Download Description of Motion in One Dimension

Topic 2: Mechanics
2.1.1
Define displacement, velocity, speed, and acceleration.
Displacement: A vector quantity that describes a position of a particle in reference to an
origin. It has both magnitude and direction.
Velocity: The total speed of a particle in a given direction.
Speed: The rate of change in position without regard to direction.
Acceleration: The rate of change of velocity.
2.1.2
Define and explain the difference between instantaneous and average values of
speed, velocity, and acceleration.
An instantaneous value is a value taken at an instant.
An average value is taken over a period of time.
2.1.3
Describe an object’s motion from more than one frame of reference.
If one car, travelling at 30 ms-1, overtakes another travelling at 25 ms-1¬, then according
to the driver of the slower car, the relative velocity of the faster car is 5 ms-1. This is
moving from one frame of reference to another. The velocities of 30 ms-1¬ and 25 ms-1¬
were from the position of a stationary observer. We moved from this frame of reference
into that of one of the drivers.
2.1.4
Draw and analyse distance-time graphs, displacement-time graphs, velocity-time
graphs and acceleration-time graphs.
Displacement-time graphs:


Gradient = velocity
Area under the graph = nothing significant
Velocity-time graphs:


Gradient = acceleration
Area under the graph = displacement
Acceleration-time graphs:


Gradient = Change in Acceleration
Area under the graph = not really significant (the amount of acceleration, total
velocity changed)
2.1.5
Analyse and calculate the slopes of displacement-time graphs and velocity-time
graphs, and the areas under velocity-time graphs and acceleration-time graphs.
Relate these to the relevant kinematic quantity. See previous pages.
2.1.6
Determine the velocity and acceleration from simple timing situations.
Velocity = Displacement/Time
Acceleration = ΔVelocity/Time
2.1.7
Derive the equations for uniformly accelerated motion. Let
t
a
u
v
s
be
be
be
be
be
time for the body accelerates
acceleration
the initial speed
the final speed
the distance travelled by object in time t
First Formula v = u+at
As the change of velocity, plus the initial equals to the final velocity.
Second Formula s=ut+1/2at2
As the displacement is the area under the velocity-time graph.
Image:V-t.jpg
Third Formula v2=u2+2as
From the first equation we can derived it as t=v-u/a. Substituting it to the second equation
we get s=uv-u/a+1/2a(v-u/a)2
Expending it we see the third formula.
2.1.8
Describe the vertical motion of an object in a uniform gravitational field.
An example of uniformly accelerated motion is the vertical motion of an object in a
uniform gravitational field. If we ignore the effects of air resistance, this is known as
free-fall. In the absence of air resistance, all falling objects have the same acceleration of
free-fall, independent of their mass.
2.1.9
Describe the effects of air resistance on falling objects.
As the object's velocity increases, the force of air resistance, which is opposite in
direction to the movement of the object, becomes greater and greater until it reaches a
value equal to the value of the force of gravity. This eliminates a net force put upon the
object, which causes it to fall at a constant velocity, no longer accelerating. This is called
terminal velocity.
2.2.1
Describe force as the cause of deformation or velocity change.
A force is recognised by the effect it produces, and can cause an object to deform (change
shape), speed up, slow down, and change direction.
2.2.2
Identify the forces acting on an object and draw free-body diagrams representing
the forces acting.
Forces should be labeled with a name or symbol – for example, weight, normal reaction,
friction, etc. Vectors should have lengths approximately proportional to their magnitudes.
Free-body diagrams are just a simple diagram of one object, for example, a book or a
person, and all of the forces acting must make contact with the object somewhere and
must be named.
Image:Freebodydiagram.png
2.2.3
Resolve forces into components.
A straight vertical or horizontal force do not have components. Diagonal forces, however,
do. A 4N force in a NW direction will have both a horizontal and vertical component. An
angle should be given, so use this in trigonometry to work out the sides (which are the
components).
Example Two forces act on particle P. Find the magnitude of the net force acting in the
horizontal direction and the magnitude of the net force acting in the vertical direction and
hence find the resultant force acting on P.
2.2.4
Determine the resultant force in different situations.
Adding of vectors is usually needed here. To find the resultant vector, join the beginning
of the first onto the end of the last. See 1.4.4 for subtracting (although all you need to do
is reverse the vector to be subtracted and add it to the other). The resultant vector may
need to be worked out through trigonometry but sometimes even just Pythagoras can be
used.
2.2.5
Describe the behaviour of a linear spring and solve related problems.
When a force is applied to a spring (through the addition of a mass or simply pulling it) a
tension force is produced. The spring increases in length. The difference between the
natural length and stretched length is called the extension of a spring.
As you pull the spring, the further you extend it, the greater the force you have to exert in
order to extend it even further.
Hooke’s Law, after Robert Hooke, states that up to the elastic limit [region of
proportionality] the extension of a spring is equal to the tension force, F. The constant of
proportionality k is called the spring constant. The SI units for the spring constant [k] are
Nm-1. F = kx
Up to the elastic limit, the force is proportional to extension, but beyond this point
proportionality is lost. If this point is passed, the spring can become permanently
deformed in such a way that when weights are removed the spring cannot go back to its
original length.
2.2.6
State Newton’s first law of motion.
An object continues in a state of rest or uniform motion in a straight line unless acted
upon by an external force.
2.2.7
Describe examples of Newton’s first law.
Examples of Newton's First Law:




A parachutist in free-fall (if the force of weight is larger than air friction the
parachutist accelerates downwards; as they get faster, air friction increases until
weight = friction. Then the parachutist is at constant velocity – the acceleration is
zero – resultant force is zero)
A car traveling (if air resistance equals the force forwards due to engine, then the
car is at constant velocity – no resulting force. If either force is larger than the
other, then there is a resultant force, and the car accelerates)
A book at rest on a table (acceleration = zero; resultant force = zero; thus it is at
rest (no resultant force)
Lifting a heavy suitcase (if the suitcase is too heavy to lift, it is not moving
therefore acceleration = zero. Thus the pull from the person plus the reaction from
ground is equal to the weight of the suitcase)
2.2.8
State the conditions for translational equilibrium.
If the resultant force on an object is zero then it is said to be in translational equilibrium,
or just equilibrium. To be in translational equilibrium, an object must be constantly at rest
or moving with uniform velocity in a straight line. There are essentially two types of
[translational] equilibrium.
Static equilibrium: Static equilibrium exists when an object is at rest. Take the example
of the book on the table.
Dynamic equilibrium: Dynamic equilibrium exists when an object is moving at constant
velocity in a straight line. Take the example of the car traveling.
2.2.9
Solve problems involving translational equilibrium.
Basically, when an object is in translational equilibrium, it means that the forces that are
acting on the object cancel each other out. They are still acting, but are equal to each
other – the net force, or resultant vector, is zero. So when a car moves at constant speed,
the force pushing it forward by the engine is equal to that of air resistance pushing it
back. The weight of the book is equal to the normal reaction from the surface of the table.
2.2.10
State Newton’s second law of motion.
Acceleration is directly proportional to the force acting and is in the same direction as
the applied force.
F = ma
But because sometimes mass of a system doesn’t remain constant (like a firework rocket,
sand falling onto a conveyer belt etc) it can be helpful to express the law in a more
general form:
F = Δ(mv)/t
Because p = mv/t, F = Δp/Δt
Thus, Force = Change of momentum over Change in time.
2.2.11
Solve problems involving Newton’s second law.
Example Problems:

The diagram shows a block of wood of mass 1kg attached via a pulley to a
hanging weight of mass .5 kg. Assuming that there is no friction between the
block and the bench and taking g to be 10 ms-2, calculate the acceleration of the
system.
The force acting on the system is the weight of the hanging mass which is 0.5g = 5N.
Using Newton’s second law F = ma, we have 5 = (1.5)a a = 3.3 Hence, the acceleration is
3.3 ms-2

A person of mass 70kg is strapped into the front seat of a car, which is travelling
at a speed of 30ms-1¬¬. The car brakes and comes to rest after travelling a
distance of 180m. Estimate the average force exerted on the person during the
braking process.
2.2.12
State Newton’s third law of motion.
To every action, there is an equal and opposite reaction
KEY POINTS ABOUT THIS LAW:

