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Review exercise 1
1 Changing the units:
diameter 7cm radius ( ) 3.5cm 0.035m
1000 2angular speed ( ) 1000 revolutions per minute radians per second
60
r
Using v r gives:
v 0.035
1000 260
3.67 ms-1 (3 s.f.)
So the speed is 3.67 m s–1 (3 s.f.)
2 a Let the tension in the string be T and let the string make an angle with the vertical.
Let OP r, then r 1.5sin
Acceleration towards the centre of the circle = r
2 1.5sin 2.72
Force towards the centre of the circle = T sin
So using F ma gives:
T sin 0.51.5sin 2.72
T 0.51.5 2.72 5.4675 5.5N (2 s.f.)
b Resolving the forces vertically R() and substituting for T gives:
T cos - 0.5g 0
T cos 0.5g
cos 0.5g
5.4675 0.8962
So cos-1 0.8962 26° (to the nearest degree)
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3 a These are the forces acting on P.
R( ) : T cos60° - mg 0
T mg
cos60° 2mg
b R( ¬) : T sin60° ma mr 2 using F ma and r is the radius of the circular motion
As r Lsin60° this gives:
2
2
sin 60 sin 60
2 2 substituting result for from part
2So
T mL
T mg gT
mL mL L
g
L
° °
a
4 The forces acting on the car are its weight mg and the normal reaction R.
R( ) : Rcos10° - mg 0
R mg
cos10°
Using F ma horizontally gives:
2 2
2
2
18sin10
18sin10 substituting for
cos10
18187.4995 190 m (2 s.f.)
tan10
mv mR
r r
mg mR
r
rg
°
°
°
°
…
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5 The forces acting on the cyclist and bicycle are their weight mg, the normal reaction R and the
friction acting down the slope.
R( ) : Rcos25° - F sin25° - mg 0
If m is the coefficient of friction between the cycle’s tyres and the track, then the maximum friction
for which the tyres do not slip is F mR 0.6R . Substituting for F gives:
cos25 0.6 sin 25 0
cos25 0.6sin 25
R R mg
mgR
° - ° -
°- °
(1)
2 2
2
2
R( ) : sin 25 cos25 using and
sin 25 0.6 cos25 as 40 and 0.6 at maximum speed40
40(sin 25 0.6cos25 )
mv vR F F ma a
r r
mvR R r F R
mvR
¬ ° °
° °
° °
(2)
Using equations (1) and (2) gives:
mv2
40(sin25° 0.6cos25°)
mg
cos25° - 0.6sin25°
v2 40g(sin25° 0.6cos 25°)
cos25° - 0.6sin25° 580.37 (2 d.p.)
v 24ms-1
(2 s.f.)
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6 a The forces acting on the ring are the the weight of the ring and tensions of equal magnitude T
along each section of string.
Note that is an an angle in a 3-4-5 triangle, with cos
3
5, sin
4
5
R( ) : T sin - mg 0
T mg
sin
5mg
4
b R( ¬) : T T cos
mv2
r using F ma and a
v2
r
T 13
5
æ
èçö
ø÷
mv2
3l as r 3l and cos
3
5
5mg
4
8
5
mv2
3l substituting for T
v2 6gl
v 6gl
c If the ring is firmly attached to the string, then the tensions in each section of the string cannot
be assumed to have the same magnitude. So the approach adopted for parts a and b would not
be valid.
7 a The forces acting on the metal ball are the weight of the ball and tension along the string.
R( ) : 3mg cosa - mg 0
cosa mg
3mg
1
3
a 70.5° (3 s.f.)
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7 b R( ¬) : 3mg sina mr2gk using F ma and a r 2
Let the length of the string be l, then from DAOB is is clear that sina
r
l
So 3mgr
l mr2gk
l 3
2k
8 a The forces acting on the particle are its weight and the normal reaction.
Let be the angle between the normal reaction and the horizontal.
Then sin
12r
r
1
2 30°
R( ) : Rsin30° - mg 0
R mg
sin30° 2mg
b R(®) : Rcos30° mx 2 using F ma and a r 2
Using the result from part a and as x r cos30° this gives:
2mg mr 2
2g
r
Time to complete one revolution
2
2r
2g
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9 a Let T1 be the tension in AP and T2 be the tension in BP. The forces acting on P are the tensions in
the two strings and its weight.
From the diagram, it can be seen that the equilateral triangle APB can be divided into two
right-angled triangles, where:
tan 60
2
3tan 60
2 2
r
h
h hr
°
°
b Resolving the forces from the diagram in part a
1 2
1 2
R( ) : cos 60 cos 60 0
2
T T mg
T T mg
° - ° -
- (1)
2 2
1 2
2
1 2
2
1 2
R( ) : sin 60 sin 60 using and
3 3 3 3 as from part
2 2 2 2
T T mr F ma a r
T T m h r h
T T mh
¬ ° °
a
(2)
Adding equations (1) and (2) gives: 2T
1 2mg mh 2 T
1 mg
1
2mh 2
Substituting for T1 in equation (2) gives: T
2
1
2mh 2 - mg
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9 c Both strings are taut, therefore T1 > 0 and T2 > 0. From part b, T1 > 0 for all values of
From the formula for T2 in part b, the condition that T
2> 0 >
2g
h
As T 2
, 2T
So >2g
h
2T
>2g
hT < 2
h
2g as T > 0 and
h
2g< 0
T < 2h
g as required
10 a The forces acting on the ring are the the weight of the ring and, as the ring is threaded on the
strong, tensions of equal magnitude T along each section of string.
R( ) : cos 0
cos
T mg
mgT
-
(1)
2 2
2
2
R( ) : sin using and
(1 sin ) substituting for from equation cos
sin(1 sin ) tan
cos cos
T T mr F ma a r
mgmr T
mgmh mh
¬
(1)
2
2
as tan by basic trigonometry
1 sin
sin
r h
g
h
æ ö ç ÷è ø
b From part a,
2 g
h
1 sinsin
æ
èçö
ø÷
g
h
1
sin1
æ
èçö
ø÷
As sin <1 it follows that
1
sin>1
Therefore 2 >g
h 2
>2g
h
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10 c Given 3g
h , then
1 sin 3
sin
g g
h h
æ ö ç ÷è ø
1 31 sin 3sin sin and so cos
2 2
2 2 3From equation ,
cos 33
mgT mg mg
(1)
11 Let a be the angle between the slant side and the axis of the cone, r the radius of the horizontal circle
with centre C and h the height of C above V. The forces acting on the particle are its weight and the
normal reaction.
From the diagram, tana
3a
4a
3
4 and tana
r
h
R( ) : sin 0
sin
R mg
R mg
aa
-
(1)
2 2R( ) : cos using and
8 8cos using
9 9
R mr F ma a r
mr g gR
a a
a
a
¬
(2)
Dividing equation (1) by equation (2) gives:
8 9tan
9 8
3 9 3 using tan
4 8 4
3
2
3 4As tan ,
4 3
4 3So 2
3 2
mrg amg
a r
a
r
ar
rh r
h
ah a
a
a
a
The height of C above V is 2a.
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12 a Let the tension in the string be T.
2 2R( ) : cos30 using and
cos30 2 cos30 using and 2 cos30 by trigonometry3 3
2 as required
3
T mr F ma a r
kg kgT m a r a
a a
kmgT
¬ °
° ° °
b Let the normal reaction be R and resolve vertically.
R( ) : R T sin30° - mg 0
R mg -2kmg
3
1
2 mg 1-
k
3
æ
èçö
ø÷
c Using the result from part b, as R > 0,1-
k
3> 0 k < 3
d Let the new tension be ¢T and let AP make an angle with the horizontal. So PX 2acos .
R( ¬) : ¢T cos mr 2 us ing F ma and a r 2
¢T cos m2acos2g
a using
2g
a and r 2acos
¢T 4mg
R( ) : ¢T sin - mg 0
sin mg
¢T
mg
4mg
1
4
Hence AX 2asin 2a 1
4
a
2
AO 2asin30° 2a 1
2 a
Therefore AX AO
2
So the point X is the midpoint of AX.
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13 a Let the tension in AP be T and in BP be S. Let ÐBAP . As the triangle APB is an equilateral
triangle ÐBAP ÐPBA .
AO
3
2l
1
2
3l
4, so from the right-angled triangle APO, cos
3l
4l
3
4
R( ) : cos cos 0
4
cos 3
T S mg
mg mgT S
- -
- (1)
2 2
22
R( ) : sin sin using and
as sin sin
T S mr F ma a r
mr rT S ml
l
¬
(2)
Adding equations (1) and (2) gives:
2T
4mg
3 ml 2 T
1
6m(4g 3l 2 )
1
6m(3l 2 4g) as required
b Subtracting equation (1) from equation (2) gives:
2S ml 2 -
4mg
3 S
1
6m(3l 2 - 4g)
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13 c From part b), 23 4
6BP
mT l g -
Since BP is taut, 0BPT �
2
3 4 06
ml g - �
23 4 0l g - �
23 4l g �
2 4
3
g
l �
14 a The forces acting on the particle are its weight, the normal reaction and friction.
R( ) : R- mg 0 R mg
R( ¬) : F mr 2 us ing F ma and a r 2
F 4ma 2
3 a s r
4
3a
As P remains at rest F Rm�
2
2
2
4 3So
3 5
4 3
3 5
9 as required
20
maR
ma mg
g
a
�
�
�
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14 b The forces now acting on the particle are its weight, the normal reaction, friction and the tension in
the elastic string.
