T is free to rotate
about a smooth fixed horizontal axis,
A uniform lamina L is formed by taking perpendicular to the plane of T, which
a uniform square sheet of material ABCD passes through the midpoint of OB. A
of side 10 cm and removing a semicircle small stud of mass _41 m is fixed at B, and T
with diameter AB from the square, as and the stud are in equilibrium with AB
shown in the figure. horizontal.
a Find, in cm to 2 decimal places, the
b Modelling the stud as a particle, find(3)
distance of the centre of mass of the the mass of T in terms of m.
lamina from the midpoint of AB. (7)
← Sections 2.4, 2.6
[The centre of mass of a uniform E/P 39 y
semicircular lamina, radius a, is at a
distance _ 34_πa_ from the centre of the A 4m
bounding diameter.]
4
The lamina is freely suspended from D
and hangs at rest. B
O 9 6m x
b Find, in degrees to one decimal
4
place, the angle between CD and (4) C 2m
the vertical.
← Sections 2.4, 2.6 The figure shows a triangular lamina
ABC. The coordinates of A, B and C
E/P 38 are (0, 4), (9, 0) and (0, −4) respectively.
Particles of masses 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
8 cm 24 cm the lamina. (4)
A XO B
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, λ).
The figure shows a template T made by b Show that k = 6. (3)
removing a circular disc, of centre X
c Calculate the value of λ. (2)
133
Review exercise 1
The combined system is freely suspended 6 cm, 8 cm and 10 cm respectively. Given
from O and hangs at rest. that the wire is uniform, find the distance
of the centre of mass of the frame from:
d Calculate, in degrees to one decimal
place, the angle between AC and the a AB (4)
(3)
vertical. b BC (2)
← Sections 2.4, 2.6 The frame is suspended from the corner
A and hangs in equilibrium.
E/P 40 4 mA O B m
c Find, to the nearest degree, the acute
angle made by AB with the downward
2a vertical. (3)
← Sections 2.5, 2.7
D 5a C 2 m E 42 B C (6 m)
A loaded plate L is modelled as a uniform 2a
rectangular lamina ABCD and three
particles. The sides CD and AD of the A 3a D (2 m)
lamina have lengths 5a and 2a respectively
and the mass of the lamina is 3m. The The figure shows four uniform rods
three particles have masses 4m, m and 2m joined to form a rectangular framework
and are attached at the points A, B and C ABCD, where AB = CD = 2a and
respectively, as shown in the figure. BC = AD = 3a. Each rod has mass
m. Particles of masses 6m and 2m are
a Show that the distance of the centre attached to the framework at points C
of mass of L from AD is 2.25a. (3) and D respectively.
b Find the distance of the centre of (2)
mass of L from AB.
The point O is the midpoint of AB. The a Find the distance of the centre of mass
loaded plate L is freely suspended from O of the loaded framework from:
and hangs at rest under gravity.
c Find, to the nearest degree, the size i AB ii AD (7)
of the angle that AB makes with the
horizontal. (3) The loaded framework is freely suspended
from B and hangs in equilibrium.
A horizontal force of magnitude P is
applied at C in the direction CD. The b Find the angle which BC makes with
the vertical. (3)
loaded plate L remains suspended from
O and rests in equilibrium with AB ← Sections 2.5, 2.7
horizontal and C vertically below B. E 43 Three uniform rods AB, BC and CA of
d Show that P = _ 54 mg. masses 2l,ml ,amndaln√d _2_ 3rmespreescptievcetliyv.eTlyhhearvoeds
(4) lengths
e Find the magnitude of the force on (4) are rigidly joined to form a right-angled
L at O.
triangular framework.
← Sections 2.4, 2.6
a Calculate, in terms of l, the distance of
E 41 A triangular frame ABC is made by the centre of mass of the framework
bending a piece of wire of length 24 cm, from:
so that AB, BC and AC are of lengths i BC ii AB (4)
134
Review exercise 1
b Calculate the angle, to the nearest AB, BC and AD are made from wire of
degree, that BC makes with the mass 0.01M kg per cm. CD is made from
vertical when the framework is freely wire of mass 0.015M kg per cm.
suspended from the point B. (2) a Find the distance of the centre of
← Sections 2.5, 2.7 mass of the framework from AB. (6)
E 44 A B The framework has mass M. A particle
of mass kM is attached to the framework
CD at C. When the framework is freely
suspended from the midpoint of BC, the
framework hangs in equilibrium with BC
horizontal.
