Post on 19-Aug-2018
TABLE OF CONTENTS 2
CHAPTER 1
Calculus in Mechanics
3 CHAPTER 2
Projectiles
4 CHAPTER 3
Turning Effect of Forces
6 CHAPTER 4
Momentum & Impulse
8 CHAPTER 5
Centre of Mass
11 CHAPTER 6
Circular Motion
13 CHAPTER 7
Elasticity
15 CHAPTER 8
Oscillatory Motion
15 CHAPTER 9
Moment of Inertia
CIE A-LEVEL FURTHER MATHEMATICS//9231
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1. CALCULUS IN MECHANICS
1.1 Acceleration is a Derivative
๐ฃ =๐๐
๐๐ก ๐ =
๐๐ฃ
๐๐ก= ๐ฃ.
๐๐ฃ
๐๐
Example:
A particle moves with acceleration, ๐ = โ2๐ฃ2 where ๐ฃ
is velocity. Initially, particle at 0 with ๐ฃ = 2
i. Find expression for ๐ฃ, in terms of ๐
ii. Find expression for ๐ฃ, in terms of ๐ก
Solution:
Part (i):
Express acceleration as a derivative with displacement
๐ฃ.๐๐ฃ
๐๐ = โ2๐ฃ2
๐ฃ. ๐๐ฃ = โ2๐ฃ2. ๐๐
Separate the variables and integrate
โซ๐ฃ
๐ฃ2. ๐๐ฃ = โซ โ2. ๐๐
ln ๐ฃ = โ2๐ + ๐
Substitute given information to find ๐
ln 2 = 0 + ๐ โด ๐ = ln 2
Rearrange found equation to make ๐ฃ the subject:
ln ๐ฃ = โ2๐ + ln 2
ln ๐ฃ โ ln 2 = โ2๐
ln (๐ฃ
2) = โ2๐
๐ฃ
2= ๐โ2๐ โด ๐ฃ = 2๐โ2๐
Part (ii):
Express acceleration as a derivative with time
๐๐ฃ
๐๐ก= โ2๐ฃ2
๐ฃโ2. ๐๐ฃ = โ2. ๐๐ก
โซ ๐ฃโ2. ๐๐ฃ = โซ โ2. ๐๐ก
Integrate the expression
โ๐ฃโ1 = โ2๐ก + ๐ Substitute given information to find ๐
โ(2)โ1 = โ2(0) + ๐ โด ๐ = โ1
2
Rearrange found equation to make ๐ฃ the subject:
โ๐ฃโ1 = โ2๐ก โ1
2
๐ฃโ1 = 2๐ก +1
2
๐ฃ =1
2๐ก +1
2
=2
4๐ก + 1
1.2 Variable Forces โข Bodies undergo variable acceleration due to the effect of
variable forces e.g. gravitational fields
โข Important to put negative sign if decelerating e.g. when
a body falls in resistive medium
โข Deriving an expression for force in terms of velocity and
displacement:
๐น = ๐๐ ๐ = ๐ฃ๐๐ฃ
๐๐ฅ
๐น = ๐๐ฃ๐๐ฃ
๐๐ฅ
โข Deriving an expression for force in terms of work done:
For an object moving from ๐ฅ1 to ๐ฅ2 the change in kinetic
energy or work done can be defined as:
๐ = โซ ๐น๐ฅ2
๐ฅ1
๐๐ฅ
โด๐๐
๐๐ฅ= ๐น
โข Deriving an expression of power in terms of force and
velocity
๐ =๐๐
๐๐ก=
๐๐
๐๐ฅร
๐๐ฅ
๐๐ก= ๐น๐ฃ
(IM) Ex: 9Misc Question 10:
A car of mass 1200kg is travelling on a straight
horizontal road, with its engine working at a constant
rate of 25kW. Given that the resistance to motion of
the car is proportional to the square of its velocity and
that the greatest constant speed the car can maintain is
50ms-1, show that 125000 โ ๐ฃ3 = 6000๐ฃ2 ๐๐ฃ
๐๐ฅ , where
๐ฃ ms-1 is the velocity of the car when its displacement
from a fixed point on the road is ๐ฅ metres. Solution:
Write down known facts:
๐ = 25000 ๐น = ๐๐ฃ2 ๐ = 1200
Find ๐:
๐น = ๐๐ฃ2 ๐
๐ฃ= ๐๐ฃ2
25000
50= ๐(502)
๐ = 0.2
Write an equation that relates the forces:
๐ธ๐๐๐๐๐ ๐๐๐๐๐ = ๐๐๐๐ ๐๐๐๐๐ + ๐ ๐๐ ๐๐ ๐ก๐๐ฃ๐ ๐๐๐๐๐ ๐
๐ฃ= ๐๐ฃ
๐๐ฃ
๐๐ฅ+ ๐๐ฃ2
25000
๐ฃ= 1200๐ฃ
๐๐ฃ
๐๐ฅ+
1
5๐ฃ2
CIE A-LEVEL FURTHER MATHEMATICS//9231
PAGE 3 OF 16
125000
๐ฃ= 6000๐ฃ
๐๐ฃ
๐๐ฅ+ ๐ฃ2
125000 = 6000๐ฃ2๐๐ฃ
๐๐ฅ+ ๐ฃ3
Hence find distance covered by the car in increasing its
speed from 30 ms-1 to 45 ms-1 by forming an integral to
define displacement in terms of velocity:
๐๐ฃ
๐๐ฅ=
125000 โ ๐ฃ3
6000๐ฃ2
๐๐ฅ
๐๐ฃ=
6000๐ฃ2
125000 โ ๐ฃ3
๐๐ฅ =6000๐ฃ2
125000 โ ๐ฃ3๐๐ฃ
๐ฅ = โซ6000๐ฃ2
125000 โ ๐ฃ3๐๐ฃ
Substitute the info we know:
๐ฅ = โซ6000๐ฃ2
