Wednesday, August 15, 2018

Unizor - Physics4Teens - Mechanics - Rotational Dynamics





Notes to a video lecture on http://www.unizor.com



Torque



So far we were mostly considering translational motion of point-objects - a motion along a straight line with or without external forces
acting upon this object. We have specified three Newton's Laws of this
motion and derived a lot of interesting facts based on these laws.



Let's recall these laws.



The Newton's First Law is the familiar Law of Inertia that states that
an object at rest stays at rest and an object in uniform motion stays in
this uniform motion, unless acted upon by unbalanced forces.



The Newton's Second Law brings quantitative relationship to vector of force (F), mass (m) and vector of acceleration (a):

F = m·a



The Newton's Third Law states that for every action there is an equal in magnitude and oppositely directed reaction.



Rotational motion obeys the rules in many respects analogous to the laws of translational motion, except we have to change linear movement to rotation, which, in essence, is an angular movement with constant radius.



Consider a point-object of some mass m connected by a weightless rigid rod of the length r to an axis, around which this object can rotate within a plane of rotation that is perpendicular to an axis of rotation.

The picture below illustrates such a movement and also indicates the position of angular velocity ω
of a rotation, which was explained earlier, when we studied the
Kinematics of rotation (we assume your familiarity with this topic and
strongly recommend to refresh it before proceeding any further).





Let's discuss the similarities and differences between translational movement along a straight line and rotation around an axis within a plane of rotation perpendicular to this axis.



Obviously, the time concept remains the same in both types of motion.



There is a clear rotational analogy to the Newton's First Law.


An object at rest stays at rest and an object in uniform rotation
stays in this uniform rotation, unless acted upon by unbalanced forces.




The Newton's Third Law is not really specific for a form of motion, so there is no need to address it separately at this moment.



The Newton's Second Law requires certain modification to be applied to rotation.



Let's address the main concept of Dynamics - the force - in connection to rotation.



The very important characteristic of force in Dynamics of translational movement along a straight line is that it can be measured by its effect on objects of certain inertial mass. Thus, a measure of force that gives linear acceleration of 1 m/sec² to an object of inertial mass of 1 kg is 1 newton. The same force applied to an object of 0.5 kg of inertial mass will cause 2 m/sec² linear acceleration.

The same force applied to different objects of the same inertial mass will cause the same linear acceleration etc.



The situation with rotational movement is not the same.



Consider a simple experiment of opening a door. If you apply a force to
open a door at its edge opposite to hinges, where the handle is usually
located, it opens relatively faster than if you apply exactly the same
force in the middle of a door. The closer a point of application of the
same force to hinges - the slower a door opens, if the force applied to
it is the same. In an extreme case, when we apply the pressure where the
hinges are, a door will not open at all.

As a continuation of this experiment, if we want to achieve the same
speed of the opening of a door by applying the force at different
distances from the hinges, we need more efforts for a point of force
application located closer to the hinges.



We can measure the force, the distance from the hinges of a point of application of this force and an angular acceleration of the door and experimentally come up with the fact that for the fixed force the angular acceleration of the door is proportional to the distance of a point of application of the force from the hinges.

Moreover, leaving the point of application of the force the same and changing the force, we can determine that angular acceleration is proportional to the force.



What it means is that in a case of rotation a force by itself does not
determine the final motion of a rotating object. It's a product of a
force F and radius to a point this force is applied r, that determines the final effect. This product τ = F·r is called the torque and it is rotational equivalent of a force in translational movement.

TranslationRotation
Force
F
Torque
τ = F·r


Returning to a picture above, we can apply some force F to
any point on the rigid rod, connecting our point-object to an axis of
rotation, directing this force perpendicularly to the rod, and observe
that the resulting angular acceleration of the object is proportional to
both the force F and the distance r from the axis to a point of application of this force, thus proportional to torque τ = F·r.



Now we have concluded that an angular acceleration α of rotational motion is proportional to a torque τ. This is analogous to linear acceleration a of translational motion being proportional to a force F. The coefficient of proportionality for translational motion is inertial mass m of an object (this is the Newton's Second Law F=m·a).



The obvious question is, what is the coefficient of proportionality between angular acceleration and torque?