b Find the value of k. (5)
← Section 2.8
G FE E/P 46 A shop sign of weight W N, that can be
modelled as a lamina and is shown in the
The figure shows the L-shaped lamina diagram below. The sign is suspended by
that is made from the rectangle ABFG two ropes that can be modelled as light
and the square CDEF. AB has length d inelastic strings.
and AG has length 3d. CD and DE have
length d. The density of the square is T1 T2
three times that of the rectangle. B
A 80 cm
Find, in terms of d, the distance of the 40 cm 80 cm
centre of mass of the lamina from:
E
a AG (4)
b GE (2)
The lamina is suspended from the point D 40 cm C
A and hangs freely in equilibrium.
c Find, to the nearest degree, the angle
that AB makes with the vertical. (3)
← Section 2.8
E/P 45 B C a Find the distance of the centre of mass
of the lamina from AE. (4)
b Find the tension in each string. (4)
4 cm 5 cm The rope attached at A will snap when
the tension in it exceeds 10W N.
A 4 cm D
The rope attached at B will snap when the
A framework is made from thin tension in it exceeds 8W N.
uniform wire of total length 20 cm. The
framework is in the shape of a trapezium A particle of weight kW is attached to
ABCD, where AB = AD = 4 cm, the sign at C. Given that neither rope
CD = 5 cm and AB is perpendicular to breaks,
BC and AD as shown in the diagram.
c find the largest possible value of k. (4)
← Section 2.8
135
Review exercise 1
A 47 A uniform solid is formed by rotating the A 50 y
E reeqguiaotnioenncylo=se√d _x_ ,btehtewxee-anxtihseacnudrtvheewliinthe
E/P
x = 4, through one complete revolution
about the x-axis. Find the distance of the
centre of mass of the solid from the R
origin O. (5) O 1 2x
← Section 3.1 The shaded region R is bounded by the
E 48 a Use integration to show that the centre caunrdvtehwe iltihneesqxua=ti1onanyd=x_ 2=1_ x_ 22 , ,tahse x-axis
of mass of a uniform semicircular shown
lamina, of radius a, is a distance
_34_πa_ from the midpoint of its straight above. The unit of length on each axis
is 1 m. A uniform solid S has the shape
made by rotating R through 360° about
edge, O. (4) the x-axis.
A semicircular lamina, of radius b with a Show that the centre of mass of S
O as the midpoint of its straight edge, is is _ 72 m from its larger plane face. (8)
removed from the first lamina.
b Show that the centre of mass of the T
resulting lamina is at a distance x¯ from H
O, where (6) 1 m
x ¯ = _34_π_ _(a_2__(+a_a_+ b__b+_) _b_2_) 2
c Hence find the position of the centre
of mass of a uniform semicircular arc
of radius a. (2) S
← Section 3.2
E/P 49 A uniform triangular lamina ABC has A sporting trophy T is a uniform solid
∠ ABC = 90° and AB = c. hemisphere H joined to the solid S. The
hemisphere has radius _12 m and its plane
a Using calculus, prove that the centre of face coincides with the larger plane face
mass of the lamina is at a distance _13 c of S, as shown above. Both H and S are
from BC. (6) made of the same material.
The diagram shows a uniform lamina in
which PQ = PS = 2a, SR = a.
The centre of mass of the lamina is G. b Find the distance of the centre of (6)
mass of T from its plane face.
SR
← Sections 3.1, 3.2
PQ
b Show that the distance of G from (5)
PS is _ 79 a. (5)
c Find the distance of G from PQ.
← Section 3.2
136
Review exercise 1
A 51 a Show, by integration, that the centre of A 53 y
mass of a uniform solid hemisphere,
E/P of radius R, is at a distance _38 R from its E/P
plane face.
The diagram shows a uniform solid top R
made from a right circular cone of base
radius a and height ka and a hemisphere Ox
of radius a. The circular plane faces of
the cone and hemisphere are coincident. The shaded region R is bounded by part
of the curve with equation y = _ 12 (x − 2)2,
V (6) the x-axis and the y-axis, as shown above.