Find T using Hooke’s law (Further Mechanics 1, Chapter 3): T
lx
l,
where l is the modulus of elesticity, x is the extension of the string and l is its natural length
So in this case, T
2mg
a
a
3
2mg
3
R( ¬) : T F mr 2 using F ma and a r 2
2 3
4ma
2mg
3 F
æ
èçö
ø÷
Note that the frictional force can act away from O against the pull of the elastic string, or
towards O through the force of the acceleration of the particle generated by the circular motion. As
P remains at rest R F Rm m- � � (where 0.6R mgm from part a).
So, from the equation for 2 , it is maximum when F mR 3mg
5
So max
2 3
4ma
2mg
3
3mg
5
æ
èçö
ø÷
3
4ma
19mg
15
19g
20a
Similarly 2 is minimum when F -mR -3mg
5
So min
2 3
4ma
2mg
3-
3mg
5
æ
èçö
ø÷
3
4ma
mg
15
g
20a
15 a i AP
x
cos, so extension in the string
x
cos- x
Using Hooke’s law: 10 10
10 cos cos
g x gT x g
x æ ö - -ç ÷è ø
(1)
2R( ) : cos 2 0 cos
Substituting for cos into equation gives:
5 10
102.5 N
4
gT g
T
T T g
gT g
-
-
(1)
ii
cos 2g
T
2g
2.5g 0.8
So arccos0.8 36.9° (3 s.f.)
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15 b
2 22R( ) : sin using and
v vT F ma a
r r¬
As cos
4
5, APO is a 3-4-5 right-angled triangle, with
sin
3
5 and tan
r
x
3
4 r
3
4x
So T sin 2v2
r
2.5g 3
5
4
3x 2v2
v2 9gx
16
v 3
4gx
16 a Let u be the speed of P when the string is horizontal.
Applying the work-energy principle, the sum of the particle’s kinetic and gravitational potential
energy remains constant, so when it reaches the horizontal the loss of kinetic energy = the gain in
potential energy.
Take the starting point as the zero level for potential energy.
So 1
2m
5gl
2-
1
2mu
2 mgl
u2 5gl
2- 2gl
gl
2
u gl
2
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16 b Let the particle move in a semi-circle about B with radius r.
Taking the line AB as the zero level for potential energy, then applying conservation of energy at
the highest point of the semi-circle:
1
2m(u
2 - v2) mgr
v2 u2 - 2gr (1)
Resolving the vertical forces at the highest point:
R( ¯) : T mg mv2
r us ing F ma and a
v2
r
T m(u
2 - 2gr )
r- mg substituting for v
2 using equation (1)
T mu
2
r- 3mg
T mgl
2r- 3mg s ubstituting for u2 using result from part a
As the string does not go slack T > 0, so
mgl
2r- 3mg > 0
mgl > 6mgr
r <l
6
As AB = l – r, this shows that AB >
5l
6
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17 a Let the speed of the particle be v when it makes an angle with the downward vertical.
Take the horizontal through B as the zero level for potential energy. Using conservation of energy:
2
2
2
1(3 ) (1 cos )
2
3 2 (1 cos )
2 cos
m gl v mgl
v gl gl
v gl gl
- -
- -
(1)
Resolving along the string:
2 2
2
R( ) : cos using and and
2 coscos substituting for using equation
3 cos (1 3cos )
mv vT mg F ma a r l
l r
mgl glT mg v
l
T mg mg mg
-
(1)
տ
as required
b The instant the string becomes slack, T = 0.
Using the expression for T from part a, this occurs when:
1 3cos 0 cos -
1
3 Substituting for cos in equation (1) gives:
v2 gl 2gl -1
3
gl
3
v gl
3
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17 c The particle now moves as a projectile under gravity. The maximum height is achieved when the
vertical component of the velocity is zero.
vy vsin(180°- )
gl
3sin
If 2< < and cos -
1
3, then sin
2 2
3
So vy
2 2
3
gl
3
Considering the particle’s vertical motion, u v
y, v 0, a -g, s h, where h is the height
above the point the string becomes slack.
Using v2 u2 - 2gh
h v
y
2
2g
8
9
gl
3
1
2g
4l
27
The height at which string becomes slack l l cos(180° - ) l(1- cos )
4l
3
So if H is the maximum height above the level of B reached by P
H
4l
3
4l
27
40l
27
18 a Take the horizontal through l as the zero level for potential energy. When angle a , the particle
is momentarily at rest. Using conservation of energy, with the loss of kinetic energy equal to the
gain in potential energy:
1
2mu
2 mgl(1- cosa ) mgl
3 a s cosa=
2
3
u2 2
3gl
u 2gl
3
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18 b Let the speed of the particle at be v.
Resolving along the string gives:
2 2
2
R( ) : cos using and and
cos
mv vT mg F ma a r l
l r
mvT mg
l
-
տ
Applying conservation of energy:
1
2mu2 -
1
2mv2 mgl(1- cos )
v2 u
2 - 2gl(1- cos ) 2gl
3- 2gl(1- cos ) using the result from part a
v2 2glcos -4gl
3
Substituting for v2 in the equation for T gives:
T mg cos 2mg cos -4mg
3
3mg cos -4mg
3
mg
3(9cos - 4)
c Maximum value of T is when cos 1
So T
max
5mg
3
Minimum value of T is when cos
2
3
min
2So
3
2 5Hence
3 3
mgT
mg mgT
� �
19 a The lengths on the diagram can be deduced from the question:
OA OB 0.8m AN 0.2m ON OA- AN 0.6m
cos
ON
OB
0.6
0.8
3
4
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19 b As the particle leaves the hemisphere at B the normal reaction R = 0.
So resolving along the radius OB gives:
2 2
2
3R( ) : cos using and and 0.8
4 0.8
30.8 0.6 5.88
4
mg mv vmg F ma a r
r
gv g
ւ
c Take the horizontal through A as the zero level for potential energy. Using conservation of energy,
the gain of kinetic energy at B equals the loss in potential energy:
1
2m5.88-
1
2mu
2 mg 0.2
u2 5.88- 0.4g 0.2g
So u 1.4
20 a This is a diagram of the problem.
Take the horizontal through A as the zero level for potential energy. Using conservation of energy,
the gain of kinetic energy at P equals the loss in potential energy:
1
2mv
2 mg(acosa - acos )
v2 2ga(cosa - cos )
b Resolving along the radius OP:
2
22
R( ) : cos
At the point when loses contact with the sphere 0
cos 2 (cos cos ) substituting for from part
33cos 2cos
2
1cos , so 60 or r
2 3
mvmg R
a
P R
mvmg gm v
a
a
a
-
-
°
a
ւ
adians
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20 c Let the speed of P when it hits the table be w. Then taking the horizontal through A as the zero
level for potential energy and using conservation of energy, the gain of kinetic energy at B equals
the loss in potential energy:
1
2mw
2 mg(a acosa )
w2 2ga 1
3
4
æ
èçö
ø÷
7ga
2
So w 7ga
2ms-1
There are alternative ways to calculate w that use projectiles, but the approach shown above is the
shortest method.
21 a Let the speed of P at point B be v. Then taking the horizontal through A as the zero level for
potential energy and using conservation of energy, the loss of kinetic energy at B equals the
gain in potential energy:
1
2mu2 -
1
2mv2 mga
1
2m
7ag
2- mga
1
2mv
2
v2 3
2ga
R( ¬) : R mv2
a
R m
a
3
2ga
3
2mg
b Let the speed of P at point C be w. Then taking the horizontal through A as the zero level for
potential energy and using conservation of energy, the loss of kinetic energy at C equals the
gain in potential energy:
21 7 1(1 cos )
2 2 2
gam mw mga - (1)
Resolving along the radius OP
2
2
2
R( ) : cos
At the point when loses contact with the sphere 0
So cos
Substituting for in equation gives:
7 cos(1 cos )
4 2
3 3cos
2 4
1cos , hence 60
2
mwmg R
a
P R
ga w
w
mga mgamga
-
°
(1)
ւ
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21 c Consider the motion of the projectile P as it moves from C to A in the horizontal direction.
The horizontal distance from C to A asin60°
The horizontal component of the speed at C, wx wcos60°
So the time t that P takes to travel from C to A is given by t asin60°wcos60°
From part b, w2 ag cos60° w ag
2
So t asin60°cos60°
ag
2 a 3
2
ag
6a
g
22 a Let the speed of the trapeze artist at the lowest point of her path be v. Then taking the horizontal
through A as the zero level for potential energy and using conservation of energy, the gain in
kinetic energy at this point equals the loss in potential energy:
1
2mv2 -
1
2m( 15)2 mg 5(1- cos60°)
v2 155g 64
v 8ms-1
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22 b Let the velocity after catching the ball be w. Then the loss in kinetic energy at point Q where the
trapeze artist becomes momentarily stationery equals the gain in potential energy from the lowest
point of the trapeze artist’s path.
1
2(60 m)w2 - 0 (60 m)g5(1- cos60°)
w2 5g 49, so w 7ms-1
So at the instant just prior to catching the ball, the trapeze artist is travelling at 8 m s–1 (part a) and
the ball is travelling at 3 m s–1 in the opposite direction, and immmediately after catching the ball
she is travelling at 7 m s–1. Using conservation of linear momentum at the instant when she catches
the ball, this gives:
60 8 ( 3 ) (60 ) 7
480 3 420 7
10 60
So 6 kg
m m
m m
m
m
-
-
c R( ) : T - 66g
66w2
r us ing F ma and a
v2
r
T 66g 66
72
5 1293.6 1300 N (2 s.f.)
