Course Name: Classical Mechanics
Course Code: SPH 3102
Course Level: Year Three
Semester: I
Credit Units: 3 CU
UNIT 1: Newtonian Mechanics
INTRODUCTION TO NEWTONIAN MECHANICS
It is assumed that the reader has been introduced to Newtonian mechanics applied to one- or twopoint objects. This chapter reviews Newtonian mechanics for motion of many-body systems as well as
for macroscopic sized bodies. Newton’s Law of Gravitation also is reviewed. The purpose of this review
is to ensure that the reader has a solid foundation of elementary Newtonian mechanics upon which to
build the powerful analytic Lagrangian and Hamiltonian approaches to classical dynamics. Newtonian
mechanics is based on application of Newton’s Laws of motion which assume that the concepts of
distance, ti me, and mass, are absolute, that is, motion is in an inertial frame. The Newtonian idea of
the complete separation of space and time, and the concept of the absoluteness of time, are violated by
the Theory of Relativity as discussed in chapter. However, for most practical applications, relativistic
effects are negligible and Newtonian mechanics is an adequate description at low velocities. Therefore,
chapters will assume velocities for which Newton’s laws of motion are applicable.
NEWTON’S LAWS OF MOTION
Newton defined a vector quantity called linear momentum P which is the product of mass and velocity.
𝑷 = 𝑚𝑟̇
Since the mass m is a scalar quantity, then the velocity vector 𝑟̇ and the linear momentum vector p are
colinear.
Newton’s laws, expressed in terms of linear momentum, are:
1. Law of inertia: A body remains at rest or in uniform motion unless acted upon by a force.
2. Equation of motion: A body acted upon by a force moves in such a manner that the time rate of
change of momentum equals the force.
𝐹=
𝑑𝑝
𝑑𝑡
3. Action and reaction: If two bodies exert forces on each other these forces are equal in magnitude and
opposite in direction.
Newton’s second law contains the essential physics relating the force F and the rate of change of linear
momentum p.
Newton’s first law, the law of inertia, is a special case of Newton’s second law in that if
𝐹=
𝑑𝑝
=0
𝑑𝑡
then is P a constant of motion.
Newton’s third law also can be interpreted as a statement of the conservation of momentum, that is, for
a two-particle system with no external forces acting,
𝐹12 = −𝐹21
If the forces acting on two bodies are their mutual action and reaction, then Equation simplifies to
𝐹12 = −𝐹21 =
𝑑𝑝1 𝑑𝑝2
𝑑
+
= (𝑝1 + 𝑝2 ) = 0
𝑑𝑡
𝑑𝑡
𝑑𝑡
This implies that the total linear momentum 𝑃 = 𝑝1 + 𝑝2 is a constant of motion. Combining Equations
and leads to a second-order differential equation.
𝐹=
𝑑𝑝
𝑑2𝑟
= 𝑚 2 = 𝑚𝑟̈
𝑑𝑡
𝑑𝑡
Note that the force on a body F, and the resultant acceleration 𝒂 = 𝑟̈ are colinear.
APPLICATIONS OF NEWTON’S EQUATIONS OF MOTION
Newton’s equation of motion can be written in the form
𝐹=
𝑑𝑝
𝑑2𝑟
= 𝑚 2 = 𝑚𝑟̈
𝑑𝑡
𝑑𝑡
A description of the motion of a particle requires a solution of this second-order differential equation
of motion. This equation of motion may be integrated to find r(t) and v(t) if the initial conditions and
the force field are known. Solution of the equation of motion can be complicated for many practical
examples, but there are various approaches to simplify the solution. It is of value to learn efficient
approaches to solving problems.
The following sequence is recommended
a. Make a vector diagram of the problem indicating forces, velocities, etc.
b. Write down the known quantities.
c. Before trying to solve the equation of motion directly, look to see if a basic conservation law applies.
That is, check if any of the three first-order integrals, can be used to simplify the solution. The use of
conservation of energy or conservation of momentum can greatly simplify solving problems.
The following examples show the solution of typical types of problem encountered using Newtonian
mechanics.