The unit of length on both axes is 1 cm.
ka A uniform solid S is made by rotating
aa R through 360° about the x-axis. Using
integration,
a calculate the volume of the solid S,
leaving your answer in terms of π (4)
b Show that the distance of the centre of b show that the centre of mass of S (7)
mass of the top from the vertex V of is _31 cm from its plane face.
the cone is
8 cm
_ (3_k_2_4_+(k_8_+ k__2+_) _3_)a_ (5) 4 cm C S
The manufacturer requires the top to AB
have its centre of mass situated at the
centre of the coincident plane faces. A tool is modelled as having two
components, a solid uniform cylinder
c Find the value of k for this (5) C and the solid S. The diameter of C is
requirement. 4 cm and the length of C is 8 cm. One
end of C coincides with the plane face
← Section 3.2 of S. The components are made of
different materials. The weight of C is
E/P 52 A bowl consists of a uniform solid metal 10W newtons and the weight of S is 2W
hemisphere, of radius a and centre newtons. The tool lies in equilibrium with
O, from which is removed the solid its axis of symmetry horizontal on two
hemisphere of radius _21 a with the same smooth supports A and B, which are at
centre O. the ends of the cylinder, as shown above.
a Show that the distance of the centre of c Find the magnitude of the force of (5)
mass of the bowl from O is _ 14_152_ a. (6) the support A on the tool.
The bowl is fixed with its plane face ← Sections 3.2, 3.4
uppermost and horizontal. It is now filled
with liquid. The mass of the bowl is M
and the mass of the liquid is kM, where k
is a constant. Given that the distance of
the centre of mass of the bowl and liquid
together from O is _ 14_78 a,
b find the value of k. (6)
← Section 3.3
137
Review exercise 1
A 54 r A b Show that the distance of the centre of
h mass of the plate from A is _ 53_a_ .
E (6)
1 h The plate, which is of mass M, has a
2 particle of mass M attached at B. The
loaded plate is suspended from C and
hangs in equilibrium.
c Prove that in this position CB makes an
angle of arctan _19 with the vertical. (4)
← Sections 3.2, 3.4
An ornament S is formed by removing a E/P 56 A closed container C consists of a thin
solid right circular cone, of radius r and uniform hollow hemispherical bowl of
height _ 21 h, from a solid uniform cylinder, radius a, together with a lid. The lid is a
of radius r and height h, as shown in the thin uniform circular disc, also of radius a.
diagram. The centre O of the disc coincides with the
centre of the hemispherical bowl. The bowl
a Show that the distance of the centre of and its lid are made of the same material.
mass of S from its plane face is _41 _70 h. (6)
a Show that the centre of mass of C is
at a distance _13 a from O. (4)
The ornament is suspended from a point
on the circular rim of its open end. It The container C has mass M. A particle
hangs in equilibrium with its axis of of mass _21 M is attached to the container
symmetry inclined at an angle α to the at a point P on the circumference of the
horizontal. Given that h = 4r, lid. The container is then placed with a
point of its curved surface in contact with
b find, in degrees to one decimal place, a horizontal plane. The container rests in
the value of α. (4)
equilibrium with P, O and the point of
← Sections 3.2, 3.4 contact in the same vertical plane.
E/P 55 a A uniform triangular lamina XYZ has b Find, to the nearest degree, the
XY = XZ and the perpendicular angle made by the line PO with the
distance of X from YZ is h. Prove, by horizontal. (5)
integration, that the centre of mass of
the lamina is at a distance _2 3_h_ from X. ← Sections 3.2, 3.4
A E/P 57
A
a
BDC θ°
α
A uniform triangular lamina ABC has
AB = AC = 5a, BC = 8a and D is the A uniform solid right circular cone has
centre of mass of the lamina. The base radius a and semi-vertical angle
triangle BCD is removed from the α, where tan α = _ 13 . The cone is freely
lamina, leaving the plate ABDC shown suspended by a string attached at a point
in the diagram. A on the rim of its base, and hangs in
138
Review exercise 1
A equilibrium with its axis of symmetry A The solid S is suspended by two vertical
making an angle of θ ° with the upward strings, one attached to the vertex and
vertical, as shown in the diagram. Find, the other attached to a point on the
to one decimal place, the value of θ. (4) bounding circular base.
← Sections 3.2, 3.4 c Given that S is in equilibrium, with
its axis of symmetry horizontal, find,
E/P 58 A uniform solid hemisphere H has base in terms of H and h, the ratio of the
radius a and the centre of its plane magnitude of the tension in the string
circular face is C. attached to the vertex to that in the
other string.
The plane face of a second hemisphere
K, of radius _2a _ , and made of the same ← Sections 3.2, 3.4
material as H, is stuck to the plane face
of H, so that the centres of the two plane E 60 r
faces coincide at C, to form a uniform
composite body S. h
a Given that the mass of K is M, show O
that the mass of S is 9M, and find, in
terms of a, the distance of the centre
of mass of the body S from C. (5)
A particle P, of mass M, is attached to A child’s toy consists of a uniform
a point on the edge of the circular face solid hemisphere attached to a uniform
of H of the body S. The body S with P solid cylinder. The plane face of the
attached is placed with a point of the hemisphere coincides with the plane face
curved surface of the part H in contact of the cylinder, as shown in the diagram
with a horizontal plane and rests in above. The cylinder and the hemisphere
equilibrium. each have radius r and the height of
the cylinder is h. The material of the
b Find the tangent of the acute hemisphere is six times as dense as the
angle made by the line PC with the material of the cylinder. The toy rests in
horizontal. (5) equilibrium on a horizontal plane with
the cylinder above the hemisphere and
← Sections 3.2, 3.4 the axis of the cylinder vertical.
