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Edexcel Mechanics 3 revision notes

Chapter 1 - Further kinematics

1.1 - Acceleration depending on time (t)
a = dv/dt
v = dx/dt
a = d²x/dt²

a = f(t)
v = ∫f(t)dt + c

v = g(t)
x = ∫g(t)dt + k

1.2 - Acceleration depending on displacement (x)
a = F(x)
dv/dt = F(x)

By the chain rule,
dy/dx = dy/du x du/dx

Therefore;
a = dv/dt = dv/dx x dx/dt

Since, dx/dt = v
a = v.dv/dx

If y = v² where v = f(x)
dy/dx = dy/dv x dv/dx = 2v.dv/dx

OR rewrite y = v² as y = v.v
dy/dx = v.dv/dx + v.dv/dx = 2v.dv/dx (by the product rule)

Hence, d(v²)/dx = 2v.dv/dx

The acceleration, therefore, may also be written;
a = d(½v²)/dx

If v = f(x) = dx/dt
1/v = 1/f(x) = 1/(dx/dt) = dt/dx
∫1/v dx = ∫1 dt = t

Reply 1

Christophicus

Chapter 2 - Elastic springs and strings

2.1 - Hooke's law
T = λx/L
T = tension in spring or string
λ = modulus of elasticity of spring or string
x = extension/compression of string or spring
L = natural length of spring/string

If a particle is attached to a string in equilibrium;
Resolve vertically
T = mg
λx/L = mg
x = mgL/λ

If a particle is attached to a string and moves;
When the particle P is in equilibrium, using Hooke's law gives;
T = λe/L
e = TL/λ
e = mgL/λ

When the particle P is at some poinr where the extension, x, is greater than mgL/λ (e) then the tension in the string will be greater than the weight of the particle and the particle will move with an acceleration a.

Using the equation of motion,
F = ma
T - mg = ma
λx/L - mg = ma
a = λx/mL - g

The resulting acceleration depends on the extension in the string and is constantly changing.

2.2 - Energy stored in an elastic string or spring
Consider a particle attached to one end of an elastic string whose other end is fixed on a smooth horizontal table. When the string is stretched beyond its natural length by pulling the particle along the table and then released, the particle will move along the table and will gain kinetic energy. As the motion is horizontal there is no change in gravitational potential energy so it follows that when stretched, the string has energy stored in it. This form of potential energy is called the elastic potential energy (E.P.E) of the string.

Work done
= area under a Tension-extension graph

Before the elastic limit of the string/spring, the gradient of the Tension-extension graph is a straight line.

So, the area under a Tension-extension graph
= ½ x Tx
= ½ x λx²/L
= λx²/2L

Therefore the work done in stretching an elastic spring with modulus λ from its natural length L to a length (L+x) is λx²/2L

if a string (or spring) which has its extension (or compression) increased from x to y will have its elastic energy increased.
This increase is given by;

Increase in E.P.E.
= final E.P.E. - initial E.P.E.
= λy²/2L - λx²/2L

2.3 - Problems involving kinetic energy, gravitational potential energy and elastic potential energy
By the work-energy principle;
The total change of the mechanical energies (that is kinetic, gravitational potential and elastic potential energies) of a system is equal to the work done by any external forces acting on a system.

mgh1 + ½mu² + λx²/2L - FRx = mgh2 + ½mv² + λy²/2L

Reply 2

Christophicus

Chapter 3 - Further dynamics

3.1 - Newton's laws for a particle moving in a straight line when the applied force is variable
Consider a particle of mass m moving under the influence of a force F which is a function of time.
F = f(t)

By Newton's second law, F = ma, gives;
ma = f(t)
a = dv/dt
m.dv/dt = f(t)
mv = ∫f(t) dt + c

Similarly, if F is a function of the displacement, that is;
F = g(x)

then using Newton's second law gives;
ma = g(x)
a = d(½v²)/dx
=> m.d(½v²)/dx = g(x)

Integrating with respect to x gives;
½mv² = ∫g(x) dx + k

Impulse and momentum
For a variable force f(t), Newton's second law, f(t) = ma can be written;
f(t) = m.dv/dt

Let the particle have a constant mass m and velocity U at time T1 and a velocity V at time T2. Then integrating with respect to t over the integral from T1 to T2 gives;
∫f(t) dt (between T1 and T2) = m∫1 dv (between V and U)
= m[v] (between V and U)
= m(V - U)
= mV - mU

mV - mU is the change of momentum of the particle. The quantity ∫f(t) dt (between T1 and T2) is the impulse of the variable force on the particle.
Thus the equation;
Impulse = change of momentum
is still valid but for a variable force the impulse is given by;
Impulse = ∫f(t) dt (between T1 and T2)