Constant Force Problems
problems having a constant force imply constant acceleration. The classic example is a block sliding on
an inclined plane, where the block of mass is acted upon by both gravity and friction. The net force F is
given by the vector sum of the gravitational force Fg, normal force N and frictional force ff
F = Fg + N + 𝑓𝑓 = ma
Taking components perpendicular to the inclined plane in the 𝑦 direction
−𝐹𝑔 cos 𝜃 + 𝑁 = 0
That is, since
𝐹𝑔 = 𝑚𝑔
𝑁 = 𝑚𝑔 cos 𝜃
Similarly, taking components along the inclined plane in the 𝑥 direction
𝐹𝑔 sin 𝜃 − 𝑓𝑓 = 𝑚
𝑑2𝑥
𝑑𝑡 2
Using the concept of coefficient of friction µ
𝑓𝑓 = 𝜇𝑁
Thus, the equation of motion can be written as
𝑚𝑔(sin 𝜃 − μcos 𝜃) = 𝑚
𝑑2𝑥
𝑑𝑡 2
The block accelerates if, sin 𝜃 > μcos 𝜃 that is,tan 𝜃 > 𝜇 . The acceleration is constant if µ and θ are
constant, that is
𝑑2𝑥
= 𝑔(sin 𝜃 − μcos 𝜃)
𝑑𝑡 2
Figure.Block of an inclined plane
Remember that if the block is stationary, the friction coefficient balances such that (sin 𝜃 − μcos 𝜃) = 0
that is, tan 𝜃 = 𝜇 . However, there is a maximum static friction coefficient 𝜇𝑠 beyond which the block
starts sliding. The kinetic coefficient of friction 𝜇𝑘 is applicable for sliding friction and usually 𝜇𝑘 < 𝜇𝑠
Another example of constant force and acceleration is motion of objects free falling in a uniform
gravitational field when air drag is neglected. Then one obtains the simple relations such 𝑣 = 𝑢 + 𝑎𝑡 ,
etc.
Linear Restoring Force
An important class of problems involve a linear restoring force, that is, they obey Hooke’s law. The
equation of motion for this case is
It is usual to define
Then the equation of motion then can be written as
which is the equation of the harmonic oscillator. Examples are small oscillations of a mass on a spring,
vibrations of a stretched piano string, etc.
The solution of this second order equation is
This is the well-known sinusoidal behaviour of the displacement for the simple harmonic oscillator.
The angular frequency 𝜔𝑜
𝜔𝑜 = √
𝑘
𝑚
Note that for this linear system with no dissipative forces, the total energy is a constant of motion as
discussed previously. That is, it is a conservative system with a total energy 𝐸 given by
The first term is the kinetic energy and the second term is the potential energy. The Virial theorem gives
that for the linear restoring force the average kinetic energy equals the average potential energy.
INERTIAL FRAMES OF REFERENCE
An inertial frame of reference is one in which Newton’s Laws of motion are valid. It is a nonaccelerated frame of reference. An inertial frame must be homogeneous and isotropic. Physical
experiments can be carried out in different inertial reference frames. The Galilean transformation
provides a means of converting between two inertial frames of reference moving at a constant relative
velocity.
Consider two frames denoted by S and S’ with (t, x, y, z) and (t’, x’, y’, z’) the coordinates in S and S’,
respectively. Without loss of generality, we assume the coordinate axes are aligned, x along x’, and so
on. Let S’ be moving relative to S in the +x-direction at a speed v, as shown in Figure.
Newtonian mechanics assumes the spacetime coordinates in S are related to those in Sby the simple
expressions.
𝑡’ = 𝑡,
𝑥’ = 𝑥 − 𝑣𝑡,
𝑦’ = 𝑦,
𝑧’ = 𝑧.
Transformations of this type are called Galilean transformations. Under this assumption, it follows
that Newton’s second law,
𝑭=
𝑑
𝒑,
𝑑𝑡
relating the applied force (F) and the momentum (p) remains invariant, and
𝐹′ =
𝑑
𝑑𝑡
𝒑′ and t ‘= t.
The time in both the S and S’ frames is assumed to be (t = t’). The Newtonian world view is that the
universe consists of three spatial directions and one-time direction. All observers agree on the time
direction up to a possible choice of units. Under these assumptions, there are no universal velocities. If
u and u’ are the velocities of a particle as measured in two frames moving with relative velocity v as
defined by Figure 1, then
𝒖’ = 𝒖 − 𝒗.
Maxwell’s electromagnetic equations, on the other hand, have a universal constant (denoted by c),
which is interpreted as the speed of light. Since this is inconsistent with Newtonian mechanics, either
Newtonian or Maxwellian mechanics would have to be modified. After carefully thinking about how
the universe would appear to an observer traveling at the speed of light, Albert Einstein decided that
Maxwell’s equations are correct to all inertial observers and the assumed transformations for
Newtonian mechanics are incorrect. The correct transformations make the speed of light the same to all
inertial observers.
Since the interval between two events of spacetime is a geometric quantity, all inertial observers
measure coordinates that preserve the value of the interval squared, (ds)2. If S and S’ are two different
inertial frames, then
𝒅𝒔′𝟐 = 𝒅𝒔𝟐 .
Thus, (ds)2 is called the square of the invariant spacetime interval.