The two forces in the pair act on different objects – this means that equal and
opposite forces that act on the same object are NOT Newton’s third law pairs.
*Not only are the forces equal and opposite, but they must be of the same type. In
other words, if the force that A exerts on B is a gravitational force, then the equal
and opposite force exerted by B on A is also a gravitational force.
2.2.13
Discuss examples of Newton’s third law.
Examples of Newton's Third Law

Forces between roller skaters (if one pushes off another, they both feel a force,
equal and opposite, but their acceleration will be different due to mass)
A roller skater pushes off from a wall (the force on the girl by the wall causes the girl to
accelerate backwards. The mass of the wall (and earth) is so large that the force on it does
not cause any acceleration)

A rocket in space propels gasses at high velocity in one direction, establishing a
force. This causes the rocket to move in the opposite direction with a force of
equal magnitude that the gasses have.
2.3.1
Define inertial mass.
Inertial mass [is the property of an object that] determines how it responds to a given
force, whatever the nature of that force. It essentially measures a body’s inertia. Different
masses have different accelerations when a force acts on them (take the first roller-skater
above).
2.3.2
Compare gravitational mass and inertial mass.
Gravitational mass [is the property of an object that] determines how much gravitational
force it feels when near another object.
Different masses have different gravitational forces acting between them. The two
concepts are different, but gravitational and inertial mass are equivalent – a body’s
gravitational mass is equal to its inertial mass. The fact that different objects have the
same value for free-fall acceleration shows this.
2.3.3
Discuss the concept of weight.
Weight is a force. The ‘weight of an object’ is a force (N). Mass and weight are often
confused. Mass is the amount of matter contained in an object, whereas weight is a force
acting on the object. However, there is ambiguity as to the definition of weight even to
physicists. It is generally defined in two ways:


(a) the gravitational force on an object, mg
(b) the reading on a supporting scale (ie, scales to ‘weigh’ yourself)
Although these two definitions are the same if the object is in equilibrium, they are
different in non-equilibrium situations.
For example: if both the object and the scale were put into a lift and the lift accelerated
upwards then the definitions would give different values. It is safer to use ‘gravitational
force’ instead of weight.
Gravitational force = mg
On the surface of the Earth, g is approximately 9.81 ms-2, so to work out ‘weight’ or
‘gravitational force’ you multiply mass in kg by 9.81 ms-2.
2.3.4
Distinguish between mass and weight.
Mass is generally defined as the amount of matter contained in a body, although this is
difficult, because what is matter, and how do we quantify it? Weight is also ambiguous,
although it is agreed that it is a force. Mass is measured in kg, for example a 7 kg block,
and weight is a force, for example, a weight of 10 N.
If an object were taken to the moon, its mass would be the same, but its weight would be
less because the gravitational forces on the moon are less than on Earth. On Earth the two
terms are muddled because they are proportional. For example, double the mass and you
double the weight. People talk about losing weight when they really want to do is lose
mass.
2.4.1
Define linear momentum and impulse.
Linear momentum is the product of mass and velocity. P = mv. Momentum is a vector,
and the units are kg ms-1.
Impulse is the change in momentum, in any situation, particularly if it happens quickly.
(∆p = F ∆t).
2.4.2
State the law of conservation of linear momentum.
The law of conservation of linear momentum states that
The total linear momentum of a system of interacting particles remains constant
provided there is no resultant external force.
2.4.3
Derive the law of conservation of momentum for an isolated system consisting of
two interacting particles.
Image:Elastic Collision.gif
2.4.4
Solve problems involving momentum and impulse.
Momentum is a vector quantity. The outcome of a collision depends on the mass of each
particle, their initial velocities, and how much energy is lost in the collision. But
whatever, the outcome, momentum is always conserved. Any predicted outcome which
violates the conservation of linear momentum will not be accepted.
Example Problems

A railway truck, B, of mass 2000kg is at rest on a horizontal track. Another truck,
A, of the same mass moving with a speed of 5 ms-1 collides with the stationary
truck and they link up and move off together. Find the speed with which the two
trucks move off and also the loss of kinetic energy on the collision.