Answer to this question will result in rotational equivalent of the Newton's Second Law.



We have experimentally established that equal torques produce equal angular accelerations.

Consider a rotation illustrated on the picture above. Assume that the point of application of force F is exactly at the point-object of mass m rotating around an axis at a distance r
from it on a rigid weightless rod. Assume further that our force acts
within a plane of rotation and directed perpendicularly to the rod.



During an infinitesimal time interval dt the motion of an object can be considered as linear and, therefore, the Newton's Second Law can be applied, giving F=m·a.

Now we can express it in terms of angular acceleration and torque as follows:

a = r·α

τ = F·r

Hence,

F·r = (m·a)·r = m·r²·α

Finally,

τ = (m·r²)·α



One more logical step is needed. We started from a force applied on an object itself at a distance r
from an axis. But we have experimentally established that equal torques
produce equal actions. It means that some other force applied to some
other point will produce the same effect, causing the same angular
acceleration, as long as the torque is the same.

So, the equality τ=(m·r²)·α is universal, regardless of point of application of force since it depends not on force, but on torque.



The above equality represents the rotational analogue of the Newton's Second Law.



Some generalization can be applied to the above.

What if the force, acting within a plane of rotation, is not perpendicular to a rod?

Obvious solution is to replace vector F
with its projection onto a line on the rotation plane that is
perpendicular to a radius and to multiply the product of two scalars F·r by a sine of an angle between corresponding vectors, effectively using a vector product Fr.

So, more general definition of a torque is a vector (or, more precisely, pseudo-vector)

τ = Fr

whose direction, like a direction of an angular acceleration, is along an axis of rotation.

So, for rotational movement the vectors of torque τ and angular acceleration α are collinear, similarly to collinearity of vectors of force F and linear acceleration a for translational movement.



We'd like to note that for purposes of simplicity in this course we will
rarely deal with forces not perpendicular to a radius of rotation.



IMPORTANT TERMINOLOGY POINTS



1. The torque τ is often called the moment of force.



2. Recall the expression tying together a torque τ and an angular acceleration α

τ = (m·r²)·α

The expression m·r² is called moment of inertia, is symbolized by letter I, which allows to specify the formula above as

τ = I·α

where I=m·r² is a moment of inertia, playing the same role in this equation as inertial mass m in the Newton's Second Law and representing resistance to a rotational force.



TranslationRotation
Force
F
Torque
τ = F·r
Acceleration
a
Angular Acc.
α = a/r
Inertial
Mass

m
Moment
of Inertia

I = m·r²
Newton's
Second Law

F=m·
Rotational
Equivalent

τ = I·α

Monday, August 13, 2018

Unizor - Physics4Teens - Mechanics - Rotational Kinematics





Notes to a video lecture on http://www.unizor.com



Rotational Kinematics



So far we were mostly considering translational motion of point-objects - a motion along a straight line with or without external forces acting upon this object.



Rotational motion obeys the rules in many respects analogous to the laws of translational motion, except we have to change linear movement to rotation.



Consider a point-object m connected by a rigid rod of the length r to an axis, around which this object can rotate within a plane of rotation that is perpendicular to an axis of rotation.

The picture below illustrates such a movement and also indicates the position of angular velocity ω of a rotation, which we will explain later.





Let's discuss the similarities and differences between translational movement along a straight line and rotation around an axis within a plane perpendicular to this axis (a plane of rotation).



The first main concept of translational motion is position or distance from the beginning of motion (for a straight line movement) as a function of time. In rotational motion its equivalent is angle of rotation from some original position as a function of time.



TranslationRotation
Distance
s(t)
Angle
φ(t)


The next concept is speed or (better) velocity of translational motion. This is a first derivative of position (or distance) by time:

v(t) = s'(t).

Its equivalent for rotational motion is angular speed, which is a first derivative of angle of rotation by time:

ω(t) = φ'(t)



While vector character of speed of translational motion is obvious and is reflected in the term velocity, vector character of angular speed is less obvious.

The angle of rotation from the first glance is a scalar function of time. But only from the first glance.