Thus, Newton’s Laws of motion are invariant under a Galilean transformation, that is, the inertial mass
is unchanged under Galilean transformations. If Newton’s laws are valid in one inertial frame of
reference, then they are valid in any frame of reference in uniform motion with respect to the first frame
of reference. This invariance is called Galilean invariance. There are an infinite number of possible
inertial frames all connected by Galilean transformations.
Galilean invariance violates Einstein’s Theory of Relativity. In order to satisfy Einstein’s postulate that
the laws of physics are the same in all inertial frames, as well as satisfy Maxwell’s equations for
electromagnetism, it is necessary to replace the Galilean transformation by the Lorentz transformation.
As will be discussed later, the Lorentz transformation leads to Lorentz contraction and time dilation
both of which are related to the parameter 𝛾 ≡ 1/√(1 − (𝑣/𝑐)2 where c is the velocity of light in
vacuum.
CONSERVATION LAWS IN CLASSICAL MECHANICS
Elucidating the dynamics in classical mechanics is greatly simplified when conservation laws are
applicable. In nature, isolated many-body systems frequently conserve one or more of the first-order
integrals for linear momentum, angular momentum, and mass/energy. Note that mass and energy are
coupled in the Theory of Relativity, but for non-relativistic mechanics the conservation of mass and
energy are decoupled. Other observables such as lepton and baryon numbers are conserved, but these
conservation laws usually can be subsumed under conservation of mass for most problems in nonrelativistic classical mechanics.
The power of conservation laws in calculating classical dynamics makes it useful to combine the
conservation laws with the first integrals for linear momentum, angular momentum, and work-energy,
when solving problems involving Newtonian mechanics. These three conservation laws will be derived
assuming Newton’s laws of motion, however, these conservation laws are fundamental laws of nature
that apply well beyond the domain of applicability of Newtonian mechanics.
THE ENERGY PRINCIPLE
Suppose a particle P of mass m moves under the influence of a force F. Then its equation of motion is
where v is the velocity of P at time t. At this stage we place no restrictions on the force F. It may depend
on the position of P, the velocity of P, the time, or anything else; if more than one force is acting on P,
then F means the vector resultant of these forces. On taking the scalar product of both sides of equation
with v, we obtain the scalar equation
And, since
This can be written in the form
𝑑𝑇
= 𝐹. 𝑣,
𝑑𝑡
1
Where 𝑚𝑣. 𝑣 = 𝑇
2
1
Kinetic energy The scalar quantity 𝑚𝑣. 𝑣 = 𝑇 = 𝟏/𝟐𝒎|𝒗|𝟐 is called the kinetic energy of the particle
2
P.
If we now integrate the equation over the time interval [t1, t2], we obtain
where T1 and T2 are the kinetic energies of P at times t1 and t2 respectively. This is the energy principle
for a particle moving under a force F.
Work done The scalar quantity
is called the work done by the force F during the time interval [t1, t2]. The rate of working of F at time t
is thus 𝐅. 𝐯.
[The SI unit of work is the joule (J) and one joule per second is one watt (W).]
Our result can now be stated as follows:
Example. A man of mass 100 kg can pull on a rope with a maximum force equal to two fifths of his own
weight. [ Take g = 10 m s−2.] In a competition, he must pull a block of mass 1600 kg across a smooth
horizontal floor, the block being initially at rest. He is able to apply his maximum force horizontally for
12 seconds before falling exhausted. Find the total work done by the man and confirm that the energy
principle is true in this case.
Solution
In this problem, the block is subjected to three forces: the force exerted by the man, uniform gravity,
and the vertical reaction of the smooth floor. However, since the last two of these are equal and
opposite, they can be ignored. The man has weight 1000 N so that the force he applies to the block is a
constant 200 N. The Second Law then implies that, while the man is pulling on the rope, the block must
have constant rectilinear acceleration 200/1600 = 1/8ms−2. Since the block is initially at rest, its velocity
v at time t is therefore v = t/8ms−1. The total work W done by the man is then given by
When t = 12 s, the block has velocity v = 12/8 = 3/2ms −1, so that the final kinetic energy of the block is
1 /2 (1600) (3/2)2 = 1800 J. Since the initial kinetic energy of the block is zero, the kinetic energy of the
block increases by 1800 J, the same as the work done by the man. This confirms the truth of the energy
principle.
ENERGY CONSERVATION
Suppose that the particle P moves along the x-axis under the force F acting in the positive x-direction.