Suppose that in the previous example that after the collision truck A and truck B
do not link and that after the collision, truck A is moving with a speed of 1ms-1
and in the same direction as prior to the collusion. Find the speed of truck B after
the collision as well as the kinetic energy lost on collision.
Elastic and inelastic collisions.
Inelastic collisions are when mechanical energy is lost when two objects collide but
momentum is conserved.
Elastic Collisions happen when there is no mechanical energy lost or momentum lost in a
collision.
In the real world mechanical energy is always lost during a collision. But some do
approximate quite well to being elastic. The collision of two pool balls is almost elastic,
as is that between two steel ball bearings.
Example A car of mass 1000 kg is parked on a level road with its handbrake on. Another
car of mass 1500kg travelling at 10ms-1 collides with the back of the stationary car. The
two cars move together after collision in the same straight line. They travel 25m before
finally coming to rest. Find the average frictional force exerted on the cars as they come
to rest.
2.5.1
Define work.
Work is done when a force moves its point of application in the direction of the force. If
the force moves at right angles to the direction of the force, then no work has been done.
Work usually involves a transfer of energy from one form to another (ie kinetic to
gravitational). The amount of energy transferred is equal to the work done. Work has
been done against a force. Work is equal to the force multiplied by the distance moved –
OR work is equal to the magnitude of the force multiplied by the displacement in the
direction of the force.
W=FΔs
Work is a scalar in Nm or Joules; work done by a system is positive; work done on a
system is negative.
2.5.2
Determine the work done by a non-constant force by interpreting a forcedisplacement graph.
A force-displacement graph in this regard will probably be one about applying a force to
a spring. The area under the graph is the work done, which is then ½ Fs. But in this case
F = ks, which means that the work done is ½ X (ks) X s = ½ ks2. The work that has been
done is stored in the spring as elastic potential energy, Eelas. So, Eelas is ½ ks2. This
idea can be extended to find the work done by any non-constant force. If we know how
the force depends on displacement then to find the work done by the non-constant force
we just calculate the area under the force-displacement graph.
2.5.3
Solve problems involving the work done on a body by a force.
Remember to use W = force X distance (this is essentially it)
Example Problems:
A force of 100 N pulls a box of weight 200 N along a smooth horizontal surface as shown
below.
Calculate the work done by the force (a) in moving the box a distance of 25 m along the
horizontal and (b) against gravity.
2.5.4
Define kinetic energy. Kinetic energy is the energy that a body possesses by virtue of its
motion. The formula is ½ mv2 – the kinetic energy a body tells us how much work the
body is capable of doing.
[edit] 2.5.5
Describe the concepts of gravitational potential energy and elastic potential energy.
If an object of mass m is lifted to a certain height h above the surface of the earth then the
work done is mgh and the object now has a potential energy equal to the work done mgh
(force = mg, displacement = h). For example, if you place an object at rest on top of a
wall it has potential to do work, because if it falls off the wall onto a nail sticking out of a
piece of wood then it could drive the nail further into the wood. Work is needed to lift the
object on to the top of the wall and we can think of this work as being ‘stored’ as
potential energy in the object. The object has gravitational potential energy by virtue of
its position in the Earth's gravitational field ∆Ep = mg ∆h. No matter where an object is
placed in the Universe it will be attracted by the gravitational force of Earth. When an
object is moved a distance h in the Earth’s gravitational field and in the direction of the
field, its change in gravitational potential energy is mgh (provided that g is constant).
Elastic potential energy is similar. Work must be done to stretch a spring or similar
object, and this work done is stored in the spring as elastic potential energy – if you let it
go or remove the mass, it will spring back, converting this elastic potential energy into
kinetic energy. Eelas = ½ ks2.
[edit] 2.5.6
State the principle of conservation of energy.
There are several ways of stating this principle:



Overall the total energy of any closed system must be constant.
Energy is neither created nor destroyed, it just changes form.
There is no change in the total energy of the Universe.
[edit] 2.5.7
List different forms of energy and describe examples of the transformation of
energy from one form into another.