125000 โ ๐ฃ3๐๐ฃ
45
30
๐ฅ = 6000 โซ๐ฃ2
125000 โ ๐ฃ3๐๐ฃ
45
30
๐ฅ = 6000 [โln(125000 โ ๐ฃ3)
3]
30
45
๐ฅ = 2125
2. PROJECTILES
2.1 Initial Velocity โข Consider the following diagram:
โข This shows that an initial speed ๐ can be defined in
vector notation: ๐ฎ = ๐ cos ๐ ๐ข + ๐ sin ๐ ๐ฃ
โข In this notation: o ๐ defines the angle of elevation from the horizontal o ๐ข defines the horizontal unit vector o ๐ฃ defines the vertical unit vector
2.2 Acceleration โข All projectiles are affected by gravity
โข Gravity effects only the vertical component of speed
โข We can thus define acceleration in vector notation:
๐ = 0๐ข โ ๐๐ฃ = โ๐๐ฃ
โข In this notation:
o ๐ is the acceleration due to gravity
o ๐ข defines the horizontal unit vector
o ๐ฃ defines the vertical unit vector
โข Note: this model assumes that air resistance is negligible
2.3 Final Velocity โข This defines the velocity of the projectile after ๐ก seconds
of travel
โข Final velocity, according to the SUVAT equations, is also
dependent on the initial velocity and any acceleration
acting on the projectile:
๐ฃ = ๐ข + ๐๐ก
โข Above is a scalar equation which can be written as a
vector equation:
๐ฏ = ๐ฎ + ๐๐ก
โข This can be further simplified into vector notation by
substituting previous equations:
๐ฏ = ๐ cos ๐ ๐ข + ๐ sin ๐ ๐ฃ โ ๐๐ก๐ฃ
๐ฏ = ๐ cos ๐ ๐ข + (๐ sin ๐ โ ๐๐ก)๐ฃ
โข Substituting any value of ๐ก into the above equation
produces the velocity of the projectile at that time
2.4 Displacement โข This defines the distance the particle has moved from
the origin
โข Displacement, according to the SUVAT equations, is
dependent on initial velocity and acceleration, or final
velocity and acceleration:
๐ = ๐ข๐ก +1
2๐๐ก2
๐ = ๐ฃ๐ก โ1
2๐๐ก2
โข These are scalar equations and can also be written as vector equations:
๐ซ = ๐ฎ๐ก +1
2๐๐ก2
๐ซ = ๐ฏ๐ก โ1
2๐๐ก2
โข Substituting previous equations into either of the above
equations defines displacement in vector notation:
๐ซ = (๐ cos ๐ ๐ข + ๐ sin ๐ ๐ฃ)๐ก โ1
2๐๐ก2๐ฃ
๐ซ = (๐๐ก cos ๐)๐ข + (๐๐ก sin ๐ โ1
2๐๐ก2) ๐ฃ
โข Substituting any value of ๐ก into the above equation
produces the displacement of the projectile at that time
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2.5 Special Events โข Time at Maximum Vertical Height: this is the time at
which the projectile reaches its max height, at a given
initial velocity and angle of elevation
o This occurs when vertical component of velocity is 0:
๐ sin ๐ โ ๐๐ก = 0
โด ๐๐๐๐๐ =๐ sin ๐
๐
โข Maximum Vertical Height: this is the maximum height
the projectile reaches during its flight
o This occurs when vertical component of velocity is 0:
๐๐๐ = ๐(๐๐๐๐๐) sin ๐ โ1
2๐(๐๐๐๐๐)2
โข Horizontal Range: this is the horizontal distance that the
projectile covers at a given initial velocity and angle of
elevation
o This occurs when the vertical displacement is 0:
๐๐ก sin ๐ โ1
2๐๐ก2 = 0
o Find ๐ก and substitute into the equation below:
๐๐ = ๐๐ก cos ๐
โข Maximum Horizontal Range: this is the MAXIMUM
possible horizontal distance that the projectile can cover
at a given initial velocity
o This occurs when the angle of elevation is 45ยฐ:
๐๐๐ = ๐๐กรโ2
2
(IM) Ex 7C: Question 9:
A ball was projected at an angle of 60ยฐ to the horizontal.