In this case, the ‘work done’ integral reduces to
where 𝑣 = 𝑥̇ is the velocity of P in the positive x-direction? For the case in which F is a force field (so
that 𝐹 = 𝐹(𝑥)), the formula for 𝑾 becomes
where 𝑥1 = 𝑥(𝑡1 ) and 𝑥2 = 𝑥(𝑡2 ). Thus, when P moves over the interval [𝑥1 , 𝑥2 ] of the x-axis, the work
done by the field F is given by
(This is a common definition of the work done by a force 𝐹. It can be used when 𝐹 = 𝐹(𝑥), but not in
general.) It follows that the energy principle for a particle moving in a rectilinear force field can be
written
Now let 𝑉(𝑥) be the indefinite integral of −𝐹(𝑥), so that
Such a V is called the potential energy∗ function of the force field F. In terms of V, the energy principle
in rectilinear motion can be written
𝑇2 + 𝑉(𝑥2 ) = 𝑇1 + 𝑉(𝑥1 )
which is equivalent to the energy conservation formula
𝑇+𝑉 = 𝐸
where 𝐸 is a constant called the total energy of the particle. This result can be stated as follows
Example
Find the potential energies of (i) the (one-dimensional) SHM force field, (ii) the (one-dimensional)
attractive inverse square force field.
Solution
(i)
The one-dimensional SHM force field is 𝐹 = −𝛼𝑥, where α is a positive constant. The
corresponding V is given by
where a, the lower limit of integration, can be arbitrarily chosen. (This corresponds to the
arbitrary choice of the constant of integration.) Note that, by beginning the integration at 𝑥 =
𝑎, we make 𝑉(𝑎) = 0. In the present case it is conventional to take 𝑎 = 0 so that V = 0 at 𝑥 =
0. With this choice, the potential energy is 𝑽 =
(ii)
𝟏
𝟐
𝜶𝒙𝟐 .
The one-dimensional attractive inverse square force field is 𝐹 = −𝐾/𝑥 2 , where 𝑥 > 0 and 𝐾
is a positive constant. The corresponding 𝑉 is given by
This time it is not possible to take a = 0 (the integral would then be meaningless) and it is
conventional to take 𝑎 = +∞; this makes V = 0 when 𝑥 = +∞. With this choice, the potential
energy is 𝑉 = −𝐾/𝑥 (𝑥 > 0).
LINEAR MOMENTUM
We begin with the definition of linear momentum for a single particle and for a system of particles.
Definition. Linear momentum If a particle has mass m and velocity 𝑣, then 𝑝, its linear momentum, is
defined to be
𝑷 = 𝒎𝒗
Figure: The multi-particle system Ssystem S consists of N particles P1, P2,.PN, of which the typical particle Pi is
labelled. The particle Pi has mass mi, position vector ri, and Velocity vi
For a multi-particle system S consisting of particles 𝑃1, 𝑃2, . . . , 𝑃𝑁 , with masses 𝑚1, 𝑚2, . . . , 𝑚𝑁 and
velocities 𝑣1, 𝑣2, . . . , 𝑣𝑁 (see Figure above, 𝑃, the linear momentum of S, is defined to be the vector sum
of the linear momenta of the individual particles, that is,
Newton’s Second Law can be written in terms of linear momentum in the form
𝑑𝑝
= 𝐹.
𝑑𝑡
The expression above can be written simply in terms of the motion of G, the centre of mass of S. Since
the position vector R of G is given by
where ri is the position vector of the particle 𝑃𝑖 , it follows that 𝑉, the velocity of 𝐺 is given by
where 𝑀 (= ∑ 𝑚𝑖 ) is the total mass of the system 𝑆. Hence
𝑃 = 𝑀𝑉
Thus the linear momentum of any system is the same as if all its mass were concentrated at its centre
of mass.
CONSERVATION OF LINEAR MOMENTUM
Suppose that 𝑆 is an isolated system, meaning that no external force acts on any of its particles. Then
𝑭, the total external force acting on 𝑆, is obviously zero. The linear momentum principle for 𝑆 then takes
the form 𝑑 𝑷/𝑑𝑡 = 0, which implies that P must remain constant. This simple but important result can
be stated as follows:
Clearly the same result applies to any system for which the total external force is zero, whether isolated
or not.
It is also possible for a particular component of 𝑷 to be conserved while other components are not. Let
n be a constant unit vector and suppose that 𝑭 · 𝒏 = 𝟎 at all times.
Then
Hence the component 𝑷 · 𝒏 is conserved. This result can be stated as follows:
Two examples of momentum conservation are as follows:
• The solar system is an example of an isolated system, being extremely remote from any other masses.
It follows that the total linear momentum of the solar system is conserved. Thus, the centre of mass of
the solar system moves with constant velocity.
•A grasshopper trying to move on a perfectly smooth horizontal table is not isolated, being subject to
gravity and the vertical reaction of the table. However, since the grasshopper is not subject to any
external horizontal force, it follows that, whatever the grasshopper tries to do, his component of total
linear momentum in any horizontal direction is conserved. His vertical component of linear momentum
is not conserved; he can leap into the air if he wishes.