Thermal energy
Can be used to boil water and produce steam. The kinetic energy of the molecules of
steam (thermal energy) can be used to rotate magnets and this rotation generates an
electric current. The electric current transfers the energy to consumers where it is
transformed into, for example, thermal and light energy (filament lamps) and kinetic
energy (electric motors).

Chemical energy
This is associated with the electronic structure of atoms and therefore with the
electromagnetic force. An example of its transformation is in combustion in which
carbon combines with oxygen to release thermal energy, light energy, and sound energy.

Nuclear energy
An example of this is the splitting of nuclei of uranium by neutrons to produce energy.

Electrical Energy

Solar Energy
Different forms of energy all fall into the category of either potential or kinetic energy
and are all associated with one or other of the fundamental forces.
[edit] 2.5.8
Define power.
Power is the rate at which energy is transferred. This is the same as the rate at which
work is done. The unit for power is Js-1 or Watt. 1W = 1 Js-1.
Power = energy transferred/Time = work done/Time
If something is moving at a constant velocity v against a constant frictional force F, the
power P needed is P = Fv
[edit] 2.5.9
Define and apply the concept of efficiency.
Efficiency is the ratio of useful energy to the total energy transferred.
The easiest equation for this is Efficiency = Wu\Wa
Wu = mgh = Useful Work Wa = Fs = Actual Work
Depending on the situation, we can categorize the energy transferred (aka work done) as
useful or not. In a light bulb, the useful energy would be light energy; the ‘wasted energy’
would be thermal energy (and non visible forms of radiant energy).
Since it is a ratio it has no units. Often it is expressed as a percentage.
[edit] 2.5.10
Solve work, energy, and power problems.
See following attached pages.
Formulas in formula booklet are:


W = Fscosθ
Power = work / time = Fv
Efficiency is not in there.
[edit] 2.6.1
Draw a vector diagram to show that the acceleration of a particle moving with
uniform speed in a circle is directed toward the centre of the circle.
Velocity changes from VA to VB; the magnitude of the velocity stays the same, but the
direction changes – thus, the particle is experiencing acceleration.
[edit] 2.6.2
State the expression for centripetal acceleration.
The expression for centripetal acceleration is
ac = v2/r
[edit] 2.6.3
Identify the force producing circular motion in various situations.
We can find the expression for the centripetal force F by using Newton’s second law, F =
ma, so that, with
ac = v2/r
we have:
F = mv2/r
where m is the mass of the particle.
In order for the particle to move in a circle a force must act at right angles to the velocity
vector of the particle and the speed of the particle must remain constant. This means that
the force must also remain constant.
The effect of the centripetal force is to produce acceleration towards the centre of the
circle. The magnitude of the particle’s linear velocity and the magnitude of the force
acting on it will determine the circular path that a particular particle describes.
These are some situations in which centripetal forces arise:




Gravitational force.
Frictional forces behind the wheels of a vehicle and the ground.
Magnetic force.
Tension force in a string whirling an object around your head.
[edit] 2.6.4
Solve problems for particles moving in circles with uniform speed.
Retrieved from "http://kstruct.com/ib_notes/Topic_2_-_Mechanics"
9.1 Dynamics
[edit] 9.1.1
On an inclined plane, application of Newton's laws is somewhat complicated. Gravity
acts downwards, but the plane exerts a force back on the object at an angle. This can be
used to build a right angled triangle, since the resultant force down the slope angles to the
normal force, and gravity will be the hypotenuse. If there is friction acting, this will be in
the opposite direction to the motion down the plane.
[edit] 9.1.2
Friction is a force which opposes motion, thus if there is no motion, then there will be no
force caused by friction. Friction can never make an object move, it can only slow an
object down, and eventually stop it.
For two solid surfaces moving over each other, the friction will be affected by the nature
(roughness etc) of the two surfaces, however the surface area and velocity of the object
do not affect the friction. There are also two types of friction of solid surfaces, static and
kinetic friction. Static friction is that which stops objects from beginning to move, and
kinetic is that which slows objects down when they are moving.
These two types are defined individually by their constants µs and µk respectively. µs <;
µk in all cases (fairly obvious if you think about it).
The force of friction caused by each type is also dependent on the normal force the
surface is applying, thus Ffr =< µsFn for objects which aren't moving, and Ffr = µsFn for
objects which are moving. In the first case, the frictional force only exists if there is a
force being applied, thus the less than or equal to sign. Friction will counteract all forces
up to this point.
Objects falling through fluid (or indeed, air) also experience friction which opposes the
force of gravity. This friction increases with velocity, and so there eventually comes a
point where the force of gravity is being counteracted by friction, and the object falls at
constant speed. This is known as terminal velocity.
[edit] 9.2 Projectile motion
[edit] 9.2.1
When objects in projectile motion are launched at an angle, the horizontal and vertical
components must first be calculated. From the vertical, we should be able to calculate it's
peak height, and the time taken to reach this height. From there, we can calculate the time
required to reach the ground and using all these times, and the horizontal component, the
horizontal distance travelled.
[edit] 9.3 Simple harmonic motion
[edit] 9.3.1
Simple harmonic motion is like that of a pendulum, where the object moves away from,
and returns to the equilibrium point, and the restoring force is proportional to the
extension (i.e. the further from the center the pendulum is, the bigger the force pulling it
back). Starting from a fully extended position, displacement follows a cos curve, velocity
a -sin curve, and acceleration a -cos curve (we're differentiating with respect to t each
time : cos -> -sin -> -cos). As can be seen from this, a displacement vs acceleration graph
will be a straight line with a negative slope (cos and -cos).
[edit] 9.3.2
The period in simple harmonic motion (SHM) is the time required for the displacement to
return to its original position (complete one cycle). The frequency is 1/period, thus the
number of cycles per second. The amplitude is the maximum displacement from the
mean (or central) position. Period is also defined as T = (2 x Pi)/w where w = 2 x π x f
Graphing displacement vs time : This graph ranges from r (the amplitude) to -r, and
follows a cos curve (assuming we're starting at the extended position). It will be a
maximum at the points of extension and zero when the object passes through the mean
position.
Graphing velocity vs time : This graph ranges from rw to -rw (where w = 2 x π x f ).
Velocity will be a maximum when the object is an the mean position, and zero when the
extension is at a maximum. Thus it will follow a sin curve, with velocity being a
maximum and when displacement is zero and vice versa.
[edit] 9.3.3
The total energy for an object in SHM is constant. When displacement is maximum, all
the energy is potential (since it is being held up, compressed or whatever). When the
displacement is zero, all the energy is kinetic (because it is moving, and at it's lowest
point/no compression or extension etc), thus the two follow parabolic curves where the
sum of the two curves is always equal.
[edit] 9.3.4
As an example, we have SHM for a mass on a spring. When the mass is pulled
downwards to an extension of x, and then released, the mass undergoes vertical
oscillations which obey SHM.