23 a Let the speed at C be v m s–1. Then taking the horizontal through B as the zero level for potential
energy and using conservation of energy, the gain in kinetic energy at point C equals the loss in
potential energy:
1
2mv2 -
1
2m 202 mg 50(1- cos60°)
v2 202 50g v2 890
So v 30ms-1 (2 s.f.)
b At C, R( ) : R- 70g
70v2
50 us ing F ma and a
v2
r
R 70g
70890
50 686 1246 1900 N (2 s.f.)
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23 c Consider motion C to D. Let the speed at D be w m s–1. Then taking the horizontal through C as the
zero level for potential energy and using conservation of energy, the loss in kinetic energy at point
C equals the gain in potential energy:
1
2m890-
1
2mw2 mg 50(1- cos30°)
w2 890-100g(1- cos30°) 759 (3 s.f.)
w 28ms-1 (2 s.f.)
d Resolving perpendicular to the slope at B just before the circular motion and just after circular
motion begins:
Before: R mg cos60° 35g
After: R - mg cos60°
m 202
50 R 35g 560
So change in R = 560 N
e Allowing for the influence of friction would mean that the skier would arrive at C with lower
speed. From the equation used in part b, this would result in a lower normal reaction.
24 a Taking the horizontal through A as the zero level for potential energy and using conservation of
energy, the loss in kinetic energy when OP makes an angle equals the gain in potential energy:
1
2m3ag -
1
2mv
2 mga(1 cos )
v2 ag(1- 2cos )
b Resolving along the radius when OP makes an angle
2
R( ) : cos
(1 2cos )cos (1 3cos )
mvT mg
a
magT mg mg
a
- - -
ւ
c The string becomes slack when T = 0, i.e. from part b when cos
1
3
Height above A a acos a
1
3a
4
3a
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24 d Using result for v2 from part a, and that cos
1
3 at point B, then 2
3
lgv at point B
Consider the vertical component of the motion of the particle under gravity
Using v2 u2 2as , where u is the vertical component of motion at B = sin ,
3
lg v is the
vertical component of motion at C = 0, a = –g and s is the vertical height of C above B
2
2 2
So 0 sin 23
8 4sin (1 cos )
6 6 6 9 27
lggs
l l l ls
-
-
The solution can also be found using conservation of energy as follows.
At C the particle has speed cos ,3
lg the horizontal component of the speed at B
Let the vertical height of C above B be h. Taking the horizontal through B as the zero level for
potential energy and using conservation of energy, the loss in kinetic energy at C equals the gain in potential energy:
21 1cos
2 3 2 3
1 8 41
6 9 54 27
lg lgm m mgh
l l lh
-
æ ö - ç ÷è ø
25 a Taking the horizontal through O as the zero level for potential energy and using conservation of
energy, the gain in kinetic energy equals the loss in potential energy:
1
2mv2 -
1
2mu2 mga sin
v2 u
2 2gasin
v2 3
2ga 2gasin
b Resolving along the radius when OP makes an angle with the horizontal:
2
R( ) : sin
3 3 3sin 2 sin 3 sin (1 2sin )
2 2 2
mvT mg
a
m ga mg mgT mg ga mg
a
-
æ ö ç ÷è ø
տ
c The string becomes slack when T = 0, i.e. from part b when 2sin -1
So sin -
1
2 210°
d From part a, v2
3
2ga 2gasin
To complete the circle, v ¹ 0 before the rod reaches the top point, i.e. 0 for 0 270v > ° °� �
But v = 0 when 3
sin 229 (3 s.f.)4
- ° , so P would not complete a vertical circle.
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25 e Using conservation of energy, as there is no change in potential energy when the particle reaches A, as A and O are on the same level, there can be no change in its kinetic energy:
So v u
3ga
2
f Using the results from parts a and c, when the string becomes slack 210° and
v2 3
2ga 2ga -
1
2
æ
èçö
ø÷
1
2ga
Its horizontal component of velocity is
1
2ga cos60°
From part e, when it reaches point A, its horizontal component of velocity is
3
2ga cosf
When P moves freely under gravity, its horizontal component of velocity will remain constant, so
3ga
2cosf
1
2ga cos60°
cosf cos60°
3
1
2 3
f 73.2° (3 s.f.)
26 a Let the velocity of the particle at C be v. Taking the horizontal through A as the zero level for potential energy and using conservation of energy, the gain in kinetic energy at C equals the loss in
potential energy:
2 2
2 2
1 1(1 cos )
2 2
2 (1 cos )
mv mu mga
v u ga
- -
- (1)
Resolving along the radius through C:
2
R ( ) : cosmv
R mga
- ւ
As the particle leaves the sphere at C, 0R so v
2 ag cos
Substituting for v2 in equation (1) gives:
ag cos u2 2ga(1- cos )
3cos u2
ag 2
cos 2
3
u2
3ag
b Using conservation of energy, the gain in kinetic energy from C to when the particle hits the
ground equals the loss in potential energy:
1
2m
9ag
2
æ
èçö
ø÷-
1
2m(ag cos ) mga(1 cos )
3
2cos
9
4-1
5
4 cos
5
6
So 34° (2 s.f.)
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 25
27 a Taking the horizontal through A as the zero level for potential energy and using conservation of energy, the loss in kinetic energy at B equals the gain in potential energy:
2 2
2 2
1 1(1 cos 60 )
2 2
3
mu mv mga
v u ga
- °
- (1)
b Resolving along the radius through B:
2
R ( ) : cos 60 mv
R mga
° (2)ւ
When u
2 6ga, v2 3ga from equation (1)
So R
mv2
a- mg cos60° 3mg -
mg
2
5mg
2
c The least value for u will be that which enables the particle to reach B, but to leave the surface at B
so that R = 0 at point B.
22
2 2
If 0 at , then from equation 2 2
7So from equation 3
2 2
7Hence
2
mg mv agR B v
a
ag agu ga u
agu
-
(2)
(1)
d The distance BC = 2 asin60° 3a
The horizontal component of the speed at B = vcos60°
For motion under gravity the horizontal velocity of the particle in travelling from B to C is
constant and using s vt gives:
3a vcos60°t t 2a 3
v
The vertical component of the speed at B = vsin60°
The vertical component of the distance the particle travels is 0 (as the particle is now at C, which is level with B)
So using s ut
1
2at2
gives:
0 vsin60°t -1
2gt
2
t 2vsin60°
g
v 3
g
So when the particle leaves the bowl at B and meets it at C
v 3
g
2a 3
v v2 2ag
So using equation (1) from part a, 2ga u2 - 3ga u2 5ga
Hence u 5ga
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 26
28 a The total mass is 3m5m lm (8 l )m
Taking moments about the y-axis:
3m 4 5m 0 lm 4 (8 l)m 2
12 4l 16 2l2l 4
l 2
b Taking moments about the x-axis:
3m 0 5m -3 lm 2 (8 l)m k
-15 4 10k substituting for l 2
-11 10k
So k = 1.1
29 The total mass is 2M xM yM (2 x y)M
Taking moments about the y-axis:
(2 ) 2 2 2 1 3
4 2 2 4 3
0
x y M M xM yM
x y x y
x y
- (1)
Taking moments about the x-axis:
(2 ) 4 2 5 3 1
8 4 4 10 3
3 2
x y M M xM yM
x y x y
x y
(2)
Subtracting equation (1) from equation (2) gives:
14 2
2
1And so from equation
2
y y
x
(2)
30 Using
mir
i r m
iåå
2 2 40.1 0.2 0.3 (0.1 0.2 0.3)
1 5 2
0.2 0.4 1.2(0.6)
0.1 1 0.6
1.8(0.6)
1.5
3
2.5
x
y
x
y
x
y
x
y
æ ö æ ö æ ö æ ö ç ÷ ç ÷ ç ÷ ç ÷-è ø è ø è ø è ø
æ ö æ ö æ ö æ ö ç ÷ ç ÷ ç ÷ ç ÷-è ø è ø è ø è ø
æ ö æ öç ÷ ç ÷
è ø è ø
æ ö æ öç ÷ ç ÷
è ø è ø
So the centre of mass is (3i + 2.5j) m
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 27
31 a Using
mir
i r m
iåå
6 0 2 32 (3 )
0 4 2
12 0 2 9 3
0 4 2 3
M M kM k Mc
k k
k c ck
æ ö æ ö æ ö æ ö ç ÷ ç ÷ ç ÷ ç ÷-è ø è ø è ø è ø
æ ö æ ö æ ö æ ö ç ÷ ç ÷ ç ÷ ç ÷- è ø è ø è ø è ø
Equating i component gives 12 2 9 3 3 as required.k k k
b Equating j component and substituting for k gives:
4- 2k 3c ck -2 6c c -
1
3
32 a The area of the lamina is 2010 200 cm2
The area of the circle is 32 9 cm2
The area of the plate is (200 - 9) cm2
Let the distance of the centre of mass of the plate from AD be x cm
As the lamina is uniform, masses are proportional to areas. By symmetry, the centre of mass of the rectangular lamina is 10 cm from AD and the centre of mass of the circle is 6 cm from AD.
Lamina Circle Plate
Area 200 9 200- 9
Distance of centre of mass from AD 10 6 x
The moment of the plate about AD equals the moment of the complete lamina less the moment of the circle that has been removed:
(200 9π) 200 10 9 6
2000 5410.658 10.7 cm (3 s.f.)