Conservative field If the field 𝑭(𝒓) can be expressed in the form∗
𝑭 = − 𝒈𝒓𝒂𝒅 𝑽,
where 𝑉(𝑟) is a scalar function of position, then 𝑭 is said to be a conservative field and the function 𝑉
is said to be the potential energy function† for 𝐹.
If 𝜓(𝑟) is a scalar field then 𝒈𝒓𝒂𝒅 𝝍 is the vector field defined by
𝑔𝑟𝑎𝑑 𝜓 = 𝜕𝜓 /𝜕𝑥 𝑖 + 𝜕𝜓 /𝜕𝑦 𝑗 + 𝜕𝜓 /𝜕𝑧 𝑘.
Thus 𝑖𝑓 𝜓 = 𝑥 𝑦 3 𝑧 5 , then 𝑔𝑟𝑎𝑑 𝜓 = 𝑦 3 𝑧 5 𝑖 + 3𝑥 𝑦 2 𝑧 5 𝑗 + 5𝑥 𝑦 3 𝑧 4 𝑘.
Example:
Show that the field 𝐹1 = −2𝑥 𝑖 − 2𝑦 𝑗 − 2𝑧𝑘 is conservative but that the field 𝐹2 = 𝑦𝑖 − 𝑥 𝑗 is not.
Solution
(i) If 𝐹1 is conservative, then its potential energy 𝑽 must satisfy
where 𝑝, 𝑞 and 𝑟 are ‘constants’ of integration, which, in this case, are functions of the other variables.
If 𝑉 really exists, then these three representations of 𝑉 can be made identical by making special choices
of the functions 𝑝, 𝑞 and 𝑟. In this example it is clear that this can be achieved by taking 𝑝 = 𝑦 2 +
𝑧 2 , 𝑞 = 𝑥 2 + 𝑧 2 , and 𝑟 = 𝑥 2 + 𝑦 2 . Hence 𝐹1 = − 𝑔𝑟𝑎𝑑(𝑥 2 + 𝑦 2 + 𝑧 2 ) and so 𝐹1 is conservative.
(iii)
If 𝑭𝟐 is conservative, then its potential energy 𝑉 must satisfy
There is no V that satisfies these equations simultaneously. The easiest way to show this is to observe
that, from the first equation, 𝜕2𝑉/𝜕𝑦𝜕𝑥 = −1 while, from the second equation, 𝜕2𝑉/𝜕𝑥𝜕𝑦 = +1. Since
these mixed partial derivatives of 𝑉 should be equal, we have a contradiction. The conclusion is that no
such V exists and that 𝑭𝟐 is not conservative.
Suppose now that the field F(r) is conservative with potential energy V(r) and let C be any path
connecting the points A and B. Then
Thus, when 𝑭 is conservative with potential energy 𝑽 ,
for any path C connecting the points A and B. The energy principle can therefore be written
Which is equivalent to the energy conservation formula
𝑇 + 𝑉 = 𝐸.
Assignment:
1. (a) Show that the uniform gravity field 𝑭 = −𝒎𝒈𝒌 is conservative with potential energy 𝑽 = 𝒎𝒈𝒛.
(b) Show that any force field of the form
𝑭 = 𝒉(𝒓)𝒓̂
(a central field) is conservative with potential energy 𝑽 = −𝑯(𝒓), where 𝐻(𝑟) is the indefinite integral
of h(r). Use this result to find the potential energies of
(i)
(ii)
the 3 − 𝐷 SHM field 𝑭 = −𝜶𝒓 𝒓̂,
the attractive inverse square field 𝑭 = −(𝑲/𝒓 𝟐)𝒓̂, where 𝛼 and 𝐾 are positive constants.
2. A particle of mass m moves along a trajectory given by.
x = x𝑜 cosω1 t and
y = y0 sinω𝑜 t.
a) Find the x and y components of the force and determine the condition for which the force is a central
force.
b) Find the potential energy as a function of x and y.
c) Determine the kinetic energy of the particle and show that it is conserved.
UNIT 2: Gravitation:
Newton’s Law of Gravitation states that each mass particle attracts every other particle in the universe
with a force that varies directly as the product of the mass and inversely as the square of the distance
between them. That is, the force on a gravitational point mass produced by a mass 𝑀𝐺
The experimental value of 𝐺 = (6.6726 ± 0.0008 × 10−11 𝑁 𝑚2 𝑘𝑔−2 )
Example.
A particle P, of mass m, and a uniform rod, of length 2a and mass M, are placed as shown in Figure.
Find the gravitational force that the rod exerts on the particle.