Work and energy
When a constant force F acts on a particle of mass m and moves it through a distance s in the direction of the force, the work done by the force is defined by;

Work done = F x s
Fs = ½mv² - ½mu²
(where u and v are the initial and final speeds respectively of the particle)

a = v.dv/dx
f(x) = ma = mv.dv/dx (1)

If the action of the force causes the particle to increase its speed from U to V while moving from the point x = x1 to the point where x = x2 then integrating (1) with respect to x gives;

∫f(x) dx (between x1 and x2) = m∫v dv = m[½v²] between V and U
= ½mV² - ½U² (2)

The RHS of (2) is the increase in the K.E. of the particle and the LHS is the work done by the variable force.

Therefore. work done = ∫f(x) dx (between x1 and x2)

3.2 - Newton's law of gravitation
The force of attraction between two bodies of masses M1 and M2 is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

F = GM1M2/d²
G is the constant of gravitation.

Units of G
F = GM1M2/d²
(where F is a function of d, the distance between the two bodies)

=> G = Fd²/[M1M2]
F is measured in Newtons, d in metres and M1 and M2 in kilograms.

So, units for G are;
Nm²kg-²
= kgms-²m²kg-²
= m³/kgs²

Hence G is 6.67 x 10^-11 m³/kgs²

Relationship between G and g
Consider a particle of mass m which is at rest on the surface of the earth.
Suppose the mass of the earth is Mekg and its radius is Rm. By Newton's law of gravitation the magnitude of the force of attraction between the particle and the earth is;

F = GmMe/R²

However, the force with which the earth attracts the particle is the weight of the particle, so;

F = mg

=> mg = GmMe/R²
=> g = GMe/R²

3.3 - Simple harmonic motion
Equation for acceleration
a = -ω²x
Always get the acceleration of the oscillating object in this form to prove that the motion is simple harmonic.

Equation for speed
v² = ω²(a² - x²)
vmax = ωa

v = speed
ω = v/a
a = amplitude of S.H.M
x = displacement moved by particle

Equations for displacement
x = asin(ωt)
- Particle starts at centre of oscillation (x = 0 when t = 0)

x = acos(ωt)
- Particle starts at either end of oscillation (x = a when t = 0)

Equations for period of motion
Period = 2π/ω = 2π/(vmax/a) = 2πa/vmax

3.4 - Horizontal oscillations of a particle attached to the end of an elastic string or spring
Consider a particle P resting on a smooth horizontal surface and attached to one end of an elastic spring whose other end is fixed to a point O on the surface. If P is pulled aside in the direction OP and then released the spring will be stretched and so will exert a tension on P resulting in an acceleration.

F = ma
-T = ma
-λx/L = ma
a = -λx/mL (1)

Equation (1) is in the form a = -ω²x
=> ω² = λ/mL

Since period = 2π/ω
The period of the particle's oscillations is 2π√[ml/λ]

String is only taut during S.H.M. When the string goes slack, T = 0 and the particle moves at a constant speed.
So using time = distance/speed you can calculate the length of time for which the string is slack etc.

3.5 - Vertical oscillations of a particle attached to the end of an elastic string or spring.
When the particle is in equilibrium;
T = mg
T = λe/L
λe/L = mg
e = mgL/λ

Hence, the total extension when the particle is displaced a distance x from the equilibrium position is; e + x = mgL/λ + x

Tension at this point is;
T = λ(mgl/λ + x)/L = mg + λx/L

Using Newton's second law, F = ma, gives;
mg - T = ma
mg - [mg + λx/L] = ma
ma = -λx/L
a = -λx/mL

String becomes slack when x = -e and the particle will then move freely under gravity until it falls back to x = -e again.

Reply 3

Christophicus

Chapter 4 - Circular motion

4.1 - Angular speed
ω = dθ/dt => θ = ωt
ω is the angular speed of the radius of a circle.
ω is measured in radians/second or revolutions/minute

2π radians = 1 revolution
1 radian = 1/2π revolutions
n rad/s = n/2π rev/s

1 minute = 60 seconds
n/2π rev/s = n/2π x 60 rev/min

Angular and linear speed
v = rω
When the radius is measured in metres and the angular speed in radians per second, the linear speed is in metres per second.

Direction of the linear velocity
The velocity of a point on the circumference of a circle is directed along the tangent to the radius of the circle at that point.