In theory, rotational motion always assumes existence of an axis of
rotation and a plane of rotation. To reflect these characteristics and a
magnitude of angular speed, an angular speed is represented by a vector
from a center of rotation along an axis of rotation perpendicularly to a
plane of rotation with a magnitude equal to a value of angular speed.



This allows to represent the rotation in its full spectrum of
characteristics - magnitude, axis, plane of rotation. The picture above
represents angular speed as a vector ω(t), which we may call angular velocity vector.



There is one more characteristic of rotational motion not yet discussed - its direction. It is also reflected in angular velocity
as a vector by its direction. In theory, we can choose two different
directions along the axis of rotation. The direction chosen is such
that, if we look from its end onto a plane of rotation, the rotation is
counterclockwise. Another interpretation of this is the "rule of the
right hand" because if you put you right hand on a plane of rotation
such that your finger go around the axis of rotation pointing to a
direction of rotation, your thumb points to a direction of the angular velocity vector.

So, angular velocity vector represents axis, plane, direction of rotation as well as magnitude of angular speed.



To be more precise, since this vector representation of angular velocity is a little unusual, it is customary to call it "pseudo-vector" instead of "vector".



During infinitesimal time interval dt an object rotating around an axis on a radius r turns by an angle dφ(t), covering the distance ds(t)=r·dφ(t) (this is the length of an arc of radius r and angle , according to a known formula of geometry).

From this follows:

ds(t)/dt = r·dφ(t)/dt or

v(t) = r·ω(t)



TranslationRotation
Speed
v(t)=s'(t)
v(t)=r·ω(t)
Angular Speed
ω(t)=φ'(t)
ω(t)=v(t)/r


The next concept is acceleration that needs its rotational analogue. Obviously, it's the first derivative of angular velocity or the second derivative of an angle of rotation by time.

Using the vector interpretation of angular velocity, we can consider angular acceleration as a vector as well. It is also directed along the axis of rotation.

During infinitesimal time interval dt an angular velocity ω(t) changes by dω(t).

From this follows relationship between linear acceleration a and angular acceleration α:

a(t) = dv(t)/dt = r·dω(t)/dt or

a(t) = r·α(t)



TranslationRotation
Acceleration
a(t)=v'(t)
a(t)=r·α(t)
Angular Acc.
α(t)=ω'(t)
α(t)=a(t)/r


Obviously, integrating the definitions of angular velocity ω and angular acceleration α for constant angular acceleration, we come up with formulas similar to those familiar from translational movement:

ω(t) = ω(0) + α·t

φ(t) = φ(0) + ω(0)·t + α·t²/2



Angular acceleration as a vector (as a pseudo-vector, to be exact), is colinear to the axis of rotation, because angular velocity is.

If rotation goes as on the above picture and the speed of rotation increases, the angular acceleration would be directed upwards, the same way as the angular velocity.

Monday, August 6, 2018

Unizor - Physics4Teens - Mechanics - Statics - Equilibrium





Notes to a video lecture on http://www.unizor.com



Equilibrium



Statics is a part of Mechanics that studies forces not as
quantitative measure of a motion they cause (that is a subject of
Dynamics), but from the more fundamental viewpoint of whether these
forces are or are not balanced, that might or might not cause the motion
of objects these forces act upon.



Historically, studies of static aspects of forces precede quantitative
studies of motion in Dynamics. People, first of all, were concerned with
how to build bridges and buildings, so that they stay in place and not
destroyed by gravity, wind or load. And a concept of equilibrium is a central point of Statics, it's goal and purpose.



Equilibrium, as it is understood in Statics, is a state of forces, that result in an object acted upon to be in the state of rest.



A person standing on a floor is at rest because two main forces that act
upon him, the gravity and the reaction of the floor are equal in
magnitude and opposite in direction, they balance each other, which
results in a state of equilibrium.



A person standing on the weights to check his weight is in a state of equilibrium
because its weight is balanced by elastic force inside the weights that
is equal in magnitude and opposite in direction to the force of
gravity. The equality in magnitude allows to measure the weight by
measuring the elasticity inside the weights.



A building is in a state of equilibrium because its each part's weight is balanced by equal in magnitude and opposite in direction force of reaction.