Solution
Consider the element [𝒙, 𝒙 + 𝒅𝒙] of the rod which has mass 𝑀 𝑑𝑥/2𝑎 and exerts an attractive force of
magnitude
on P, where R is the distance shown in Figure. By symmetry, the resultant force acts towards the
centre O of the rod and can be found by summing the components of the contributing forces in the
direction P O. Since the rod is a continuous distribution
of mass, this sum becomes an integral. The resultant force exerted by the rod thus has magnitude
𝐹 given by
Free-Fall Acceleration and the Gravitational Force
Systems with Variable Mass
Classic examples of systems with variable mass are the rocket, nuclear fission and other modes of
nuclear decay.
Consider the problem of rocket motion in a gravitational field. When there is a vertical gravitational
external field the vertical momentum is not conserved due to both gravity and the ejection of rocket
propellant. In a time 𝑑𝑡 the rocket ejects propellant with exhaust velocity relative to the rocket of 𝑢.
Thus, the momentum imparted to this propellant is
𝒅𝒑𝒑 = −𝒖𝒅𝒎𝒑
Therefore, the rocket is given an equal and opposite increase in momentum
𝒅𝒑𝑹 = +𝒖𝒅𝒎𝒑
In the time interval 𝑑𝑡 the net change in the linear momentum of the rocket plus fuel system is given
by
𝑑𝑝 = (𝑚 − 𝑑𝑚𝑝 )(𝑣 + 𝑑𝑣) + 𝑑𝑚𝑝 (𝑣 − 𝑢) − 𝑚𝑣 = 𝑚𝑑𝑣 − 𝑢𝑑𝑚𝑝
The rate of change of the linear momentum thus equals
𝐹𝑒𝑥
𝑑𝑚𝑝
𝑑𝑝
𝑑𝑣
= 𝑚
−𝑢
𝑑𝑡
𝑑𝑡
𝑑𝑡
Consider the problem for the special case of vertical ascent of the rocket against the external
gravitational force 𝐹𝑒𝑥 = −𝑚𝑔. Then
−𝑚𝑔 + 𝑢
𝑑𝑚𝑝
𝑑𝑣
=𝑚
𝑑𝑡
𝑑𝑡
This can be rewritten as
−𝑚𝑔 + 𝑢𝑚̇ 𝑝 = 𝑚𝑣̇
The second term comes from the variable mass. But the loss of mass of the rocket equals the mass of the
ejected propellant. Assuming a constant fuel burn then 𝑚̇ = −𝑚̇ 𝑝 = −𝛼
Where 𝛼 > 0 Then the equation becomes
Since
Then
Inserting this in the above equation gives
Integration gives
But the change in mass is given by
That is
Thus
Note that once the propellant is exhausted the rocket will continue to fly upwards as it decelerates in
the gravitational field.
Energy Considerations in Planets and Satellite Motion [important~!]
Assignment:2
QN. Derive the expression for Gravitational Potential Energy.
(submit )
Qn. Do Escape Speed of a Rocket?
Ans
Unit 3: Dynamics of Systems:
▪
Dynamics of systems of particles,
▪
centres of mass of systems of particles, relativistic mass and momentum.
▪
Rotational kinematics and dynamics.
▪
Equilibrium of extended rigid bodies.
▪
Lorentz force under electric magnetic and gravitation fields.
▪
Application of uniform fields to spectrographs, cyclotrons and Hall Effect. (10hours)
Particle dynamics is concerned with the problem of calculating the motion of a particle that is acted
upon by specified forces. The entirety of particle dynamics is based on the Second Law
𝑚𝑎 = 𝐹1 + 𝐹2 +··· + 𝐹𝑁,
where 𝐹1, 𝐹2, . . . , 𝐹𝑁 are the various forces that are acting on the particle. The typical method of
solution is to write the Second Law in the form
𝑚 𝑑𝑣/ 𝑑𝑡 = 𝐹1 + 𝐹2 +··· + 𝐹𝑁 ,
Which is a first order ODE for the unknown velocity function 𝑣(𝑡) and is called the equation of
motion of the particle. If the initial value of v is given, then equation above can often be solved to
yield 𝒗 as a function of the time 𝒕. Once 𝒗 is determined (and if the initial position of the particle is
given), the position vector 𝒓 of the particle at time 𝑡 can be found by solving the first order ODE
𝑑𝑟/𝑑𝑡 = 𝑣.
If the rectilinear motion takes place along the z-axis, the equation of motion reduces to the scalar
equation
where 𝑣 is the (one-dimensional) velocity of the particle and 𝐹(𝑧) is the (one-dimensional) force field,
both measured in the positive z-direction. First, we consider the problem of vertical motion of a
particle under uniform gravity with no air resistance.
Example
A particle is projected vertically upwards with speed 𝑢 and moves in a vertical straight line under
uniform gravity with no air resistance. Find the maximum height achieved by the particle and the
time taken for it to return to its starting point.