4.2 - Acceleration in circular motion
The acceleration of a particle moving on a circular path centre O of radius r is given by;

a = rω² or a = v²/r

and is directed towards the centre O of the circle.

4.3 - Uniform motion of a particle moving in a horizontal circle
Using F = ma,
F = mv²/r as a = v²/r
OR
F = mrω² as a = rω²

Situations include, the conical pendulum and motion on a banked surface. See M3 book for more detail.

4.4 - The motion of a particle in a vertical circle
The components of the acceleration in a vertical circle are;
rω² or v²/r along the inward radius and dv/dt along the tangent.

By the work-energy principle;
work done by the forces acting on a particle = change in mechanical energy of the particle

But generally, if no forces are acting on the particle, then;
mgh1 + ½mu² = mgh2 + ½mv²

Motion of a particle on a fixed vertical circle
mgh1 + ½mu² = mgh2 + ½mv²
0 + ½mu² = mgr - mgrcos θ + ½mv²
½mu² - ½mv² = mgr - mgrcos θ

Dividing through by m and multiplying by 2 gives;
u² - v² = 2gr - 2grcos θ
v² = u² - [2gr - 2grcos θ]
v² = u² - 2gr + 2grcosθ

Motion of a particle which can leave the circular path
A particle which cannot leave its vertical circular path (for example a particle on a rod) will describe complete circles provided its velocity at the highest point of the circle is greater than or equal to zero.

A particle which can leave its vertical circular path (for example a particle on a string) will do so when the force towards the centre of the circle becomes zero, e.g. the tension, the reaction.

Reply 4

Christophicus

Chapter 5 - Statics of rigid bodies

5.1 - Centre of mass of a uniform plane lamina
x-bar = (∫xy dx)/(∫y dx) both integrals between the x-values x1 and x2 and where y = f(x)

y-bar = (½∫y²dx)/(∫y dx) both integrals between the x-values x1 and x2 and where y = f(x)

5.2 - Centre of mass of a uniform solid body
1 - The centre of mass of a solid body is the point at which the weight acts.
2 - The weight of a uniform solid body is evenly distributed throughout its volume.
3 - The centre of mass of a uniform solid body must lie on any axis of symmetry.
- The centre of mass of a uniform solid body must lie on any plane of symmetry.

Use of symmetry
Uniform solid sphere - The sphere has an infinite number of planes of symmetry, so the centre of mass of the sphere is directly in the centre.
Uniform solid right circular cylinder - The line through the plane face perpendicular to this face is an axis of symmetry.
Uniform solid right circular cone - The axis of a solid right circular cone is an axis of symmetry of the cone.
Uniform solid hemisphere - The line the centre of the plane face perpendicular to the plane face is an axis of symmetry.

Centres of mass of solids of revolution
x-bar = (∫πy²x dx)/(∫πy² dx) both integrals between the x-values x1 and x2 and where y = f(x)

Centres of mass of surfaces of revolution
When part of a curve is rotated through 360 degrees about a fixed line we obtain a surface known as a surface of revolution. Examples include a hollow cone and a hemispherical shell. See M3 book for detailed diagrams and examples.

5.3 - Simple cases of equilibrium of rigid bodies
A rigid body is in equilibrium if;
- the vector sum of the forces acting is zero, that is the sum of the components of the forces in any given direction is zero.
- the algebraic sum of the moments of the forces about any given point is zero.

Suspension of a body from a fixed point
- A rigid body hangs in equilibrium with its centre of mass vertically below the point of suspension.

Equilibrium of bodies on a horizontal plane
1) If the line of action of the weight lies inside the area of contact, then the body is in equilibrium.

2) If the line of action of the weight lies outside the area of contact, then it is not possible to have equilibrium as the reaction must act somewhere in the area of contact. In this case the body will topple.

3) When the line of action of the weight passes through either end points of the area of contact. The body is in limiting equilibrium.

Equilibrium of bodies on an inclined plane
Same principles as previous section.

5.4 - Slipping and toppling
1) When the body is on the point of sliding, FR = µR
2) When the body is on the point of toppling, the reaction acts at the point about which the body will turn.

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C.O.M standard results for uniform bodies
Solid hemisphere, radius r - 3r/8 from centre
Hemispherical shell, radius r - ½r from centre
Circular arc, radius, angle at centre 2θ - rsinθ/θ from centre
Sector of circle, radius r, angle at centre 2θ - 2rsinθ/3θ from centre
Solid right circular cone, height h - 3h/4 from vertex OR h/4 from base
Conical shell, height h - 2h/3 from vertex OR h/3 from base

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Mechanics M3 FORMULAE GIVEN IN EXAM
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Reply 5