200 9
x
x
- -
-
- …
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32 b Let the angle between AB and the vertical be a. When the plate is suspended freely from A, its
centre of mass G is vertically below the point of suspension A. The distance of G from AD was
found in part a and the distance of G from AB is 5 cm by symmetry.
tana 5
x
5
10.658 0.469 (3 s.f.)
a 25° (to the nearest degree)
33 a The area of rectangle ABDE is 26 8 48a a a
The centre of mass of rectangle is 4a from X
The area of ∆BCD is
1
2 6a 4a 12a2
The centre of mass of the triangle is
1
3h
1
3 4a from the base BD of the triangle.
The area of lamina ABCDE is 48a2 12a2 60a2
Lamina Rectangle Triangle
Area 60a2 48a2 12a2
Displacements from X GX 4a 4
3a-
Taking moments about X, the rectangle is on one side of X, taken as positive, and the triangle is on
the other side of X, taken as negative:
2 2 2 3 3 3
3
2
460 48 4 12 192 16 176
3
176 44 as required
60 15
a GX a a a a a a a
aGX a
a
æ ö - - ç ÷è ø
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33 b With the particle at C, taking moments about X:
Mg GX lMg 4a
M g 44
15a l M g 4 a substituting for GX from part a
l 44
15 4
11
15
There is a vertical force at B but, as the line of action of this force passes through X, its moment
about X is zero.
34 a Let the distance of the centre of mass, G, of the loaded plate from AB and BC be x cm and y cm
respectively.
Taking B as the origin and using
mir
i r m
iåå
0 0 2 22 3 4 10 (1 2 3 4 10)
2 0 0 2
2420
20
l l l xM M M M M M
l l l y
l x
l y
æ ö æ ö æ ö æ ö æ ö æ ö ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷
è ø è ø è ø è ø è ø è ø
æ ö æ öç ÷ ç ÷
è ø è ø
So distance from AB x
24l
20
6l
5
b From working in part a, distance from BC y
20l
20 l
c When the framework hangs freely from D, its centre of mass G is vertically below D. Let the angle
made by DA with the downward vertical be a. Then from the diagram in part a:
tana 2l - y
2l - x
2l - l
2l - 6l5
l4l5
5
4
a 51° (to the nearestdegree)
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 30
35 a The area of rectangle ABCD is 3a 2a 6a2
The area of triangle BCE is
1
2a 2a a2
The area of lamina ABCD is 6a2 - a2 5a2
Let the distance of the centre of mass of the lamina, G, from AD be x cm. The distance of the
centre of mass of the rectangle from AD is 1.5a. Consider the centre of mass of the triangle, .G ¢
If Y is the midpoint of BE then CG¢ : G¢Y 2 :1.
Using similar triangles,
CX
XE
CG¢G¢Y
2
1
As CE a, XE
1
3a , so ¢G is
1
3a from BE and
3a-
1
3a
8
3a from AD.
Rectangle Lamina Triangle
Area 6a2 5a2 a2
Distance from AD
3
2a
x
8
3a
The moment of the lamina about AD is the moment of the complete rectangle less the moment of
the triangle which has been removed from the rectangle:
5a2 x 6a2 3
2a - a2
8
3a
5a2x 9a
3 -8
3a
3 19
3a
3
x 19
15a
b Let N be the mid-point of AB. Taking moments about N:
Mg3
2a - x
æ
èçö
ø÷ mg
3
2a
m 2
3aM
3
2a - x
æ
èçö
ø÷
2
3aM
3
2a -
19
15a
æ
èçö
ø÷
2
3
7
30M
7
45M
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 31
36 a Let the distance of the centre of mass of the decoration, G, from B be x cm.
The area of the smaller circle is 102 100
The area of the larger circle is 202 400
The area of the decoration is 100 400 500
As the card is uniform, the masses of the decoration and circles are proportional to their areas.
Decoration Small circle Large circle
Area 500 100 400
Distance from B (cm) x 30 0
Taking moments about B:
500 x 100 30 400 0
x 3000
500 6cm
b When the decoration is freely suspended from C, its centre of mass G is vertically below C.
Let the angle between AB and the vertical be a. Then from the diagram drawn for part a:
tana AC
GA
10
10 (20- x )
10
24
5
12
a 22.6° (1 d.p.)
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 32
37 a Let the distance of the centre of mass, G, of the lamina from the midpoint of AB be x cm.
The area of the semicircle is 21 25
52 2
So the area of the lamina L is 25
1002
-
From the question, the distance of the centre of mass of the semicircle from the midpoint of AB
is 4 20
as the radius is 5cm in this case.3 3
aa
Square Semicircle Lamina L
Area 100 25
π2
25
100 π2
-
Distance from midpoint of AB (cm) 5 20
3π x
Taking moments about the midpoint of AB:
100 -25
2
æ
èçö
ø÷x 100 5-
20
3
25
2 500 -
250
3
1250
3
x 2500
3(200 - 25 6.86 (2 d.p.)
b When L is suspended freely from D, the centre of mass G hangs vertically below D.
The downward vertical is drawn in the diagram. Let the angle between CD and the vertical be a.
tana GN
DN
(10- x )
5
10 - 6.86
5 0.628
a 32.1° (1 d.p.)
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 33
38 a Let the distance of the centre of mass of T from A be x cm, i.e. AG x .
The area of the smaller circle is 2 64
The area of the larger circle is 2 5762 The area of T is 576 64 512 -
As the lamina is uniform, the masses of T and the large circle are proportional to their areas.
T Small circle Large circle
Area 512 64 576
Distance from A (cm) x 16 24
Taking moments about A:
512 576 24 64 16
576 24 64 1625cm
512
x
x
-
-
b Let C be the midpoint of OB and the mass of T be M.
12cm, 36 25 11cmBC CG AC AG - -
Taking moments about C
Mg 111
4mg12
M 3m
11
39 a Using
mir
i r m
iåå
0 9 04 6 2 (4 6 2)
4 0 4
5412
8
xm m m m
y
x
y
æ ö æ ö æ ö æ ö ç ÷ ç ÷ ç ÷ ç ÷-è ø è ø è ø è ø
æ ö æ öç ÷ ç ÷
è ø è ø
So x
54
12
9
2 and y
8
12
2
3
So the cooordinates of the centre of mass of the particles without the lamina are 9 2
,2 3
æ öç ÷è ø
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 34
39 b The uniform lamina is a triangle so its centre of mass is (3, 0). Its mass is km.
The combined system of the lamina and the three particles has centre of mass (4,l).
Its mass 4m 6m 2m km (12 k)m
Using
mir
i r m
iåå and the result from part a to treat the three particles as a single entity:
92
23
3 412 (12 )
0
54 3 4(12 )
8
m km k m
kk
l
l
æ ö æ ö æ ö ç ÷ ç ÷ ç ÷
è ø è øè øæ ö æ ö
ç ÷ ç ÷è ø è ø
So equating the i components gives:
54 3k 48 4k
k 6
c Equating the j components from the vector equation in part b and substituting for k gives:
8 (12 k)l 18l
l 8
18
4
9
d Let the angle between AC and the vertical be a.
tana 4
l 4
4
9 9
a 83.7° (1 d.p.)
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 35
40 a Let the distances of the centre of mass of L, say G, from AD and AB be x and y respectively.
The mass of L is 3 4 2 10 .m m m m m
Plate Rectangle Particles
L ABCD A B C
Mass 10m 3m 4m m 2m
Distance from AD x 2.5a 0 5a 5a
Distance from AB y a 0 0 2a
Taking moments about AD:
10m x 3m 2.5a m 5a 2m5a 22.5ma
x 22.5a
10 2.25a as required
b Taking moments about AB:
10m y 3m a 2m 2a 7ma
y 7a
10 0.7a
c When L is freely suspended from O, the centre of mass G of the complete system hangs vertically
below O. Let a be the angle between OA and the vertical.
tana y
2.5a - x
0.7a
0.25a 2.8
a 70° (to the nearestdegree)
So the angle that AB makes with the horizontal is (90- 70)° 20° (to the nearest degree)
d The total weight 10mg of the loaded plate L acts vertically through the centre of mass G, so from
part a the force has a perpendicular distance from O of 2.5a - x 2.5a - 2.25a 0.25a
The force P acts horizontally through C, so this force has a perpendicular distance from O of 2a
The loaded plate is in equilibrium, so taking moments about O:
P 2a 10mg 0.25a
P 2.5mg
2
5
4mg as required
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 36
40 e The forces acting on the system are the horizontal force at C, the weight of the loaded plate
through G and the reaction force at O. Let the horizontal and vertical components of the force
acting on L at O be X and Y respectively.
R(®) X P 5
4mg
R() Y 10mg
Let the magnitude of the force acting on L at O be R.
2
2 2 2 2 2 25 1625(10 )
4 16
1625 5 65
4 4
R X Y mg mg m g
R mg mg
æ ö ç ÷è ø
41 a As the wire is uniform, each side has a mass proportional to its length and the centre of mass
of each side is at the middle of the side. The triangle has sides in the ratio of 3 : 4 : 5 so it is a
right-angled triangle.
Taking B as the origin and axes along AB and BC:
0 4 46 8 10 (6 8 10)
3 0 3
7224
48
x
y
x
y
æ ö æ ö æ ö æ ö ç ÷ ç ÷ ç ÷ ç ÷
è ø è ø è ø è ø
æ ö æ öç ÷ ç ÷
è ø è ø
So distance from AB x
72
24 3cm
b 48
Using vector calculation in part , distance from 2cm24
BC y a
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41 c When the frame hangs freely from A, its centre of mass G is vertically below A.