Since the spring obeys Hooke's law ( F = kx, where k is the spring constant ) we know
that the restoring force = kx. Newton's second law. We also have F = ma, so combining
the two we get a = kx/m. This fits the SHM formula a = -w2x producing a = -k/m x (if
acceleration is in the same direction as x, i.e. away from mean position) Therefore, w2 =
k
/m and so, T = 2 x π x (√(m/k)) since T = (2 x π)/w (this is as given in the data book).
If the elastic limit of the spring is exceeded, then it will no longer follow Hooke's law,
and no longer follow SHM.
[edit] 9.3.5
Assume a mass is on a pendulum, with string length l, and is displaced by x. This forms
an angle of Ø between the mean position of the string, and the displaced position. The
force of gravity towards the origin = mg sinØ ( this can be shown with a right angle
triangle, mg is the hypotenuse, mg sinØ is the side at a tangent to the curve, because it is
a similar triangle to the one made by the mass, the origin and the top of the string ). For
small angles Ø, sin Ø = x/l(If we assume once again, that the curve between the origin and
the mass is actually a straight line, which is valid for small Ø). Thus, the force towards O
= mgx/l. Since F = ma, acceleration towards O = gx/l. Thus acceleration in x direction
(away from O) = -gx/l. So as above, w2 = g/l, and so, T = 2 x Pi x (square root of l/g) (once
agian, this is given in the data book).
[edit] 9.4 Circular motion
[edit] 9.4.1
Angular displacement : The angle from the centre of a circle (in radians) around which
an object has moved. It's symbol is Ø, and it is defined as Ø = s/rwhere s is the length
travelled around the circumference, and r is the radius. It is equivalent to displacement in
linear motion.
Angular velocity : Uses the symbol w. angular displacement/time (or w = ΔØ/Δt). Its linear
equivalent is velocity. This is no longer included in the syllabus.
[edit] 9.4.2
s = rØ : The length covered on the circumference is the radius x the angle (in radians)
covered. This allows the angular displacement to be converted into a length.
v = rw : The velocity of an object on the circumference is the radius times the angular
velocity (as defined above, in radians s-1)
[edit] 9.4.3
That an object rotating in a circle has an acceleration towards the centre can be shown by
taking two vectors at tangents to different points on the circle, and subtracting one from
the other. The resulting vector will be directed towards the centre of the circle.
If we take two such vectors (of length V) from close together, so the angular
displacement of their starting points is small, and draw them from the same origin point,
then the vector between their ends is the centripetal acceleration. Using similar triangles,
take the angle between the two vectors to be Ø (the same as the angular displacement),
and let the vector between them be ΔV. If ac is centripetal acceleration, then it equals
ΔV
/Δt. From the triangle, Ø = ΔV/V, so ΔV = VØ. Subbing into the acceleration formula, we
get ac = VØ/Δt. From above, w = Ø/Δt, thus Ø = wt, and so ac = vw. This can be rearranged
by subbing v = rw to get ac = rw2 = V2/r (as in the data book).
[edit] 9.4.4
For an object to travel in a circular path, constant acceleration is required. F=ma
(newton's second law) in circular motion F = |mac| = |m v2/r| = |4 x π2 x mr/T2| = |mrw2|.
[edit] 9.4.5
Bodies in circular motion include planets orbiting ( gravity provides ac ), buckets being
spun around on ropes ( rope provides ac ) and cars going around banked curves ( The
'down the slope' component of the normal force provides ac ). Problems involving these
can then be solved by finding ac and then working back through the above equations.
[edit] 9.4.6
Masses moving in a vertical circle and affected by gravity : At every point in the circle,
the force keeping the object moving in a circle (ie towards the center) must be constant,
and can be calculated with the equations above. At the top of the circle, the force total
keeping the object in a circle is Fc - Fg (since gravity is providing the force, Fc is
smaller). At the bottom, Ft = Fc + Fg. At the sides, Ft = Fc (since no component of gravity
is acting towards the center).
[edit] 9.4.7
Centripetal force does not change the kinetic energy of a body because it only affects the
direction of the velocity, not the magnitude, and Ek = 1/2mv2 is only conserved with the
magnitude.
[edit] 9.5 Universal gravitation
[edit] 9.5.1
Newton's law of universal gravitation : F = (Gm1m2) / (R2), where G is the gravitational
constant (on earth, 6.67 x 10-11 N m2 kg-2), m1 and m2 are the two masses in question, and
R is the distance between the centers of the two masses. This must be applied both as a
source of centripetal and gravitational force (in the pulling objects down sense). (This
equation is in the data book)
[edit] 9.5.2