One second later another ball was projected from the
same point at an angle of 30ยฐ to the horizontal. One
second after the second ball was released, the two balls
collided. Show that the velocities of the balls were
12.99ms-1 and 15ms-1. Take the value of ๐ to be 10ms-2. Solution:
Visualise the scenario:
At zero time:
At ๐ก = 1:
At ๐ก = 2:
Let velocity for first ball be ๐1 and second ball be ๐2
Displacement from origin is equal for both at impact
โด ๐ซ1 = ๐ซ2
๐ซ1 = (๐1(2) cos 60)๐ข + (๐1(2) sin 60 โ 5(2)2)๐ฃ
๐ซ2 = (๐2(1) cos 30)๐ข + (๐2(1) sin 30 โ 5(1)2)๐ฃ
Equate horizontal components:
โด 2๐1 cos 60 = ๐2 cos 30
๐1 =โ3
2๐2
Equate vertical components and substitute info above:
โด 2๐1 sin 60 โ 20 = ๐2 sin 30 โ 5
(2 (โ3
2) (
โ3
2) โ
1
2) ๐2 = 15
๐2 = 15ms-1
Find ๐1
๐1 =โ3
2(15) = 12.99ms-1
3. TURNING EFFECTS OF FORCES
3.1 Moment of a Force Moment of a force = |๐ |ร๐
|๐ |: magnitude of the force
๐: perpendicular distance from pivot to point of force
โข Units: Newtonmetre (Nm)
โข Moments are vector quantities and act clockwise or
anticlockwise around pivot
โข Clockwise is generally considered as positive direction
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โข Principle of Moments: when a system is in equilibrium
the sum of anticlockwise moments is equal to the sum of
clockwise moments
anti-clockwise moments = clockwise moments
(IM) Ex 10A: Question 4a:
The diagram shows an aerial view of a revolving door.
Four people are exerting forces of 40N, 60N, 80N and
90N as shown. Find the distance ๐ฅ if the total moment
of the forces about ๐ is 12Nm
Solution:
Use the sum of turning forces equation:
clockwise moments + anticlockwise moments = 12Nm
Find clockwise moments:
(80ร1.6) + (60ร0.8) = 128 + 48 = 176
Find anticlockwise moments:
โ((90ร1.6) + (40ร๐ฅ)) = โ(144 + 40๐ฅ)
Substitute back into formula and solve for ๐ฅ:
176 + (โ(144 + 40๐ฅ)) = 12
176 โ 144 โ 40๐ฅ = 12
๐ฅ = 0.5m
3.2 Unknown Forces โข Magnitude of forces may not always be given
โข Eliminate an unknown/unwanted force by making the
point on which it acts the pivot
(IM) Ex 10A: Question 11:
A uniform plank is 12m long and has mass 100kg. It is
placed on horizontal ground at the edge of a cliff, with
4m of the plank projecting over the edge.
i. How far out from the cliff can a man of mass
75kg safely walk?
ii. The man wishes to walk to the end of the
plank. What is the minimum mass he should
place on the other end of the plank to do this?
Solution :
Part (i)
Draw up diagram of given scenario:
Use the principle of moments and solve for ๐ฅ:
anti-clockwise moments = clockwise moments
100๐ร2 = 75๐ร๐ฅ
โด ๐ฅ =100๐ร2
75๐= 2.67
Part (ii)
Draw up diagram of given scenario:
Minimum mass therefore maximum distance from
pivot
Use the principle of moments and solve for ๐ฅ:
anti-clockwise moments = clockwise moments
(100๐ร2) + (๐๐ร8) = 75๐ร4
โด ๐ =(75๐ร4) โ (100๐ร2)
๐ร8= 12.5
3.3 Forces in Different Directions โข Forces may act at an angle to the plane
โข Equilibrium maintained using components of forces
(IM) Ex 10A: Question 9:
A uniform ladder of mass 20kg and length 3m rests
against a smooth wall with the bottom of the ladder on
smooth horizontal ground and attached by means of a
light inextensible string, 1m long, to the base of the wall
i. Find the tension in the string.
ii. If the breaking strain of the string is 250N, find
how far up the ladder a man of mass 80kg can
safely ascend.
CIE A-LEVEL FURTHER MATHEMATICS//9231
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Solution:
Part (i)
Draw up diagram of given scenario:
Find the angle of elevation, ๐, from the horizontal:
cos ๐ =1
3
๐ = 70.5ยฐ
Use the principle of moments and solve for ๐:
1.5ร20๐ร cos 70.5 = 3ร๐ร cos 19.5
๐ = 34.7 Part (ii)
In above scenario assume that the man can take any
position on the ladder, call it ๐ฅ
Use the principle of moments and solve for ๐ฅ:
anti-clockwise moments = clockwise moments
((1.5)20๐ + (๐ฅ)80๐) cos 70.5 = (3)(250) cos 19.5
๐ฅ = 2.32
4. MOMENTUM & IMPULSE
4.1 Momentum Momentum = ๐๐ฃ
โข Conservation of linear momentum: The total
momentum of a system in a particular direction remains
constant unless an external force is applied
Momentum Before Collision = Momentum After Collision
For example: find the speed after the collision
Momentum before = (6ร4) + (โ4ร2) = 16
Momentum after = 6ร๐ฃ = 6๐ฃ
16 = 6๐ฃ
๐ฃ = 2.67 ms-1
4.2 Impulse Impulse = change of momentum = ๐๐ฃ โ ๐๐ข
Impulse = force ร time
Impulse of a variable force:
Impulse = โซ ๐น ๐๐ก๐ก
0
(IM) Ex 13A: Example 4:
Steve kicks a ball mass 0.8kg along the ground at a
velocity of 5ms-1 towards Monica. She kicks it back
towards him, lofting it so that it leaves her foot at 8ms-1
and with an elevation of 40ยฐ to the horizontal. Find the
magnitude and direction of the impulse from her kick. Solution:
Sketch a diagram of what is happening
Find the initial momentum
๐ฅ-direction = 0.8รโ5 = โ4 Resolving the final velocity ๐ฅ-direction = 0.8ร8ร cos 40 = 4.903 ๐ฆ-direction = 0.8ร8ร sin 40 = 4.114 Find the change in momentum i.e. impulse Impulse = (4.903๐ข + 4.114๐ฃ) โ (โ4๐ข) = 8.903๐ข + 4.114๐ฃ Hence find the magnitude and direction
Magnitude: โ8.9032 + 4.1142 = 9.808
Direction: tan ๐ =4.114
8.903 hence ๐ = 24.8ยฐ
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4.3 Newtonโs Law of Restitution When two objects collide directly,
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐
๐ด๐๐๐๐๐๐โ ๐๐๐๐๐= ๐
โข The constant ๐ is called the coefficient of restitution for
the objects and takes a value between 0 and 1
o ๐ = 0; totally inelastic impact, no rebounding
o ๐ = 1; perfectly elastic impact, speeds unchanged
In reality, perfect elasticity does not occur hence
0 โค ๐ < 1
4.4 Oblique Impacts Impact between a smooth sphere and a fixed surface:
โข The impulse on the sphere acts perpendicular to the
surface, along line of impact
โข Newtonโs law of restitution applies to the component of
the velocity of the sphere along line of impact
โข The component of the velocity of the sphere along plane
of impact is unchanged.