Let the acute angle made by AB with the downward vertical be a.
tana x
6- y
3
6 - 2
3
4
a 37° (to the nearest degree)
42 a Let the distance of the centre of mass of the loaded framework, say G, from AB and AD be x and
y respectively. As the rods that make up the framework are uniform, the centre of mass of each
rod is at its midpoint.
Taking A as the origin and axes along AB and AD:
0 1.5 1.5 3 3 36 2 (1 1 1 1 6 2)
0 2 2
3012
16
a a a a a xm m m m m m m
a a a a a y
a x
a y
æ ö æ ö æ ö æ ö æ ö æ ö æ ö ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷
è ø è ø è ø è ø è ø è ø è ø
æ ö æ öç ÷ ç ÷
è ø è ø
i Distance of the centre of mass of the loaded framework from AB x
30
12a
5
2a
ii Distance of the centre of mass of the loaded framework from AD y
16
12a
4
3a
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 38
42 b Let a be the angle BC makes with the vertical (see diagram for part a).
43
52
22 2 2 4tan
3 5 15
15 (to the nearest degree)
a aa y
x aa
a
--
°
43 a Let the distance of the centre of mass of the loaded framework, say G, from BC and AB be x and
y respectively. As l
2 l 2 (l 2)2 , the angle at B is a right angle. As each rod is uniform, the
centre of mass of each rod is at its midpoint.
Taking B as the origin and axes along AB and BC:
0 0.5 0.52 3 (1 2 3)
0.5 0 0.5
2.56
2
l l xm m m m
l l y
l x
l y
æ ö æ ö æ ö æ ö ç ÷ ç ÷ ç ÷ ç ÷
è ø è ø è ø è ø
æ ö æ öç ÷ ç ÷
è ø è ø
i Distance of the centre of mass of the framework from BC x
2.5l
6
5l
12
ii Distance of the centre of mass of the loaded framework from AB y
2l
6
l
3
b When the framework changes freely from A, its centre of mass G is vertically below A. The
vertical line has been drawn in the diagram (see part a). Let the angle that BC makes with the
vertical be a.
3
512
1 12 4tan 0.8
3 5 5
39 (to the nearest degree)
l
l
y
xa
a
°
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 39
44 a Let G be the origin so that GE lies on the x-axis and AG lies on the y-axis.
The coordinates of the centre of mass of the rectangle ABGF are 3
, .2 2
d dæ öç ÷è ø
The coordinates of the centre of mass of the square CDFE are 3
, .2 2
d dæ öç ÷è ø
Rectangle ABGF has area 3d2, square CDFE has area d2, but the square has density three times
that of the rectangle. So the mass ratios of the rectangle, square and lamina are 3d2, 3 × d2 = 3d2 and 6d2 respectively.
2 2 2
3
2
3
3
2 23 3 6
3
2 2
66
6
d d
xd d d
d d y
xdd
yd
æ ö æ öç ÷ ç ÷ æ öç ÷ ç ÷
æ öç ÷
ç ÷ç ÷ ç ÷ è øç ÷ ç ÷è ø è ø
æ
è ø
öç ÷è ø
3
2
6Distance of the centre of mass of the framework from .
6
dAG x d
d
b
3
2
6Distance of the centre of mass of the framework from .
6
dGE y d
d
c The centre of mass of the lamina lies at (d, d), which are the coordinates of point C. If the lamina
is freely suspended from point A, the centre of mass will align itself vertically below point A, so that AC will be the new vertical.
Let the angle that AB makes with the vertical be , then .CAB Ð
2
tan 2
63 (to the nearest degree)
CB d
AB d
°
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 40
45 a Let N be the foot of the perpendicular from D to BC and NC = p cm.
The total length of the frame is AB BC CD AD 20 4 BC 5 4 20 BC 7cm
Hence NC = BC – BN = 7 – 4 = 3
Let the distance of the centre of mass of the frame from AB be x cm
Calculating the mass of each section of wire from the densities given in the question:
AB BC CD DA
Mass (kg) 0.04M 0.07M 0.075M 0.04M
Distance from AB (cm) 0 3.5 5.5 2
Taking moments about AB:
(0.04 0.07 0.075 0.04) 0.04 0 0.07 3.5 0.075 5.5 0.04 2
0.73753.2777 3.28cm (3 s.f.)
0.225
M x M M M M
x
…
b The weight of the framework acts through the centre of mass, which is (3.5- x )cm from the
vertical through the midpoint of BC.
Taking moments about the midpoint of BC:
(3.5 ) 3.5
3.5 3.5 3.27770.0635 (3 s.f.)
3.5 3.5
Mg x kMg
xk
-
- -
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 41
46 a Let the midpoint of BC be G and the midpoint of EG be F. Divide the lamina into three sections, a rectangle ABEG, a square CDFG and a triangle EDF.
Let A be the origin so that AB lies on the x-axis and AE lies on the y-axis.
The coordinates of the centre of mass of the rectangle ABEG are (40, 20).-
The coordinates of the centre of mass of the square CDFG are (60, 60).-
The coordinates of the centre of mass of the triangle EDF are found by taking the average of the
coordinates of its three vertices, in this case (0,-40) , (40,-40) and (40,-80) , hence it is:
0 40 40
3,-40 -80 - 40
3
æ
èçö
ø÷
80
3,-
160
3
æ
èçö
ø÷
Rectangle ABEG has area 3200 cm2, square CDFG has area 1600 cm2, and triangle EDF has
area 800 cm2.
Using
mir
i r m
iåå
80
40 60 33200 1600 800 (3200 1600 800)
20 60 160
3
5600736 000
3
608 000
3
x
y
x
y
æ öç ÷æ ö æ ö æ ö
ç ÷ç ÷ ç ÷ ç ÷- - ç ÷è ø è ø è ø-ç ÷è ø
æ öæ ö ç ÷
ç ÷ è øç ÷ç ÷ç ÷-ç ÷è ø
The distance of the centre of mass of the lamina from AE 736 000
43.8cm (3 s.f.)3 5600
x
b 1 2
R( ) : T T W (1)
Taking moments about A:
2
2
2
80 0
736 00080
16800
736 230.548 N (3 s.f.)
1344 42
W x T
T W
T W W W
-
1 2
19From equation 0.452 N (3 s.f.)
42T W T W W - (1)
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 42
46 c 1 2
R( ) : T T W kW (2)
Taking moments about A:
2
2
2
80 80 0
736 00080 80
16 800
23
42
W x kW T
T W kW
T W kW
-
2
23So if 8 8
42
238
42
7.45 (3 s.f.)
T W W kW W
k
k
-
� �
�
�
1 2
1
19From equation
42
10
T W kW T W
T W
-
(2)
�
So the largest value of k satisfying T1 ⩽ 10W and T2 ⩽ 8W is k = 7.45 (3 s.f.)
47 The centre of mass lies on the axis of symmetry, y = 0.
To find the x coordinate, using
x y2xdxòy2 dxò
, and substituting y x gives:
4 42 3130 0
4 421
2 00
d 64 88
3 3d
x x xx
xx x
òò
So the coordinates of the centre of mass of the solid are 8
, 03
æ öç ÷è ø
, a distance of 8
3 from O.
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 43
48 a Let the straight edge lie along the y-axis. Then the centre of mass lies on the x-axis from
symmetry.
If P has coordinates (x, y) and the elemental strip PQRS has width d x then its area is 2 .y xd
The mass M of the lamina = 21,
2a where is the mass per unit area of the lamina.
Let x be the distance of the centre of mass from O, then M x 2xydx
0
a
ò
As the boundary of the semicircle has the equation x
2 y2 a
2, then y (a
2 - x2)
12
312 22 2 2 2 3
00
323
212
2 2So 2 ( ) d ( )
3 3
4 as required
3
aa
M x x a x x a x a
a ax
a
- - -
ò
b Using the result from part a and letting x be the distance of the centre of mass of the resulting
lamina from O:
Shape Mass Distance of centre
of mass from O
Semicircle radius a
1
2a2
4a
3
Semicircle radius b
1
2b2
4b
3
Resulting
1
2(a2 - b2) x
Taking moments about O:
2 2 2 2
3 3 2 2 2 2
2 2
1 1 4 1 4( )
2 2 3π 2 3π
4 ( ) 4 ( )( ) 4 ( ) as required
3 ( ) 3 ( )( ) 3 ( )
a ba b x a b
a b a b a ab b a ab bx
a b a b a b a b
- -
- -
- -
c As b ® a, the area becomes a circular arc and from the equation found in part b
x ®4
3
(a2 a2 a2 )
(a a)
4
3
3a2
2a
2a
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49 a Let B be the origin so that BC lies on the x-axis and AB lies on the y-axis. Let the equation of the
line AC be y c- mx , so that the coordinates of the triangle are (0, 0), (0, c) and ,0c
m.