Gravitational field strength is the gravitational force per unit mass at a particular point in
the gravitational field. In simple terms, it's the force on one kg of mass, which is 9.8 N on
Earth.
[edit] 9.5.3
Beyond the surface of the the mass, the force of gravity decreases in a parabolic curve
(because of the R2 term it obeys the inverse square law). Inside the mass is messy to work
out, but luckily it isn't required.
[edit] 9.5.4
Gravitational potential ( with the symbol V ) is defined as the potential energy per unit
mass at a point in the field. V approaches 0 as the distance between the masses
approaches infinity. V = -Gm/r where m is the mass of the earth (or whatever) and r is the
distance from the center of the Earth. (given in the data book)
[edit] 9.5.5
Escape velocity is the speed at which an object must leave the planet's surface to
completely escape its gravitational field. For this, the object must be given enough kinetic
energy to go from the surface, where V = -Gm/r x (mass of object) to where V = 0. Thus,
1
/2m(Vescape)2 - GMm/r = 0 rearranges to Vescape = (2GM/r)^1/2. Note, this one is not in the
data book, so you can either memorise it, or remember how to derive it.
[edit] 9.6 Momentum and energy
[edit] 9.6.1
The work done on a force displacement graph is the area under the graph. This would
normally be found by integrating, but since there's no calculus in this course, in any given
question you'll be able to break the area up into triangles and rectangles to find the area.
[edit] 9.6.2
For a linear spring, the force to extend (or compress) the spring is directly proportional to
the displacement. If no force acts on it, the spring will naturally return to its equilibrium
position. Unless, of course, the spring is displaced beyond it's elastic, in which case it's
broken and probably no longer interesting.
[edit] 9.6.3
Energy (elastic potential) is stored in a spring by compressing or extending it. When
released, the spring will convert this energy into other forms (kinetic, thermal, sound
etc.).
[edit] 9.6.4
Deriving conservation of momentum for two bodies from Newton's laws.
First assume there are two objects, of mass m1 and m2 travelling towards each other at
velocities v1 and v2. First we write Newton's second law in the form 'force is the rate of
change of momentum' (F = ΔP/Δt), then multiply both sides by t to get Ft = ΔP.
This can then be applied to the first mass giving the equation Ft = m1v'1 - m1v1. Using
Newton's third law, we know that the force on m2 will be equal and opposite and so the
second mass. -Ft = m2v'2 - m2v2.
These two can then be combined to form m1v'1 - m1v1 = -(m2v'2 - m2v2) which rearranges
to m1v1 + m2v2 = m1v'1 + m2v'2, which is the law of conservation of momentum for two
masses.
[edit] 9.6.5
In two dimensions, the law of conservation of momentum applies in both senses, so the
momentum value must be broken up into its vector components. The total amount of
vertical (or north/south or whatever) momentum will be conserved, and will the total
amount of horizontal momentum. Therefore, a problem involving this can be treated as
two separate one dimensional conservation of momentum problems.
[edit] 9.7 Rotational motion of a rigid body
[edit] 9.7.1
Angular acceleration is the rate of change of angular velocity : α = Δw/Δt. This is no
longer included in the syllabus.
Torque is the rotational equivalent of force : T = I x α (moment of inertia x angular
acceleration).
Moment of inertia is the rotational equivalent of mass. It is defined as the sum of (mr2)
for every point mass in the system where m = mass and r = the distance form the axis of
rotation. This allows dumbbell type arrangements to be calculated, but the following
shape formulae should also be known



A 'hoop' or cylinder rotating around the center of the hoop : I = MR2
A solid disk rotating around the center of the disk : I = 1/2MR2
A solid sphere rotating about any axis running through the centre : I = 2/5MR2
Angular momentum is the equivalent of linear momentum : L = Iw (angular momentum
= moment of inertia x angular velocity).
[edit] 9.7.2
All of the quantities and equations in rotational motion are equivalent to others in linear
motion. The tables below illustrate this equivalence.
Linear quantity Rotational quantity
Displacement : s Angular displacement : Ø
Velocity : v
Angular velocity : w
Acceleration : a Angular acceleration : α
Force : F
Torque : T
Mass : m
Moment of inertia : I
Linear Equation
Rotational equation
s
Velocity : v = /t
Rotational velocity : w = Ø/t
Acceleration : a = v/t Angular acceleration : α = w/t
Force : F = ma
Torque : T = I x α
Work done : W = Fs Work done : W = TØ
KE : E = 1/2mv2
KE : E = 1/2Iw2
Momentum : p = mv Angular momentum : L = Iw
All these can be applied as appropriate to problems.
Retrieved from "http://kstruct.com/ib_notes/Topic_9_-_Mechanics"