Impact between two spheres:
โข Impulse affecting each sphere also acts along line of
impact
โข Components of velocities of spheres along plane of
impact unchanged
โข Newtonโs law of restitution applies to components of the
velocities of the spheres along line of impact
โข The law of conservation of momentum applies along line
of impact
{S06-P02} Question 1:
A ball drops vertically onto a smooth plane inclined to
the horizontal at an angle ๐ผ. It hits the plane with
speed 8ms-1 and rebounds horizontally. The coefficient
of restitution between the ball and the plane is 1
3. Find
the value of ๐ผ and speed with which the ball
rebounds. Solution:
Sketch a diagram of what is happening
Resolving the initial velocity Along the plane 8 sin ๐ผ Along the line 8 cos ๐ผ Resolving the final velocity Along the plane ๐ฃ cos ๐ผ Along the line ๐ฃ sin ๐ผ The velocity along the plane so can form the equation:
8 sin ๐ผ = ๐ฃ cos ๐ผ 8
๐ฃsin ๐ผ = cos ๐ผ
Use velocities along line with coefficient of restitution eqn.
๐ =1
3=
๐ฃ sin ๐ผ
8 cos ๐ผ
cos ๐ผ =3๐ฃ
8sin ๐ผ
Equate equations and find ๐ฃ 3๐ฃ
8sin ๐ผ =
8
๐ฃsin ๐ผ
๐ฃ =8โ3
3 ms-1
Use initial equation and find ๐ผ sin ๐ผ
cos ๐ผ=
๐ฃ
8
๐ผ = 30ยฐ
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5. CENTRE OF MASS โข Centre of Mass: centre of gravity of the system when it
is placed in a gravitational field such that each part of
system is subject to the same gravitational acceleration
โข Centroid: geometrical centre; coincides with the centre
of mass when the object is made of a uniformly dense
material
5.1 Finding Centre of Mass โข We can find the centre of mass by taking moments
โข If each mass ๐๐ has position vector ๐ซ๐ then the position
vector of the centre of mass ๏ฟฝฬ ๏ฟฝ is
๏ฟฝฬ ๏ฟฝ = โ ๐๐๐ซ๐
โ ๐๐
โด (๐1 + ๐2 + โฏ ) (๏ฟฝฬ ๏ฟฝ๏ฟฝฬ ๏ฟฝ
) = ๐1 (๐ฅ1
๐ฆ1) + ๐2 (
๐ฅ2
๐ฆ2) + โฏ
(IM) Ex 11A: Question 8:
A light triangular framework ๐ด๐ต๐ถ has ๐ด๐ต = 4.6cm,
๐ด๐ถ = 6.3cm and angle ๐ต๐ด๐ถ = 68ยฐ. Masses of 3kg, 6kg
and 8kg are placed at ๐ด, ๐ต and ๐ถ respectively. The
framework is suspended from ๐ด. Find the angle which
๐ด๐ต makes with the vertical. Solution:
Draw diagram of scenario:
Take ๐ด as the origin and find position vectors of ๐ต and
๐ถ using simple Pythagoras theorem:
Substitute into the equation and solve for coordinates:
(3 + 6 + 8) (๏ฟฝฬ ๏ฟฝ๏ฟฝฬ ๏ฟฝ
) = 3 (00
) + 6 (1.74.3
) + 8 (6.30
)
(๏ฟฝฬ ๏ฟฝ๏ฟฝฬ ๏ฟฝ
) = (3.5731.505
)
Draw diagram as described by scenario:
Find ๐ using Pythagoras:
๐ = 68 โ tanโ1 (1.5
3.6)
๐ = 45.2ยฐ
5.2 Centre of Mass of Rigid Bodies 1-Dimensional Objects:
โข Uniform rod: centre of mass lies at midpoint of the rod
2-Dimensional Objects:
โข Uniform Rectangular Lamina: centre of mass is at the
intersection of diagonals
๐บ = (๐๐๐๐๐๐๐๐ก ๐๐ ๐ด๐ต๐๐๐๐๐๐๐๐ก ๐๐ ๐ถ๐ท
)
โข Uniform Circular Lamina: centre of mass is at the centre
of the circle
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โข Uniform Triangular Lamina:
๐ด = (๐ฅ1
๐ฆ1) , ๐ต = (
๐ฅ2
๐ฆ2) , ๐ถ = (
๐ฅ3
๐ฆ3)
โด ๐บ = (
1
3(๐ฅ1 + ๐ฅ2 + ๐ฅ3)
1
3(๐ฆ1 + ๐ฆ2 + ๐ฆ3)
)
โข Uniform Semicircular Lamina:
โ =4๐
3๐
3-Dimenionsnal Objects:
โข Uniform Solid Prism: centre of mass lies at the centre of
mass of the cross-section and in the midpoint of length
โข Uniform Cone: centre of mass lies in the centre of the
base and height in ratio 3:1 to its height
๐๐บ: ๐บ๐ถ = 3: 1
โข Uniform Hemisphere: centre of mass lies in the centre of
the base and height 3
8 of the radius
โ =3๐
8
5.3 Composite Bodies โข Split composite body into simple geometrical shapes
โข Find the centre of mass of each shape individually
โข By taking moments with vectors, find the centre of mass
of the composite body
โข If the separate geometrical shapes have different
densities, use ๐ร๐ instead of just ๐
โข Therefore in general:
(๐ฃ1๐1 + ๐ฃ2๐2 + โฏ ) (๏ฟฝฬ ๏ฟฝ๏ฟฝฬ ๏ฟฝ
) = ๐ฃ1๐1 (๐ฅ1
๐ฆ1) + ๐ฃ2๐2 (
๐ฅ2
๐ฆ2) + โฏ
โข For a lamina with a hole in it, find the centre of mass of
the lamina as a whole, then find the centre of mass of
the hole and use moments to find the centre of mass of
the shaded part
In these cases:
๐๐ โ๐๐๐๐ (๐ฅ๐ โ๐๐๐๐
๐ฆ๐ โ๐๐๐๐) = ๐ (
๏ฟฝฬ ๏ฟฝ๏ฟฝฬ ๏ฟฝ
) โ ๐โ๐๐๐ (๐ฅโ๐๐๐
๐ฆโ๐๐๐)
(IM) Ex 11B: Question 7:
The diagram shows a box consisting of a cylinder of
diameter 60cm and height 80cm, with a hollow
cylindrical interior and a hollow hemispherical cap. The
thickness of the wall, cap and base is 10cm throughout.
Find the height of the centre of mass of the empty box
above its base.
CIE A-LEVEL FURTHER MATHEMATICS//9231
PAGE 10 OF 16
Solution:
Draw diagram in 2-D since 3-D not required:
Centre of mass of outside:
For the outer semicircle:
โ =3๐
8
3ร30
8= 11.25
โด โ๐๐๐โ๐ก = 80 + 11.25 = 91.25
Then use moments to calculate total outer:
72000๐ (3040
) + 18000๐ = 90000๐ (๏ฟฝฬ ๏ฟฝ1
๏ฟฝฬ ๏ฟฝ1)
(๏ฟฝฬ ๏ฟฝ1
๏ฟฝฬ ๏ฟฝ1) = (
3050.25
)
Centre of mass of inside:
For the inner semicircle:
โ =3๐
8
3ร20
8= 7.5
โด height = 80 + 7.5 = 87.5
Then use moments to calculate the total inner:
28000๐ (3045
) +16000
3๐ (
3087.5
) =100000
3๐ (
๏ฟฝฬ ๏ฟฝ2
๏ฟฝฬ ๏ฟฝ2)
(๏ฟฝฬ ๏ฟฝ2
๏ฟฝฬ ๏ฟฝ2) = (
3051.8
)
Centre of mass of the object:
Use the formula:
๐๐ โ๐๐๐๐ (๐ฅ๐ โ๐๐๐๐
๐ฆ๐ โ๐๐๐๐) = ๐ (
๏ฟฝฬ ๏ฟฝ๏ฟฝฬ ๏ฟฝ
) โ ๐โ๐๐๐ (๐ฅโ๐๐๐
๐ฆโ๐๐๐)
Substitute the given scenario into these variables:
๐๐๐ข๐ก๐๐ (๏ฟฝฬ ๏ฟฝ1
๏ฟฝฬ ๏ฟฝ1) = ๐ (
๏ฟฝฬ ๏ฟฝ๏ฟฝฬ ๏ฟฝ
) โ ๐๐๐๐๐๐ (๏ฟฝฬ ๏ฟฝ2
๏ฟฝฬ ๏ฟฝ2)
90000๐ (30
50.25) =
170000
3๐ (
๏ฟฝฬ ๏ฟฝ๏ฟฝฬ ๏ฟฝ
) โ100000
3๐ (
3051.8
)
(๏ฟฝฬ ๏ฟฝ๏ฟฝฬ ๏ฟฝ
) = (30
49.34)
โด height of centre of mass above its base is 49.34 cm
5.4 Sliding and Toppling โข Sliding: when the resultant force on the object parallel
to the plane of contact becomes non-zero, that is, the
limiting friction force is exceeded by the other forces,
the object will slide.
โข Toppling: when total moment of the forces acting on the
object becomes non-zero, the object will topple over.