If a point on the line has coordinates (x, y), then an elemental horizontal strip of width d y at that
point has area is xd y
The mass M of the triangular lamina =
1
2c
c
m
c22m
, where is the mass per unit area
Let y be the distance of the centre of mass from BC, then M y xyd y
0
c
ò
2
2 2 3
0 0 00
3 3
2
( )So given
1 1d ( ) d ( )d
2 2 3
2 1 1
2 3 3
cc c c
c yx
m
cy xy y c y y y cy y y cy y
m m m m
cy c c
c
-
æ ö - - -ç ÷ è ø
æ ö - ç ÷è ø
ò ò ò
b Divide the lamina into two sections, a rectangle and a triangle. Let be the mass per unit area of
the lamina. Let the point P be the origin, so that PQ is the x-axis and SP is the y-axis, and the
coordinates of G, the lamina’s centre of mass be (x , y ) .
The mass of the rectangle is 2a2 and its centre of mass is , .
2
aa
æ öç ÷è ø
The mass of the triangle is a
2 and its centre of mass is
4a
3,
2a
3
æ
èçö
ø÷, using the result from part a.
Using
mir
i r m
iåå
2 2 2 2
4
32 (2 )2
2
3
7
33
8
3
aa
xa a a a
a ya
a
x
a y
æ öæ ö ç ÷ æ öç ÷ ç ÷ ç ÷ç ÷ç ÷ ç ÷ è øç ÷è ø
è ø
æ öç ÷ æ ö
ç ÷ ç ÷ç ÷ è øç ÷è ø
The distance of the centre of mass G of the lamina from PS 7
9
ax
c The distance of the centre of mass G of the lamina from PS 8
9
ay
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50 a Use
x y2xdxòy2 dxò
, and substitute 2
1
2y
x
x xy2 dx
1
2
òy2 dx
1
2
ò
x 14
x-4 dx1
2
ò14
x-4 dx1
2
ò
x-3 dx1
2
òx-4 dx
1
2
ò
-1
2x-2
1
2
-1
3x-3
1
2
3 1-2 - 2-2 2 1
-3 - 2-3
3
2
3
4
8
7
9
7
The centre of mass is 9
m7
from the y-axis, hence 9 2
1 m7 7
æ ö- ç ÷è ø
from the larger plane face.
b Let be the mass per unit volume of the hemisphere H and the solid S.
Then the mass of the S = 2
2 22 4 3
1 11
1 7d d 1
4 12 12 8 96y x x x x
- - æ ö - - ç ÷ è øò ò
The mass of the hemisphere and its distance of its centre of mass from its plane face can be found
from the standard formulae, using r = 0.5.
Shape Mass Distance of centre of mass from
trophy’s plane face
Solid, S
796
1-
2
7
5
7
Hemisphere, H
23
1
2
æ
èçö
ø÷
3
12
1
3
8
1
2
19
16
Trophy, T
7
96
1
12
æ
èçö
ø÷
532
x
Taking moments around the trophy’s plane face (the bottom of the trophy as shown):
532
x 12
19
16
796
5
7
x 32
5
19
192
5
96
æ
èçö
ø÷
32
5
29
192
29
30 0.967 m (3 s.f.)
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51 a A hemisphere is generated when a semicircle is notated through 180° about the x-axis.
Divide the hemisphere into circular discs, with each disc having mass y2d x and centre of mass
at a distance x from O.
2 2 2
0 0
2 2 2
0 0
2 2 4 4 4
0
3 32 3
0
d ( )dSo
d ( )d
1 1 1 13 1 32 4 2 4
1 2 4 8133
R R
R R
R
R
xy x x R x xx
y x R x x
R x x R RR R
R RR x x
-
-
- - --
ò òò ò
b Using the standard formulae for a cone:
Shape Mass Distance of centre of mass
from V
Cone
1
3a2ka
3
4ka
Hemisphere
2
3a3
ka
3
8a
Top
1
3a3(k 2) x
Taking moments about V:
1
3a3(k 2)x
1
3a3k
3
4ka
æ
èçö
ø÷
2
3a3 ka
3a
8
æ
èçö
ø÷
(k 2)x 3
4k 2a 2ka
3a
4
x (3k 2 8k 3)a
4(k 2)
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51 c The manufacturer’s requirement is that x ka
Hence from part b 3k
2 8k 3
4(k 2) k
3k 2 8k 3 4k 2 8k
k 2 3, so k 3
52 a Using the standard results for a hemisphere:
Shape Mass Ratio of masses Distance of centre
of mass from O
Large hemisphere
2
3a
3 8
3
8a
Small hemisphere
2
3
a
2
æ
èçö
ø÷
3
1
3
16a
Remainder
2
3
7a3
8 7 x
Taking moments about O:
7x 83
8a -1
3
16a
x 1
7
45
16a
45a
112
b Using the result from part a:
Mass ratios Distance of centre
of mass from O
Bowl M 45
112a
Liquid kM 3
16a
Bowl + liquid (k + 1)M
17
48a
Taking moments about O:
(k 1) M 17
48a M
45
112a kM
3
16a
k17
48-
3
16
æ
èçö
ø÷
45
112-
17
48
8
48k
45
112-
17
48
k 645
112-
17
48
æ
èçö
ø÷
6(2160 -1906)
5376
254
896
2
7
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53 a V y2
0
1
ò dx 4
(x - 2)4
0
1
ò dx substituting y 1
2(x - 2)2
2
5 5 3
0
32 8( 2) ( 2) cm
20 20 20 5V x
- - -
b Let x be the distance of the centre of mass of S from its plane face
To find, using M x y2xdxò , and substituting 2 21 8
( 2) and from part 4 5
y x M
- a
gives:
x
5
32(x - 2)4 xdx
0
2
ò
Integrate using the substitution u x - 2
x 5
32u4(u 2)du
-2
0
ò 5
32u5 2u4 du
-2
0
ò
5
32
1
6u6
2
5u5
-2
0
5
32
2
5 25 -
1
626æ
èçö
ø÷
5
32
64
5-
64
6
æ
èçö
ø÷
101
5-
1
6
æ
èçö
ø÷
10
30
1
3
The centre of mass lies on the axis of symmetry at a distance of 1
cm3
from the plane base.
c Let the reaction force at A be A, and at B be B.
Taking moments about B:
18 2 10 4
3
240
118 593
8 24 12
A W W
WW W
A
æ ö-ç ÷è ø
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54 a Using the formulae for standard uniform bodies:
Shape Mass Mass ratios Distance of centre
of mass from base
Cylinder r 2h 1 2
h
Cone
1
3r 2 h
2
1
6
h-1
4
h
2
æ
èçö
ø÷
7h
8
Ornament
5
6r 2h
5
6 x
Take moments about O, the centre of plane base:
5
6x 1
h
2-
1
6
7h
8
5
6x
h
2-
7h
48
17h
48
x 17h
48
6
5
17h
40
b The centre of mass G of the ornament will be directly below the point of suspension.
tana h-
17h
40r
23h
40r
As h 4r, this gives tana 23r
10r 2.3
a 66.5° (1 d.p.)
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55 a Let O be the midpoint of YZ. Let OZ be the x-axis and OX the y-axis. The triangle is isosceles and
the y-axis is the line of symmetry.
The equation of the line XZ is y h- mx x
h- y
m
The area of the elemental strip of width d y is 2xd y
Let y be the distance of the centre of mass from O, then:
20
0 0
0 00
2 3 3
0
22
0
d2 d d
2 d dd
1 1
12 3 61 3122
hh h
h hh
h
h
h yy yyx y yh y ym
yh yx y h y yy
m
hy y h
h
hhy y
-æ öç ÷ -è ø -æ ö -ç ÷
è ø
- -
òò òò òò
So distance from 2
3 3
h hX h -
b Let E be the midpoint of BC, then BE = 4a and AE = 3a, as DABE is a 3 : 4 : 5 triangle,
and DE = a, from part a.
Let be the mass per unit area of the lamina.
Shape Mass Distance of centre
of mass from A
∆ABC 212 a 2a
∆BCD 4a2
22
3
aa
Plate ABDC 8a2 x
Taking moments about A:
8a2x 12a2 2a - 4a2 8a
3
æ
èçö
ø÷
24a2 -32a
3
3
40a3
3
So x
40a
8 3
5a
3
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55 c Let be the angle between CB and the vertical.
From part b, the distance GE is 5 4
33 3
a aa -
Taking moments about C:
Mg 8asin Mg4a
3cos - 4asin
æ
èçö
ø÷
12 Mgasin 4
3Mgacos
sincos
tan 4
36
1
9
So arctan1
9
So CB makes an angle1
arctan9
æ öç ÷è ø
with the vertical.
56 a Let be the mass per unit area of the material and x the distance of the centre of mass of the
closed container from O. Using the formulae for standard uniform bodies:
Shape Mass Distance of centre of
mass from O
Circular disc a2 0
Hemispherical bowl 2a2
1
2a
Closed container 3a2 x
Taking moments about O:
2 23π 0 2π2
3
aa x a
ax
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56 b The container is resting in equilibrium, so the weight of P acts to one side and the weight of C
balances on the other side.
Taking moments about O:
Mg a
3sin
1
2Mg acos
So tan 3
2
56° (to the nearest degree)
57 Let V be the vertex of the cone and O be the centre of its base.
Let G be the position of its centre of mass.
From ∆VAO, tana
OA
OV
a
h, where h is the height of the cone.
So tana 1
3
a
h h 3a
So from the standard result for a uniform solid right circular cone OG 1
4h
3a
4
Then from ∆GAO
tan AO
GO
a
3a
4
4
3
53.1° (1 d.p.)
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58 a The hemisphere K has mass M. The hemisphere H has double the radius, so it has 8 times the
volume. Its mass is 8M. Therefore the composite body S has mass M 8M 9M .