An airplane, flying horizontally on its route, is in a state of vertical equilibrium
because its weight is balanced by the lifting power of the air under
its wings, where the air pressure is larger because of the wings' shape.



Let's study forces applied to the same object and balance each other causing this object to be in equilibrium.

As we know, force is a vector. All forces acting on the same point-object can be added as vectors, using the rules of the Vector Algebra.

If the result is a null-vector, we have an equilibrium.



So, if there are N forces Fi (where 1 ≤ i ≤ N) acting simultaneously on the same point-object, the condition of equilibrium is

ΣFi = 0 .



Let's consider a few examples.



1. An object of weight W is hanging on three threads as pictured below.



Thread a is horizontal, thread b is at angle φ to horizon.

The system is in equilibrium.

What is the magnitude of tension forces Ta and Tb on threads a and b?



Solution

There are three forces acting at the point where all threads are connected:

vertical down - the weight W of the object;

tension Ta horizontally to the left along the a thread;

tension Tb at angle φ to horizon along the b thread.

Since the system is in equilibrium, this point where threads come
together does not move in any direction. In particular, it does not move
in a vertical, nor horizontal direction.

This is sufficient to determine magnitude of tension vectors Ta and Tb.

In the horizontal direction the force of gravity is irrelevant, so the
only two forces acting on out object in horizontal direction are tension
Ta acting to the left and a projection of tension Tb on the horizontal line acting to the right.

That gives the first equation about magnitudes of tension forces:

Ta = Tb·cos(φ)

Now consider the vertical direction. The thread a is irrelevant. So, the only two forces acting vertically, are weight W pulling down and a vertical component of Tb that is equal to Tb·sin(φ) pulling upwards.

This gives the second equation

Tb·sin(φ) = W

From these equations we derive:
Tb = W/sin(φ)

After which we can find Ta:

Ta = W·cos(φ)/sin(φ)



2. Two objects of mass M (larger, on an inclined plane) and m (smaller, hanging freely over the edge of an inclined plane) are connected with a thread that goes over a pulley.



What is the angle of an inclined plane φ for this system to be in equilibrium?

Ignore the friction.



Solution

Let's represent the weight of an object on an inclined plane as a sum of two forces:

one is perpendicular to the plane and balanced by a plane's reaction;

another is parallel to the plane and balanced by a tension of a thread.

To be in equilibrium, the force of weight of an object hanging freely
over the edge of an inclined plane must be equal in magnitude to a
tension of a thread (same thread, same tension as before).

Therefore, a component of the weight of an object on a plane that goes
parallel to a plane must be equal in magnitude to a weight of another
object.

M·g·sin(φ) = m·g

From this we derive

sin(φ) = m/M

φ = arcsin(m/M)

Monday, July 30, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Rockets and Gravitation





Notes to a video lecture on http://www.unizor.com



Rocket in Gravitational Field



In the previous lectures we examined the motion of a rocket with no
external forces (like gravity or drag) acting on it and came with the
Rocket Equation, stating that

IF

m(t) is the mass of a rocket (including propellant) at time t and

V(t) is its speed in some inertial reference frame
(related to stars, for example, and positioned in such a way that a
rocket moves along one axis in a positive direction) and

m(t+Δt) is the mass after time interval Δt, during which a rocket was throwing propellant with constant (relatively to a rocket) effective exhaust speed ve and

V(t+Δt) is its speed after time interval Δt in the same inertial reference frame

THEN

the maximum increment of the rocket's speed

ΔV=V(t+Δt)−V(t) during this interval of time Δt is

ΔV = −ve·ln[m(t)/m(t+Δt)]



The equation above should be interpreted as the vector equation.

If inertial frame of reference is directed in such a way that the rocket
moves along one axis in positive direction and the exhaust is directed
backwards relative to a rocket's movement, the ve is negative. The mass during this process decreases, so m(t) is greater than m(t+Δt) and the logarithm is positive. This results in the positive ΔV, that is a rocket accelerates.

If the exhaust is directed forward relative to a rocket's movement, the ve is positive, ΔV is negative and a rocket decelerates.



Now let's add gravity as an external force that acts on a rocket.