Solution
Let 𝑣 be the upwards velocity of the particle after time 𝑡 , as shown in figure below.
Then the scalar equation of motion takes on the form,
Since the drag force D is absent. A simple integration gives
𝑉 = − 𝑔𝑡 + 𝐶,
where 𝐶 is the integration constant, and, on applying the initial condition 𝑣 = 𝑢 when 𝑡 = 0, we
obtain 𝐶 = 𝑢. Hence the velocity 𝑣 at time t is given by
𝑣 = 𝑢 − 𝑔𝑡.
To find the upward displacement z at time t, write
𝑑𝑧/ 𝑑𝑡 = 𝑣 = 𝑢 − 𝑔𝑡.
A second simple integration gives
𝑧 = 𝑢𝑡 − 1 /2 𝑔𝑡 2 + 𝐷,
Where D is a second integration constant, and, on applying the initial condition 𝑍 = 0 when 𝑡 = 0,
we obtain D= 0, hence the upward displacement of the body at time t is given by
The maximum height Zmax is achieved when
𝑑𝑧
𝑑𝑡
= 0, that is, when 𝑣 = 0, Thus Zmax is achieved when
𝑡 = 𝑢/𝑔 and is given by
The particle returns to 0 when 𝑧 = 0, that is, when
Thus, the particle returns after a time 2𝑢/𝑔.
QN:
A particle 𝑃 of mass 𝒎 moves under the gravitational attraction of a mass 𝑴 fixed at the origin 𝑂.
Initially P is at a distance a from O when it is projected with speed u directly away from O. Find the
condition that P will ‘escape’ to infinity.
CENTRE OF MASS
we can now introduce the notion of the centre of mass of a collection of particles.
Suppose we have a system S of particles 𝑃1, 𝑃2, . . . , 𝑃𝑁 with masses 𝑚1, 𝑚2, . . . , 𝑚𝑁, and position
vectors 𝑟1, 𝑟2, . . . , 𝑟 𝑁 respectively. Then the centre of mass of S is defined as follows:
Definition Centre of mass ` the centre of mass of S is the point of space whose position vector R is
defined by
Where 𝑀 is the total mass of 𝑆.
The centre of mass of a system of particles is simply a ‘weighted’ mean of the position vectors of the
particles, where the ‘weights’ are the particle masses. Centre of mass is an important concept in the
mechanics of multi-particle systems.
Example
Find the Centre of mass of:
(i)
(ii)
A pair of particles of different masses
Three identical particles
Solution
(i)
For a pair of particles P1, P2, the vector position of the Centre of mass is given by
𝑚1 𝑟1 + 𝑚2 𝑟2
𝑅=
𝑚1 + 𝑚2
It follows that the centre of mass lies on the line 𝑃1𝑃2 and divides this line in the ratio 𝑚2 ∶ 𝑚1.
(ii)
For three identical particles 𝑃1, 𝑃2, 𝑃3, the position vector of the centre of mass is given by
It follows that the centre of mass lies at the centroid of the triangle 𝑃1𝑃2𝑃3.
QN;
Find the position of the centre of mass of a uniform semi-circular lamina.
Suppose that the lamina has radius a and occupies the region A shown in Figure.
The lamina has rotational symmetry about the y-axis so that G must lie on the y-axis, as shown. It
remains to find its y-coordinate 𝑌 which, from the definition is given by
Where M is the mass of the lamina. However, since the lamina is a continuous distribution of
mass, this sum must now be interpreted as an integral.
𝑀𝑑𝐴
Consider an element of area 𝑑𝐴 of the lamina: this has mass 1
2
becomes
𝜋𝑎2
and the above equation now
where y is the y-coordinate of the element. On ‘summing’ these contributions (and cancelling by
M) we find that
A solid hemisphere of radius a occupies the region ν shown.
where the integral is taken over the region 𝐴 occupied by the lamina. This double integral is most
easily evaluated using standard polar coordinates.
In these coordinates∗ 𝑑 𝐴 = (𝑑𝑟)(𝑟 𝑑𝜃 ) = 𝑟 𝑑𝑟 𝑑𝜃, and 𝑦 = 𝑟 𝑠𝑖𝑛 𝜃. The ranges of integration
for 𝑟, 𝜃 are 0 ≤ 𝑟 ≤ 𝑎 and 0 ≤ 𝜃 ≤ 𝜋. We therefore obtain
Hence the Centre of mass of the semi-circular lamina is on the y-axis a distance 4𝑎/3𝜋 from the
origin.
Relativistic mass and momentum
Using the classical definition of momentum, that is 𝒑 = 𝒎𝒖, the linear momentum is not
conserved using the above relativistic velocity transformations if the mass is a scalar quantity.