Shape Mass Distance of centre of
mass from C
H 8M 3
8
a
K M 3
16
a-
S 8M M x
Taking moments about C:
9M x 8M 3a
8- M
3a
16 M
45a
16
x 45a
916
5a
16
b The composite body with particle P attached rests in equilibrium, with C above the point of
contact with the plane, X.
Mg acosa 9Mgx sina 95
16Mg sina substituting result for x from part a
tana 16
45
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59 a A line R
y R xh
- rotated around the x-axis with generate a solid cone with its base on the y-axis.
The volume of an elemental strip of width d x is 2 .x xd
Let x be the distance of the centre of mass of the solid come from its base, then:
22 2
2 2 3220 00
2 2 22
2 2
0 200
2 2 2 3 2 4
2 22
0
2 2 2 32
2
2
0
d 2 dd
d 2 dd
2 1 2 1
2 3 4 2 3 4
11 1
33
13
12 4
h hh
hhh
h
h
R R Rx R x x R x x x xxy x h h hxR RRy x R x x xR x xh hh
R x R x R xR h
h h
R x R x R hR xh h
hh
æ ö - - ç ÷ è ø æ ö - -ç ÷è ø
æ ö- - ç ÷ è ø æ ö - ç ÷- è ø
ò òòò òò
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59 b Using the result from part a:
Shape Mass Mass ratio Distance from base of
centre of mass
Large cone
1
3a2 H H
H
4
Small cone
1
3a2h h
h
4
Solid S
1
3a2(H - h) H – h x
Taking moments about the base:
(H - h)x H H
4- h
h
4
x 1
4
(H 2 - h2 )
(H - h)
1
4
(H - h)(H h)
(H - h)
1
4(H h)
So the distance from the vertex is H -
1
4(H h)
3
4H -
1
4h
1
4(3H - h) as required
c Let the tension in the string attached to the vertex be T1 and tension in the other string be T2.
Taking moments about the centre of mass G:
2 1
11 4
12 4
1( ) where (3 ) from part
4
( )
(3 ) 3
T x T H x H x H h
H hT x H h
T H x H h H h
- - -
- - -
b
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60 a Let be the mass per unit volume of the material of the cylinder and x the distance of the centre
of mass of the toy from O. Using the formulae for standard uniform bodies:
Shape Mass Mass ratio Distance of centre
of mass from O
Hemisphere 32
π 63
r 4r 5
8
r
Cylinder 2πr h h 2
hr
Toy 2π (4 )r r h 4r h x
Taking moments about O:
(4r h)x 4r 5r
8 h
h
2 r
æ
èçö
ø÷
x 5r 2 h2 2rh
2(4r h)
h2 2hr 5r 2
2(h 4r) as required
b If the toy remains in equilibrium when resting on any point of the curved surface, its centre of
mass must be in the centre of the hemisphere’s flat surface, so x r.
Hence from part a h2 2hr 5r 2
2(h 4r) r
h2 2hr 5r 2 2rh8r 2
h2 3r 2
h 3r
61 a Let x be distance of the centre of mass from AB on the axis of symmetry in the direction away
from O. Using the formulae for standard uniform bodies:
Shape Mass Distance of centre
of mass from AB
Hemisphere M 3
8r
Cone m 3
4r-
Toy m + M x
Taking moments about AB:
(m M )x 3
8Mr -
3
4mr
3r
8( M - 2m)
x 3( M - 2m)
8( M m)r as required
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61 b Let D be the point on the axis of symmetry vertically above B when the toy is placed on a
horizontal surface. The toy will not remain in equilibrium if its centre of mass is not on the line
segment OD, i.e if x > CD .
From the diagram: tana
r
3r
CD
rCD
1
3r
So x >CD 3( M - 2m)
8( M m)r >
1
3r
9( M - 2m) > 8( M m)
M > 26m as required
62 a The centre of mass lies on the axis of symmetry OX. Let the distance of the centre of mass of S
from O be x and let it be at point G.
Using
x y2xò dx
y2 dxò
and substituting y
2 rx gives:
x rx
2dx
0
y
òrxdx
0
y
ò
13rx3 0
y
12rx
2 0
y
1
3r 4
1
2r3
2
3r
b The solid will not topple providing G is vertically above a point on the plane face in contact with
the inclined plane. The maximum value of a is when G is directly above N, the edge of the solid.
tana r
r -2
3r
r 1
3r 3
a 72° (to the nearest degree)
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 58
63 a Using
y 12
y2 dxòydxò
with y sin x gives:
y 12sin2 xdx
0
òsin xdx
0
ò
14
1- cos2xdx0
òsin xdx
0
ò
1
4
x - 12sin2x 0
-cos x 0
1
4
(11)
8
b When S is on the point of toppling, G is above O. Let A be the point midway along the base of S.
2 2tan 4
2
76 (to the nearest degree)
y
°
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 59
64 a Let be the mass per unit volume of the material and x be the distance of the centre of mass of
the solid S from O. Using the formulae for standard uniform bodies:
Shape Mass Mass ratios Distance of centre
of mass from O
Cylinder
(2a)2 3
2a
æ
èçö
ø÷ 6
3
4a
Hemisphere
2
3a
3 2
3
3
8a
Solid, S
6a3 -2
3a3æ
èçö
ø÷
16
3 x
Taking moments about O:
16
3x 6
3
4a -
2
3
3
8a
9
2a -
1
4a
17
4a
x 51a
64 0.797a (3 s.f.)
b On the point of toppling: G is above S – the lowest point on the bottom circular face.
Let X be the centre of the base of the cylinder.
2 64 2 128tan substituting value for from part
3 96 51 45
2
70.6 (3 s.f.)
SX ax
XGa x
a
a
--
°
a
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64 c There are three forces acting on S, its weight, the reaction force and friction.
When S is on the point of sliding F mR 0.8R
Resolving perpendicular to the plane
R( ) : cos 0 cos
Resolving along the plane
R( ) : sin 0 sin
Using 0.8 gives
sin 0.8 cos tan 0.8
So 38.7 (3 s.f.)
R mg R mg
F mg F mg
F R
mg mg
-
-
°
տ
ր
65 a Let be the mass per unit volume of the material and x be distance of the centre of mass of the
bowl B from O. Using the formulae for standard uniform bodies:
Shape Mass Mass ratio Distance of centre
of mass from O
Large hemisphere
2
3(6a)
3 144a3 27
3
8 6a
Small hemisphere
2
3(2a)
3 16
3a
3 1
3
8 2a
Bowl B
2
3(6
3 - 23)a
3 416
3a
3 26 x
Taking moments about O:
26x 27 3
8 6a -1
3
8 2a
243a
4-
3a
4 60a
x 30a
13
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65 b Let y be distance of the centre of mass of the bowl B from O.
Shape Mass Mass ratio Distance of centre
of mass from O
Bowl, B
416
3a3
52 30
13
a
Cylinder 24a3 9 6a + 3a
Combined solid, S
488
3a3
61 y
Taking moments about O:
61y 52 30a
13 9 9a 120a 81a
y 201
61a
c Let N be the point on the plane vertically below the centre of mass of the combined solid G, and
let X be the centre of the cylindrical base of S.
NX GX tan12° 12a-201
61a
æ
èçö
ø÷tan12°
531
61a tan12° 1.85a (3 s.f.)
The solid will topple when point N is outside the base circle; as NX < 2a, S will not topple.
© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free. 62
66 a Let be the mass per unit volume of the material and x be the distance of the centre of mass of
the solid from O. Using the formulae for standard uniform bodies:
Shape Mass Mass ratio Distance from O of
centre of mass
Hemisphere
2
3a3 2
3a
8
Cone
1
3a
3 1
a
4
Solid
1
3a
3 1 x
Taking moments about O:
1 x 23
8a-1
a
4
x a
2
b As the solid is resting in equilibrium, its centre of mass G is above the point of contact N between
the solid and the plane.
From DOGN , and using the result from part a:
sina
a
2a
1
2a
6
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66 c There are three forces acting on the solid, its weight, the reaction force and friction.
For limiting equilibrium, when the solid is about to slip .F Rm
Resolving the forces perpendicular and parallel to the inclined plane:
R( ) : sin
R( ) cos
As , sin cos
sintan
cos
1tan
6 3
1 1So , hence is thesmallest valueof
3 3
F mg
R mg
F R mg mg
a
a
m a m aa
m m aa
a a
m m
ր
տ
� �
�
67 a Let be the mass per unit volume of the material and x be distance of the centre of mass of the
bollard from O. Using the formulae for standard uniform bodies:
Shape Mass Mass ratio Distance from O of
centre of mass
Cone
1
3(3r)2 h 3h
3
4h
Cylinder (4r)2 r 1 6 r
h
r
2
Bollard (16r 3h)r 2 16 3r h x
Taking moments about O:
(16r 3h)x 3h3h
416r h
r
2
æ
èçö
ø÷
9h2
416rh 8r 2
x 32r 2 64rh 9h2
4(16r 3h)
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67 b When the bollard is on the point of toppling, its centre of mass G lies above N a point on the plane
which is at the lowest point on its base.