There are two cases:

(a) when a rocket is launched from a planet to an orbit, gravity acts
against its movement, thus requiring extra effort by an engine to
overcome the gravity;

(b) when a rocket returns back to a planet for soft landing, gravity
acts in the direction of its movement, but we have to decelerate a
rocket using propellant exhausted also in the same direction, so it also
requires extra effort by an engine.

So, no matter how rocket moves in the gravitational field, an engine
should work harder to either launch it to an orbit or to slow it down
for soft landing.



We will consider the launching from the Earth case only.



Let's follow the same logic as in case of a rocket moving in empty space
with no forces involved and add the effect of gravity in the equation
of conservation of momentum.



1. At moment t the momentum of an entire system of a rocket with its propellant was equal to m(t)·V(t).



2. During the next time interval dt the rocket has exhausted m(t)−m(t+dt)=−dm(t) of propellant with constant relatively to a rocket speed ve. Since a rocket moves in some inertial system with speed V(t) and propellant moves relatively to a rocket with constant speed ve, the speed of propellant in the inertial system equals to ve+V(t). This resulted in the momentum of exhausted propellant at moment t+dt to be dm(t)·[ve+V(t)].

This equation should be interpreted in the vector form. When a rocket
accelerates, velocity vector of its movement and velocity vector of
exhausted propellant are opposite in their directions.



3. A rocket with remaining propellant at moment t+dt has mass m(t+dt)=m(t)+dm(t), velocity V(t+dt)=V(t)+dV(t) and momentum

[m(t)+dm(t)]·[V(t)+dV(t)]



4. When rocket leaves the planet, the force of gravity F=m(t)·g acts against its movement. During time dt it reduces the impulse of a rocket by

dt = m(t)·g·dt.



Now we are ready to apply the Law of Conservation of Momentum.

Item 1 above describes the momentum of the system at time t.

At the moment t+dt the momentum of the system is a combination of the momentum of the exhausted propellant during time dt
(see item 2 above) plus the momentum of the remaining rocket mass (see
item 3 above) plus impulse of the gravitational force (see item 4
above).



Equalizing these two momentums at time t and t+dt, according to the Law of Conservation of Momentum, we get the following equation:

m(t)·V(t) = −dm(t)·[ve+V(t)] + [m(t)+dm(t)]·[V(t)+dV(t)] + m(t)·g·dt.



We would like to express the dependency between rocket's speed and the
way it exhausts its propellant without mentioning the time parameter.
This can be done by using the following:

m'(t) = dm(t)/dt (by definition of the derivative and differential)

From this:

dt = dm(t)/m'(t)



The rocket equation above can be simplified. After this the equation looks like

0 = −ve·dm(t) + m(t)·dV(t) + dV(t)·dm(t) + m(t)·g·dt

Ignoring an infinitesimal of a higher order dV(t)·dm(t), the resulting equation looks like

m(t)·dV(t)+m(t)·g·dt=ve·dm(t)



Divide both parts by m(t) and take into consideration that dm(t)/m(t) = d[ln(m(t))]. Then the differential equation of a rocket in the gravitational field looks like

dV(t) + g·dt = ve·d[ln(m(t))]

Replacing dt with its equivalent dm(t)/m'(t), we obtain an equivalent equation

dV(t)+g·dm(t)/m'(t) = ve·dm(t)



Expression on the right is positive because

(a) m(t) is a decreasing function,

(b) ln(m(t)), therefore, is also a decreasing function,

(c) differential of a decreasing function d[ln(m(t))] is always negative,

(d) ve is negative since it is a vector
directed against the movement of a rocket, which we consider as moving
to a positive direction of a coordinate axis,

(e) product of two negative values is positive.



As is obvious from this equation, unless dt is less than ve·d[ln(m(t))], the rocket will not move from the launching pad.