(IM) Ex 11: Example 12:
A prism of mass ๐, having a cross-section as shown,
rests in a rough horizontal plank ๐๐. The coefficient of
friction between the plank is 0.4. The end ๐ of the plank
is gradually raised until the equilibrium is broken. Will
the prism slide or topple?
Solution:
First find the angle needed to slide:
Draw diagram at hypothetical angle
Resolve forces horizontal to slope:
๐น โ ๐๐ sin ๐ = 0
โด ๐น = ๐๐ sin ๐
Resolve forces vertical to slope:
๐ โ ๐๐ cos ๐ = 0
โด ๐ = ๐๐ cos ๐
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Form a relationship by dividing the unknowns: ๐น
๐ = tan ๐
Substitute ๐น = 0.4๐ into the equation
tan ๐ = 0.4
๐ = tanโ1 0.4 = 21.8
Now find angle needed to topple:
Find centre of mass by using composite bodies rule:
๐ (๏ฟฝฬ ๏ฟฝ๏ฟฝฬ ๏ฟฝ
) =2
3๐ (
0.150.15
) +1
3๐ (
0.10.4
)
(๏ฟฝฬ ๏ฟฝ๏ฟฝฬ ๏ฟฝ
) = (0.130.23
)
Prism topples when centre of mass is vertically above
point ๐ด
Calculate this required angle using Pythagoras:
tan ๐ =0.13
0.23
๐ = 29.7
From this we can see that ๐ < ๐, thus when the
equilibrium is broken, the object will slide.
6. CIRCULAR MOTION
6.1 The Basics โข Consider the following diagram:
โข Particle ๐ moves in a circular path (the red line)
โข Angular Displacement:
๐ยฐ
โข Angular Velocity:
๐ =2๐
๐
Where ๐ is the time period for one complete revolution
โข Liner Displacement:
๐ = ๐๐
โข Liner Velocity:
๐ฃ = ๐๐
And is always tangential to the circle
โข Acceleration:
๐ = ๐2๐ =๐ฃ2
๐
Acts towards the centre of the circle
IM-Ex.16C: Question 8:
A particle of mass 3kg is placed on a rough, horizontal
turntable and is connected to its centre by a light,
inextensible string of length 0.8m. The coefficient of
friction between the particle and the turntable is 0.4.
The turntable is made to rotate at a uniform speed. If
the tension in the string is 50N, find the angular speed
of the turntable. Solution:
Draw diagram of scenario:
Consider the resultant force:
๐ท = ๐ + ๐น
3๐ = 50 + ๐๐
3๐ = 50 + 0.4(๐๐)
3๐ = 50 + 0.4(3)(9.81)
โด ๐ = 20.6 ๐๐ โ2
From this we can find the angular speed of the particle
๐ = ๐2๐
๐2๐ = 20.6
๐ = โ20.6
0.8= 5.07 ๐๐๐๐ โ1
The angular speed of the particle is equal to that of the
turntable.
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6.2 Conical Pendulum โข In this scenario a body is attached to a fixed point by
string and travels in a horizontal circle below that point
{S10-P51}: Question 3:
A particle of mass 0.24kg is attached to one end of a
light inextensible string of length 2m with the other
attached to a fixed point. The particle moves with
constant speed in a horizontal circle. The string makes
an angle ๐ with the vertical, and the tension in the
string is ๐N. The acceleration of the particle is 7.5ms-2.
i. Show that tan ๐ = 0.75 & find the value of ๐
ii. Find the speed of the particle.
Solution:
Part (i):
Simplify the diagram of the scenario:
Resolve forces vertically:
๐ cos ๐ = 0.24๐
Resolve forces horizontally:
๐ sin ๐ = 0.24๐
Form a relationship by dividing the two equations: ๐ sin ๐
๐ cos ๐=
0.24ร7.5
0.24ร10
โด tan ๐ = 0.75
Find ๐ and substitute into original equation to find ๐:
๐ = 36.9ยฐ
๐ =0.24ร10
cos 36.9= 3๐
Part (ii):
Simple algebraic manipulation and Pythagoras:
๐ =๐ฃ2
๐
๐ฃ = โ๐๐ = โ7.5ร2 sin 36.9 = 3 ๐๐ โ1
6.3 Banked Curves โข Curved sections of public roads/railway tracks banked,
enabling cars/trains to travel more quickly around curves
โข Normal reaction between the vehicle and the track has a
horizontal component when track is banked
โข This component helps to provide central force needed to
keep vehicle travelling in a circle
(IM) Ex 16D: Question 10:
A car is travelling round a curve of radius 150m banked
at 25ยฐ to the horizontal. The coefficient of friction
between the wheels and the road is 0.4.
a) What is the ideal speed of the car round the
curve: that is, no lateral frictional force?
b) What is the maximum safe speed of the car?