Let X be the centre point of the base, so GX OX - x
As h 4r, OX h r 5r
And from part a: x 32r 2 256r 2 144r 2
4(28r)
432r
112
27r
7
So GX 5r -27r
7
8r
7
From DGNX : tana NX
GX
4r
8r
7
28
8 3.5
a 74° (to the nearest degree)
68 a Let be the mass per unit volume of the material and x be the distance of the centre of mass of
the tree from O. Using the formulae for standard uniform bodies:
Shape Mass Mass ratio Position of centre of
mass from O
Cylinder r 2h 3 2
h
Cone
1
3(2r)2 h 4
4
h-
Tree
r 2h 14
3
æ
èçö
ø÷ 7 x
Taking moments about O:
7x 3h
2- 4
h
4
h
2
x h
14
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68 b When the tree is on the point of toppling, its centre of mass G lies above S, a point on the desk
which is at the lowest point on its cylindrical base. GX can be found using the result in part a.
tana r
h- h14
7
26
r 7
26
13h
14
æ
èçö
ø÷
1
4h
69 a Using the formulae for standard uniform bodies:
Shape Mass Distance of centre of mass from O
Hemisphere, H 2M 3
8h r
Cylinder, C 3M
h
2
Body 5M x
Taking moments about O:
5M x 2M h3
8r
æ
èçö
ø÷ 3M
h
2 2h
3
4r
3h
2
7h
2
3r
4
x 14h 3r
20
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69 b When the body is on the point of toppling, its centre of mass G lies directly above a point on the
plane which is at the lowest point on its cylindrical base.
From the diagram: tana
r
x
20r
14h 3r
As tana 4
3, this gives
20r
14h 3r
4
3
60r 56h12r
48r 56h
h 48
56r
6
7r
70 a Let be the mass per unit volume of the material in the cylinder and x be the distance of the
centre of mass of the top from O. Using the formulae for standard uniform bodies:
Shape Mass Mass ratio Position of centre of mass
from O
Cylinder 36r3
1 2r
Cone 36r3 1 –r
Toy 72r3 2 x
Taking moments about O:
2x 1 2r 1 (-r) r
x r
2
b In ∆AGX, AX is the radius 3r of the cylinder and X is the centre of the base of the cylinder.
tan AX
GX
AX
OX -OG
3r
4r - x
3r
4r - r2
6
7
47.5° (1 d.p.)
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70 c The toy will not topple if G (the centre of mass) is vertically above a point between B and V.
Let Y be the point directly above B on the axis of symmetry of the toy.
tana 3r
4rfrom DVOB and tana
OY
OB
OY
3rfrom DBOY
So OY
3r
3
4OY
9r
4
But OG r
2so OG < OY
This means that the centre of mass is above the face of the body in contact with the place, so the
toy will not topple.
71 a Total mass 1.5
0
3d
(1 )(2 )h
h h
ò
Rewrite
3
(1 h)(2 h) as
1
1 h-
1
2 h
æ
èçö
ø÷ so:
1.5 1.5
00
1 13 d 3 ln(1 ) ln(2 )
1 2
53(ln 2.5 ln 3.5 ln 2) 3ln 1.07kg (2 d.p.)
3.5
m h h hh h
æ ö - - ç ÷ è ø
-
ò
b Let h be the distance of the centre of mass of the rod from its base.
Taking moments about the base of the rod:
1.5 1.5
0 0
1.52
1.5
0
0
5 3 2 13ln d 3 d
3.5 (1 )(2 ) 2 1
(2 )3 2 ln(2 ) ln(1 ) 3 ln
1
493(ln 4.9 ln 4) 3ln
40
hh h h
h h h h
hh h
h
æ ö -ç ÷ è ø
-
-
ò ò
So 4940
53.5
ln0.57 m (2 d.p.)
lnh
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72 a Total mass m
5
1 4x2dx
0
2.4
ò 51
1 4x2dx
0
2.4
ò 51
2arctan 2x
0
2.4
So m 2.5arctan4.8 3.41kg (2 d.p.)
b Let x be the distance of the centre of mass of the oar from its end.
Taking moments about the end of the oar:
2.5arctan4.8 x 5x
1 4x2dx
0
2.4
ò 5
8ln(1 4x2 )
0
2.4
5
8ln24.04
So 2 ln 24.04
0.58m (2 d.p.)8 arctan 4.8
x
73 a Since the mass per unit length of the rod increases as the distance from A increases, the centre of mass of the rod will be closer to B than to A.
b Total mass m
(10 kx)dx0
20
ò 10x k
2x2
0
20
200 200k
As m = 750, this gives 200 200 750k
k
11
4
c Let x be the distance of the centre of mass of the oar from A.
Taking moments about A:
2020
2 3
00
11 11 22 000 28 000750 10 d 5 2000
4 12 3 3
28 000 112 112m
750 3 3 3 9
x x x x x x
x
æ ö ç ÷ è ø
ò
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74 Let the length of the rod be L and its mass be M. Let x be the distance of the centre of mass of the oar from A.
M (8 x2 )dx0
L
ò 8x x3
3
0
L
8LL3
3
M x (8 x2 )xdx0
L
ò 4x2 x
4
4
0
L
4L2 L
4
4
If the tension in the string at A is T then the tension in the string at B is 2T.
3
3
R( ) : 2 83
8
3 9
LT T Mg L g
L LT g
æ ö ç ÷
è ø
æ ö ç ÷
è ø
Taking moments about A:
4
2
3
2
4 24
28
Mg x LT
LL g LT
LT L g
æ ö ç ÷
è ø
æ ö ç ÷
è ø
Equating the two expressions for T gives:
8L
3
L3
9
æ
èçö
ø÷g 2L
L3
8
æ
èçö
ø÷g
2
3
L2
72 L2 48
L 48 4 3 m
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Challenge
1 The two particles have equal but opposite velocities just before collision at point A. Let their velocities at point A be u and –u respectively.
After collision, P1 travels three times the distance travelled by P2 before colliding at point B. Therefore, the velocities after collision are –3v and v respectively.
Using conservation of momentum:
mu 2m(-u) m(-3v) 2mv
u v
Coefficient of restitution:
1
3 2
u ue
u u
- -
- -
2 a Let the instantaneous speed at any point be w and the normal reaction between the ball and the ring
is N. Then resolving towards the centre of the circle:
N
mw2
R using Newton’s second law
So the frictional force opposite to the direction of motion is given by:
F m
mw2
R
Resolving in the direction of motion:
m
dw
dt -m
mw2
R
Let v be the velocity at time t, so integrating:
2 0
0
1d d
1
1 1
1 1
v t
u
vt
u
w tw R
tw R
tv u R
R u tt
v u R uR
uRv
R u t
m
m
m
m m
m
-
- -
- -
ò ò
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Challenge
2 b From part a
v dx
dt
uR
R umt
The ball completes one revolution in time t when x 2R
1dx0
ò uR
R umtdt
0
t
ò 20
0.510tdt
0
t
ò
2ln(0.510t) 0
t
2(ln(0.510t)- ln0.5)
2ln(1 20t)
20t e2 -1
t e2 -1
20 0.191s (3 s.f.)
3 a R() : T
1T
2T
3cos45° 4Mg
3R( ) : cos 45 0T® °
Therefore T1 + T2 = 4Mg (1)
Taking moments about B:
2 1
5 10 20T T Mg
2 12 4T T Mg (2)
Subtracting (1) from (2) gives T1 = 0 and T2 = 4Mg
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Challenge
3 b First find the centre of mass of the lamina. Let A be the origin. Using geometry, the coordinates of the points on the lamina are:
A(0,0), B(10,0), C(5,0), D15
2,-5 2
æ
èçö
ø÷, E
5
2,-5 2
æ
èçö
ø÷
The x-coordinate of the centre of mass of the lamina is x 5 by symmetry.
The lamina is composed of three equivalent triangles ACE, CED, BCD, each with height 5 2 and
triangle CED has twice the density as the other two triangles.
So the y-coordinate of the centre of mass of the lamina is given by:
5 2 10 2 5 24 2
3 3 3
30 2 5 2
12 2
M y M M M
y
æ ö æ ö æ ö - - -ç ÷ ç ÷ ç ÷ç ÷ ç ÷ ç ÷
è ø è ø è ø
- -
If a mass of 10M is attached to the lamina at B, while AB remains horizontal the position of the
new centre of mass is:
510
14 4 105 20
2
60
7
5 2
7
xM M M
y
x
y
æ öæ ö æ öç ÷ ç ÷ ç ÷ç ÷-è ø è øç ÷
è ø
æ öç ÷æ öç ÷ç ÷ç ÷è ø -ç ÷è ø
The new centre of mass will align itself so that it is vertically below point A.
So angle made by AB with the vertical in this new position is 5 2
arctan 6.72 (3 s.f.)60
æ ö
°ç ÷ç ÷è ø
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Challenge
4 Consider a cylindrical disc at a distance x cm from the base of the hemisphere. The thickness of the
disc is d x and the radius of the disc is 122 2)(r x- so that the volume of the disc is 2 2( ) .r x xd -
Let m be the total mass of the hemisphere:
2 2 2 2 2 3 4
00
3 4
5 2 5( )(5 2)d 2
2 3 4
4 5
3 4
rr
m r x x x r x r x x x
r r
- - -
ò
Let x be the distance of the centre of mass of the hemisphere from its plane face. Taking moments about the face:
3 4 2 2 2 2 2 3 4 5
00
4 5
2
2
4 5 5 1( )(5 2) d
3 4 3 2
2
2 3
26 82 3
4 5 16 15
3 4
rr
r r x r x x x x r x r x x x
r r
r rr r
xr r
æ ö - - -ç ÷ è ø
ò
When the hemisphere is on the point of tipping, tan(arctan 2)r
x
26 8So 2
16 15
12 16 16 15
4 cm
r rr
r
r r
r
æ öç ÷è ø