We can simplify this launching condition, using the following:

d[ln(m(t))] = [m'(t)/m(t)dt

This allows to express this condition as

g is less than ve·[m'(t)/m(t)]
or

ve·m'(t) is greater than g·m(t)



Integrating the differential equation of a rocket in the gravitational field on the interval Δt of time from the beginning of engine's work tbeg to the end of this period tend, we obtain the equation for an increment of the rocket's speed during this interval:

V(tend)−V(tbeg)+g·(tend−tend)=

= ve·
[ln(m(tend))−ln(m(tbeg))]=

= ve·
ln[m(tend)/m(tbeg)]=

= −ve·
ln[m(tbeg)/m(tend)]



That is,

ΔV(t) = −ve·ln[m(tbeg)/m(tend)] − g·(tend−tend)



The last equation does not take into consideration that the force of
gravity decreases with height. It's relatively precise only in the
beginning of the rocket's movement. Obviously, taking this factor into
consideration will complicate the calculations.



The next complication is the drag of the atmosphere, which is not that
important in theory, but for practical matters must be taken into
consideration.



Another important factor of launching is the planet's rotation. If we
launch a rocket to the East, the Earth's rotation helps to achieve
required speed.



All these and other complications make rocket science a rather involved theory.

Friday, July 27, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Rocket Calculations





Notes to a video lecture on http://www.unizor.com



Rocket Calculation 1



Let's use the rocket equation

ΔV = −ve·ln[m(tbeg)/m(tend)]

to calculate how much propellant must be taken by a rocket to reach an orbit.



Here ve is effective exhaust speed, m(tbeg) - mass of a rocket in the beginning of a time period during which a rocket's engine is working, m(tend) - mass of a rocket at the end of this period of acceleration or deceleration.



Recall that the minus sign in this equation signifies the vector
character of the movement: positive direction of the exhausted
propellant (that is, the same as the movement of the rocket) causes
negative increment in rocket's speed - deceleration, while the negative
direction of exhausted propellant (that is, opposite to the movement of a
rocket) causes increase in rocket's speed - acceleration.



Contemporary rocket engine can have a very high effective exhaust
velocity. The speed of about 4km/sec is mentioned in a few sources we
are familiar with. So, we can assume that

ve=4000m/sec.



An international Space Station's speed is about 7.8km/sec, as was calculated in one of the previous lectures on gravity.

Assuming that the initial speed of a rocket is zero, the increment of speed of a rocket must be

ΔV = 7800m/sec



From this follows that

ln[m(tbeg)/m(tend)] =

= 7800/4000 = 1.95




Therefore,

m(tbeg)/m(tend) ≅ 7



So, the mass of a rocket at start is 7 times greater than its mass at
the end of its acceleration. For example, to launch 1,000 kg of useful
equipment and/or passengers to an International Space Station we will
need 6,000 kg of fuel.





Rocket Calculation 2



We still assume that

ve=4000m/sec.



A rocket that goes far from Earth needs about 11.2km/sec speed to escape Earth gravity.

Assuming that the initial speed of a rocket is zero, the increment of speed of a rocket must be

ΔV = 11200m/sec



From this follows that

ln[m(tbeg)/m(tend)] =

= 11200/4000 = 2.8




Therefore,

m(tbeg)/m(tend) ≅ 16



So, the mass of a rocket, that is supposed to leave the Earth's gravity,
at start is 16 times greater than its mass at the end of its
acceleration. For example, to launch 1,000 kg of useful equipment and/or
passengers to Mars we will need 15,000 kg of fuel.

Wednesday, July 25, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Rocket Equation





Notes to a video lecture on http://www.unizor.com



Ideal Rocket Equation

(Tsiolkovsky's Equation)



The most important difference between the motion of a rocket and motions
analyzed before in this course is that the rocket propels itself by
throwing back part of its own mass (propellant), thus becoming lighter.
Its mass is variable. Before, in most cases, we were dealing with motion
of objects of some specific mass, not changing during the movement.



In this lecture we will analyze the motion of an ideal rocket that
throws back propellant with constant (relatively to the rocket) speed.
The formula that will be derived was suggested by Russian mathematician
K.E.Tsiolkovsky and is historically named after him, though he was not
the first to derive it.



In particular, we will analyze the dependency between loss of the total
mass of a rocket, throwing propellant backward in the absence of any
external forces, and its gain in speed caused by this process.