E 59 a Prove, by integration, that the position a Show that the distance d of the centre
of the centre of mass of a uniform of mass of the toy from its lowest
solid right circular cone is one quarter point O is given by
of the way up the axis from the base.
(8) d = _ h_2 _2+_(h_2_h+ _r_4+_r_)5 _r_ 2 (7)
A solid is formed by removing a solid When the toy is placed with any point
cone of height h and radius a from a solid of the curved surface of the hemisphere
cone of height H and radius a. The axes resting on the plane it will remain in
of the two cones coincide. equilibrium.
b Show that the centre of mass of the b Find h in terms of r. (3)
remaining solid S is a distance
_14 (3H − h) ← Sections 3.3, 3.4
from the vertex of the original cone.
(10)
139
Review exercise 1
A 61 O A The solid is placed with its plane face on
a plane which is inclined at an angle α to
E the horizontal. The plane is sufficiently
rough to prevent S from sliding. Given
3r that S does not topple,
AB b find, to the nearest degree, the (4)
maximum value of α.
← Sections 3.2, 3.5
2r E 63 y Figure 1
A child’s toy consists of a uniform solid 1
hemisphere, of mass M and base radius
r, joined to a uniform solid right circular R
cone of mass m, where 2m , M. The cone
has vertex O, base radius r and height 3r. O πx
Its plane face, with diameter AB, coincides
with the plane face of the hemisphere, as A uniform lamina occupies the region R
shown in the diagram above. bounded by the x-axis and the curve
y = sin x, 0 < x < π, as shown in Figure 1.
a Show that the distance of the centre of a Show, by integration, that the
mass of the toy from AB is y-coordinate of the centre of mass of
the lamina is _π8 _
_ 38_((M_M__−_+_2m_m_)_) r (6)
(5)
The toy is placed with OA on a horizontal R
surface. The toy is released from rest and Figure 2
does not remain in equilibrium.
b Show that M . 26m (4) θ°
← Sections 3.3, 3.4 A uniform prism S has cross section R.
The prism is placed with its rectangular
E/P 62 y face on a table which is inclined at an
angle θ to the horizontal. The cross
R section R lies in a vertical plane as shown
in Figure 2. The table is sufficiently rough
Or x to prevent S sliding. Given that S does
not topple,
The diagram shows the region R bounded b find the largest possible value of θ. (3)
by the curve with equation y2 = rx, where
r is a positive constant, the x-axis and the ← Sections 3.2, 3.5
line x = r. A uniform solid of revolution
S is formed by rotating R through one
complete revolution about the x-axis.
a Show that the distance of the centre of
mass of S from O is _ 23 r. (6)
140
Review exercise 1
A 64 Oa A 6a O 2a
E/P
3 a
2
2a 6a
A uniform solid cylinder has radius 2a 2a
and height _ 32 a. A hemisphere of radius a
is removed from the cylinder. The plane The bowl B is fixed to a plane face of a
face of the hemisphere coincides with the uniform solid cylinder made from the
upper plane face of the cylinder, and the same material as B. The cylinder has
centre O of the hemisphere is also the radius 2a and height 6a and the combined
centre of this plane face, as shown in the solid S has an axis of symmetry which
diagram above. The remaining solid is S. passes through O, as shown.
a Find the distance of the centre of (6) b Show that the centre of mass of S
mass of S from O.
is _ 26_011_ a from O. (6)
The lower plane face of S rests in
equilibrium on a desk lid which is The plane surface of the cylindrical base
inclined at an angle θ to the horizontal. of S is placed on a rough plane inclined
Assuming that the lid is sufficiently rough at 12° to the horizontal. The plane is
to prevent S from slipping, and that S is sufficiently rough to prevent slipping.
on the point of toppling when θ = α,
c Determine whether or not S will (4)
b find the value of α. (3) topple.