This problem originates from the fact that both 𝒙 and 𝒕 have non-trivial transformations and thus
𝒖 = 𝒅𝒙/𝑑𝑡 is frame dependent.
Linear momentum conservation can be retained by redefining momentum in a form that is
identical in all frames of reference, that is by referring to the proper time τ as measured in the rest
frame of the moving object. Therefore, we define relativistic linear momentum as
But we know the time dilation relation
Note that the 𝛾𝑢 in this relation refers to the velocity between the moving object and the frame;
this is quite different from the
which refers to the transformation between the two frames of reference. Thus, the new relativistic
definition of momentum is
𝒑≡𝒎
𝑑𝒙
𝒅𝝉
= 𝒎𝜸𝒖
𝒅𝒙
𝒅𝒕
= 𝜸𝒖 𝑚𝒖
The relativistic definition of linear momentum is the same as the classical definition with the rest
mass m replaced by the relativistic mass 𝛾𝑚.
Rotational Kinematics and Dynamics
►Angular Displacement, Velocity, and Acceleration Review
- A rigid object rotating about a fixed axis through O perpendicular to the plane.
- An arbitrary point P is at a fixed distance r from the origin and rotates about it in a circle or
radius r. -The relationship between the arc length s, radius r and angle θ is
𝜽 = 𝑺/𝒓
- θ is the ratio of an arc length and the radius of the circle with the unit radian, where one radian
is the angle subtended by an arc length equal to the radius of the arc
1rad = 57.3° = 0.159revs
- Angular speed is defined as 𝑤 =
∆𝜃
∆𝑡
and angular acceleration is defined as 𝛼 =
∆𝑤
∆𝑡
When rotating about a fixed axis, every particle on a rigid object rotates through the same angle
and has the same angular speed and the same angular acceleration
►Rotational Kinematics: Rotational Motion with Constant Acceleration
When an object’s rotational motion is under constant angular acceleration, the kinematic
relationships can be expressed as
►Rotational Energy
- The rotational kinetic energy of a rotating rigid object is expressed as
1
KR =𝐾𝑅 = 𝐼𝑤 2 , where I = ∑mi ri²: rotational inertia(=moment of inertia)
2
1
*~Compare with 𝐾 = 𝑚𝑣 2
2
- The moment of inertia (or rotational inertia) is a measure of the resistance of an object to changes
in its rotational motion, just as mass is a measure of the resistance of an object to changes in its
linear motion.
- Note that mass is an intrinsic property of an object, where I depends on the physical
arrangement of that mass
►Calculation of Rotational of Inertia ‘I’ (revisit)
I = ∑mi ri²: moment of inertia
- The moment of inertia for a point particle of mass ‘m’is expressed as
I = mr² (Figure1) where r is the perpendicular distance from the axis of rotation to the body.
- The moment of inertia for a point particle of mass ‘m’is expressed as
I = 2mr² (Figure2)
For a continuous body the moment of inertia can be generalized to an integral over the mass
density of the body.
where r is perpendicular to the rotation axis. The definition of the moment of inertia allows
rewriting the angular momentum about a symmetry axis Lsym in the form
where the moment of inertia Isym is taken about the symmetry axis and assuming that the angular
velocity of rotation vector is parallel to the symmetry axis.
Example:
Consider that the door has width a and height b and assume the door thickness is negligible with
areal density. Assume that the door is hinged about the y axis. The mass of a surface element of
dimension at a distance from the rotation axis is 𝑑𝑚 = 𝜎𝑑𝑥𝑑𝑦 thus the mass of the complete door
is 𝑀 = 𝜎𝑎𝑏. The moment of inertia about the axis is given by
Equilibrium of extended rigid bodies
A rigid body is said to be in equilibrium if both the resultant force and resultant couple moment
about any point O are equal to zero,
These equations are the necessary and sufficient conditions for equilibrium.
PROOF
Summing the moments in the equivalent resultant force & couple moment system about some
other point, such as A, we have
For equilibrium, the sum of the couple moments must be zero
Since r ≠ 0, the above equation is satisfied only if
𝐹𝑅 = 0
(𝑀𝑅 )0 = 0
which are the equilibrium conditions given in Eqn1.
EXAMPLE
Given;
Determine the horizontal and vertical components of reaction on the beam caused by the pin at B
and the rocker at A.
• Neglect the weight of the beam.
SOLUTION
LORENTZ FORCE
ASSIGNMENT 3.
Discuss the Application of uniform fields to spectrographs, cyclotrons and Hall Effect.
(10marks). Kindly include relevant diagrams, formulas,
(make groups of 5-8 members, prepare power point. To be presented in class during lecture
time)