c) What is the minimum safe speed of the car? Solution:
Part (a):
Draw diagram of scenario:
Resolve forces vertically:
๐ cos 25 = ๐๐
Resolve forces horizontally:
๐ sin 25 =๐๐ฃ2
๐
Form a relationship by dividing the two:
tan 25 =๐ฃ2
150๐
๐ฃ = 26.2 ๐๐ โ1
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Part (b):
At max speed, car about to slip upwards โด frictional
force would act towards the centre of the circle
Resolve forces vertically (direction of gravity):
๐๐ + 0.4๐ sin 25 = ๐ cos 25
Resolve forces horizontally (direction of centripetal):
๐๐ฃ2
๐= 0.4๐ cos 25 + ๐ sin 25
Pull out common factors from both sides:
๐(๐) = ๐ (cos 25 โ 0.4 sin 25)
๐ (๐ฃ2
150) = ๐ (0.4 cos 25 + sin 25)
Form a relationship by dividing these equations:
๐ฃ2
150๐=
0.4 cos 25 + sin 25
cos 25 โ 0.4 sin 25
๐ฃ = 36.9 ๐๐ โ1 Part (c):
At min speed, car about to slip downwards โด frictional
force would act away from the centre of the circle
Resolve forces vertically (direction of gravity):
๐๐ = ๐ cos 25 + 0.4๐ sin 25
Resolve forces horizontally (direction of centripetal):
๐๐ฃ2
๐= ๐ sin 25 โ 0.4๐ cos 25
Pull out common factors from both sides:
๐(๐) = ๐ (cos 25 + 0.4 sin 25)
๐ (๐ฃ2
150) = ๐ (sin 25 โ 0.4 cos 25)
Form a relationship by dividing these equations:
๐ฃ2
150๐=
sin 25 โ 0.4 cos 25
cos 25 + 0.4 sin 25
๐ฃ = 9.06 ๐๐ โ1
6.4 Circular Motion with Non-Uniform Speed For a particle moving with non-uniform speed:
โข The speed of the particle is |๐ฏ| = ๐๏ฟฝฬ๏ฟฝ
where ๏ฟฝฬ๏ฟฝ is not constant
โข The particle has a tangential component of acceleration
of ๐๏ฟฝฬ๏ฟฝ
โข The particle has a radial component of acceleration of
โ ๐๏ฟฝฬ๏ฟฝ2, which is directed towards the centre of the circle
6.5 Motion in a Vertical Circle โข When moving in a vertical circle a body is in circular
motion with non-uniform speed
โข When body cannot leave circle:
e.g. body rotating on end of rigid rod
o If energy of system is sufficient, the body rotates in a
complete circle
o If energy insufficient, the body cannot reach the
highest point of the circle and so oscillates between
two symmetrical points at each of which its speed is
instantaneously zero
โข When body can leave circle:
e.g. body rotating on end of string
o If energy of system is sufficient, the body rotates in a
complete circle
o If energy of system is so low that body cannot rise
beyond level of centre of the circle, it oscillates
between two symmetrical positions, at each of which
its speed is instantaneously zero
o If energy of system is such that the body can rise
above the level of the centre of the circle without
being enough to carry it completely around the circle,
it will leave the circle and go into projectile motion
โข Solve question by using the conservation of energy
7. ELASTICITY โข Springs can be compressed and stretched
โข Elastic strings can only stretch; they go slack
7.1 Hookeโs Law ๐ = ๐๐ฅ
o ๐ is the magnitude of tension
o ๐ฅ is the extension or compression
o ๐ is the spring constant a.k.a. stiffness
โข Combining spring constants ๐
o Springs in parallel:
๐๐ = ๐1 + ๐2 + โฏ
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o Springs in series:
1
๐๐=
1
๐1+
1
๐2+ โฏ
7.2 Modulus of Elasticity
๐ =๐๐ฅ
๐
o ๐ is the modulus of elasticity
o ๐ is the natural length
โข Relating spring constant ๐ and modulus of elasticity ๐:
๐ = ๐๐
7.3 Scenarios โข If a mass is hanging at one end of a spring
with the other attached to a fixed point, the
tension in the spring must be equal to the
weight of the object
โข If two springs are attached between two fixed points and
both are extended, the tension in both must be the
same in order for there to be no net force overall
โข If two springs are attached to fixed points and an object
in the middle, tensions and frictional force must act in
such way that overall force on mass = 0
โข If a spring is attached to an object on a rough surface,
the frictional force acts in the direction opposing tension
in the spring (preventing it to return to its original shape)
โข If a mass is on an incline and held at rest by a
spring, the tension in the spring must be
equal to the component of the weight
parallel to the slope
7.4 Elastic Potential Energy Work done in stretching a string or spring is given by
๐ =1
2๐๐ฅ2
Also gives work done in compressing a spring
โข For a scenario, form equation by the conservation of
energy and can include e.p.e, k.e, g.p.e and work done
against friction
{S10-P51}: Question 3:
A particle ๐ of mass 0.35 kg is attached to mid-point of
a light elastic string of natural length 4m. Ends of the
string are attached to fixed points ๐ด & ๐ต. ๐ hangs in
equilibrium 0.7m vertically below mid-point ๐ of ๐ด๐ต.