Given instead that the coefficient of ← Sections 3.2, 3.5
friction between S and the lid is 0.8, and
that S is on the point of sliding down the E/P 66
lid when θ = β,
c find the value of β. (3)
← Sections 3.2, 3.5 O
E/P 65 The diagram shows a cross section of a
solid formed by the removal of a right
6a O 2a circular cone, of base radius a and height
a, from a uniform solid hemisphere of
A uniform solid hemisphere, of radius 6a base radius a. The plane bases of the
and centre O, has a solid hemisphere of cone and the hemisphere are coincident,
radius 2a, and centre O, removed to form both having centre O.
a bowl B as shown above. a Show that G, the centre of mass of
the solid, is at a distance _a2 _ from O. (5)
a Show that the centre of mass of B
is _ 310_3 a from O. (6)
141
Review exercise 1
A A The cylinder has radius 4r and height r.
OG The cone has vertex O, base radius 3r and
height h.
a Show that the distance from O of the
centre of mass of B is
_ 3_2_r 4_2(_+1_66_r_4 +_r h_3_+h_)_9 _h_2 (6)
α The bollard is placed on a rough plane
which is inclined at an angle α to the
The second diagram shows a cross section horizontal. The circular base of B is in
of the solid resting in equilibrium with contact with the inclined plane. Given
a point of its curved surface in contact that h = 4r and that B is on the point of
with a rough inclined plane of inclination toppling,
α. Given that O and G are in the same b find α, to the nearest degree. (4)
vertical plane through a line of greatest ← Sections 3.2, 3.5
slope of the inclined plane, and that OG E 68
is horizontal,
b show that α = _ π6_ (4)
Given that α = _ π6_ , h
c find the smallest possible value of
the coefficient of friction between
the solid and the plane. (4) 2r O
← Section 3.5 G
h
E 67 O
h r
3r A model tree is made by joining a
r uniform solid cylinder to a uniform solid
cone made of the same material. The
4r centre O of the base of the cone is also
the centre of one end of the cylinder, as
An experimental plastic traffic bollard shown in the diagram. The radius of the
B is made by joining a uniform solid cylinder is r and the radius of the base
cylinder to a uniform solid right circular of the cone is 2r. The height of the cone
cone of the same density. They are joined and the height of the cylinder are each h.
to form a symmetrical solid, in such a The centre of mass of the model is at the
way that the centre of the plane face of point G.
the cone coincides with the centre of a Show that OG = _ 11_4 h.
one of the plane faces of the cylinder, as (8)
shown in the diagram.
142
Review exercise 1
A The model stands on a desk top with its A 70 4r
plane face in contact with the desk top.
The desk top is tilted until it makes an E/P
angle α with the horizontal, where
tan α = _ 276_ . The desk top is rough enough O
to prevent slipping and the model is
about to topple. 4r
A 3r
b Find r in terms of h. (4) A toy is formed by joining a uniform
solid right circular cone, of base radius 3r
← Sections 3.2, 3.5 and height 4r, to a uniform solid cylinder,
also of radius 3r and height 4r. The cone
E/P 69 H and cylinder are made from different
materials, and the density of the cone is
Ch three times the density of the cylinder.
The plane face of the cone coincides with
rO a plane face of the cylinder, as shown in
the diagram. The centre of this plane face
A body consists of a uniform solid is O.
circular cylinder C, together with a
uniform solid hemisphere H which is a Find the distance of the centre of (6)
attached to C. The plane face of H mass of the toy from O.
coincides with the upper plane face of C,
as shown in the diagram. The cylinder C The point A lies on the edge of the plane
has base radius r, height h and mass 3M. face of the cylinder which forms the base
The mass of H is 2M. The Sections O is of the toy. The toy is suspended from A
the centre of the base of C. and hangs in equilibrium.
a Show that the distance of the centre of b Find, in degrees to one decimal
mass of the body from O is place, the angle between the axis
_1_4_h2_0_+_ 3__r of symmetry of the toy and the (4)
(5) vertical.
The body is placed with its plane face on a The toy is placed with the curved surface
rough plane which is inclined at an angle α of the cone on horizontal ground.
to the horizontal, where tan α = _ 43 .
The plane is sufficiently rough to prevent c Determine whether the toy will (4)
slipping. Given that the Sections is on the topple.
point of toppling,
← Sections 3.3, 3.4, 3.5
b find h in terms of r. (4) E 71 A non-uniform rod is 1.5 m long. At any
distance h cm from its base, the mass per
← Sections 3.3, 3.5 unit length of the rod is given by
m = _( 1__+__h_)3_( 2__+__h_) kg m−1. Find:
a the mass of the rod (4)
b the distance of the centre of mass (4)
of the rod from its base.
← Sections 3.3, 3.4, 3.5
143
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