i. Find tension in the string and hence show that
the modulus of elasticity of the string is 25N
๐ is now held at 1.8m vertically below ๐, and released
ii. Find speed with which ๐ passes through ๐ Solution:
Part (i):
Find the extension of the string by finding length ๐ด๐
and ๐๐ต using right angled triangles
๐ด๐ = ๐๐ต = โ2.42 + 0.72 = 2.5m
2.5 + 2.5 โ 4 = 1m
The vertical component of tension in the string is
equal to the weight as the system is in equilibrium
2๐ cos ๐ = ๐๐
2๐ร (0.7
2.5) = 0.35ร10
๐ = 6.25
Find the modulus of elasticity by using info calculated
๐ =๐
๐๐ฅ
๐ =6.25ร4
1= 25
Part (ii):
Find the spring constant ๐
๐ =๐
๐=
25
4= 6.25
As before, find the extension of the string
๐ด๐ = ๐๐ต = โ2.42 + 1.82 = 3m
3 + 3 โ 4 = 2m
Form an equation by the conservation of energy,
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e.p.e = gain in k.e. + gain in g.p.e. + e.p.e at ๐
(1
2ร6.25ร22) = (
1
2ร0.35ร๐ฃ2) + (0.35ร10ร1.8)
+ (1
2ร6.25ร0.82)
12.5 = 0.175๐ฃ2 + 6.3 + 2
๐ฃ2 = 24
๐ฃ = ยฑ4.90 ms-1
8. OSCILLATORY MOTION
8.1 Simple Harmonic Motion โข A particle undergoes simple harmonic motion if its
acceleration is directed towards a fixed point and is
proportional to the displacement of the particle from
that point
๏ฟฝฬ๏ฟฝ = โ๐2๐ฅ
8.2 Solutions of the Equation โข Depending upon the scenario the equation of motion
can have different solutions
โข When particle starts at max. displacement from origin:
๐ฅ = ๐ cos ๐๐ก
โข When particle starts at zero displacement from origin:
๐ฅ = ๐ sin ๐๐ก
โข When particle starts at intermediate point from origin:
๐ฅ = ๐ cos(๐๐ก + ๐ผ)
Note: ๐ outside is different from ๐ผ inside
8.3 Velocity โข Velocity for simple harmonic motion can be found by
differentiating the above solutions
Displacement (๐) Velocity (๏ฟฝฬ๏ฟฝ or ๐)
๐ cos ๐๐ก โ๐๐ sin ๐๐ก
๐ sin ๐๐ก ๐๐ cos ๐๐ก
๐ cos(๐๐ก + ๐ผ) โ๐๐ sin(๐๐ก + ๐ผ)
โข If given the displacement at a point, the velocity can be
found by:
๐ฃ2 = ๐2(๐2 โ ๐ฅ2)
8.4 Period and Frequency โข Oscillation of SHM are periodic and have a time period:
๐ =2๐
๐
โข This can determine the frequency of oscillations:
๐ =1
๐=
๐
2๐
8.5 Motion Approximation โข In a simple pendulum, a
bob, attached to a rod or
string, moves from one side to
the other through the
equilibrium position
โข Assume that:
o String or rod is inextensible
o String or rod has no mass
o Bob is a particle
o No air resistance
o No frictional forces
o Angle of swing is small
โข This motion can then be modelled by SHM equations:
o ๐ is the length of the string or rod
o ๐ is the mass of the bob
o ๐ is the angular displacement
o ๐๏ฟฝฬ๏ฟฝ is tangential acceleration
โข Resultant force in the tangential direction is:
โ๐๐ sin ๐
โข Using Newtonโs second law we can derive an equation
relating the acceleration and displacement:
โ๐๐ sin ๐ = ๐๐๏ฟฝฬ๏ฟฝ
๏ฟฝฬ๏ฟฝ = โ๐
๐sin ๐
โข When ๐ is small, sin ๐ โ ๐, causing the equation to be:
๏ฟฝฬ๏ฟฝ = โ๐
๐๐
โข This is an SHM equation where:
๐2 =๐
๐
9. MOMENT OF INERTIA
9.1 Important Definitions and Formulae โข Moment of Inertia: a quantity that expresses a body's
tendency to resist angular acceleration; the sum of the
products of the mass of each particle in a body with the
square of its distance from the axis of rotation
๐ผ = โ๐๐๐๐2
โข Rotational Kinetic Energy: the kinetic energy of an
object due to its rotation; is part of the total kinetic
energy of the system
๐ธ๐ =1
2๐ผ๏ฟฝฬ๏ฟฝ2
where ๏ฟฝฬ๏ฟฝ is the angular velocity; can be written as ๐
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โข Moment of Couple: the rate of change of angular
momentum; better known as torque
๐ถ = ๐ผ๏ฟฝฬ๏ฟฝ
where ๏ฟฝฬ๏ฟฝ is the angular acceleration
โข Moment of Momentum: is the rotational analog of
linear momentum; better known as angular momentum
๐ = ๐ผ๏ฟฝฬ๏ฟฝ
9.2 Parallel & Perpendicular Axes Theorems โข Parallel Axis Theorem:
๐ผ๐ = ๐ผ๐บ + ๐๐2
โข Perpendicular Axis Theorem:
๐ผ๐ง = ๐ผ๐ฅ + ๐ผ๐ฆ
โข When solving problems, the axis required may not be
the same as that given in the standard results. By
combination of symmetry and the theorems above, it is
possible to solve and find the moment of inertia
o E.g. the axis could be perpendicular to a point. For this,
find the moment from the centre and then use the
parallel theorem to solve
o E.g. the axis may not be perpendicular. Use the result
of the perpendicular axis, symmetry and the
perpendicular axis theorem to solve
9.3 Energy in a Rotating Body โข Total Energy: sum of rotational
kinetic energy of the body and
any gravitational potential energy
๐ธ๐ =1
2๐ผ๏ฟฝฬ๏ฟฝ2 + ๐๐โ(1 โ cos ๐)
where ๐ is the mass of the body,
โ is the height above original
position and ๐ is the angle that a
line connecting the centre of gravity and point of
rotation makes with the horizontal