The final formula we will derive states that

IF

m(t) is the mass of a rocket (including propellant) at time t and

V(t) is its speed in some inertial reference frame
(related to stars, for example, and positioned in such a way that a
rocket moves along one axis in a positive direction) and

m(t+Δt) is the mass after time interval Δt, during which a rocket was throwing propellant with constant (relatively to a rocket) effective exhaust speed ve and

V(t+Δt) is its speed after time interval Δt in the same inertial reference frame

THEN

the maximum increment of the rocket's speed

ΔV=V(t+Δt)−V(t) during this interval of time Δt is

ΔV = −ve·ln[m(t)/m(t+Δt)]



The equation above should be interpreted as the vector equation.

If inertial frame of reference is directed in such a way that the rocket
moves along one axis in positive direction and the exhaust is directed
backwards relative to a rocket's movement, the ve is negative. The mass during this process decreases, so m(t) is greater than m(t+Δt) and the logarithm is positive. This results in the positive ΔV, that is a rocket accelerates.

If the exhaust is directed forward relative to a rocket's movement, the ve is positive, ΔV is negative and a rocket decelerates.



Here is how to derive this formula.

First of all, let's recall that in the absence of external forces the
combined momentum of motion of a system of objects is constant.
This is the Law of Conservation of Momentum.

Let's apply this law to our situation.



Assume, we are comparing the momentum of the system at times t and infinitesimally incremented t+dt.



1. At moment t the momentum of an entire system of a rocket with its propellant was equal to m(t)·V(t).



2. During the next time interval dt the rocket has exhausted m(t)−m(t+dt)=−dm(t) of propellant with constant relatively to a rocket speed ve. Since a rocket moves in some inertial system with speed V(t) and propellant moves relatively to a rocket with constant speed ve, the speed of propellant in the inertial system equals to ve+V(t). This resulted in the momentum of exhausted propellant at moment t+dt to be dm(t)·[ve+V(t)].

This equation should be interpreted in the vector form. When a rocket
accelerates, velocity vector of its movement and velocity vector of
exhausted propellant are opposite in their directions.



3. A rocket with remaining propellant at moment t+dt has mass m(t+dt)=m(t)+dm(t), velocity V(t+dt)=V(t)+dV(t) and momentum

[m(t)+dm(t)]·[V(t)+dV(t)]



Now we are ready to apply the Law of Conservation of Momentum.

Item 1 above describes the momentum of the system at time t.

At the moment t+dt the momentum of the system is a combination of the momentum of the exhausted propellant during time dt (see item 2 above) plus the momentum of the remaining rocket mass (see item 3 above).



Equalizing these two momentums at time t and t+dt, which is the consequence of the Law of Conservation of Momentum, we get the following equation:

m(t)·V(t) =

= −
dm(t)·[ve+V(t)] +

+ [m(t)+dm(t)]·[V(t)+dV(t)]



This can be simplified. After this the equation looks like

0 = −ve·dm(t) + m(t)·dV(t) +

+
dV(t)·dm(t)




The last member in this equation dV(t)·dm(t) is an infinitesimal of the higher order that we can remove from this equation, and the resulting equation looks like

m(t)·dV(t) = ve·dm(t)



Divide both parts by m(t) and take into consideration that dm(t)/m(t) = d[ln(m(t))]. Then our equation looks like

dV(t) = ve·d[ln(m(t))]



Integrating this on the interval Δt of time from the beginning of engine's work tbeg to the end of this period tend, we obtain the equation for an increment of the rocket's speed during this interval:

V(tend) − V(tbeg) = ΔV(t) =

= ve·
[ln(m(tend))−ln(m(tbeg))]=

= ve·
ln[m(tend)/m(tbeg)]=

= −ve·
ln[m(tbeg)/m(tend)]



That is,

ΔV(t) =−ve·ln[m(tbeg)/m(tend)]



So, the increment of a rocket's speed during a period of Δt=tend−tbeg,
when its engine works, exhausting the propellant, equals to a product
of the speed of exhausted propellant times a logarithm of a ratio of the
rocket's mass at the beginning of this period to its mass at the end of
it.



The minus in front of a formula is very important. This is a vector
equation and, if the exhaust is directed back relatively to rocket's
movement (that is, ve is negative), the
increment of a rocket's speed is positive, a rocket accelerates; if,
however, the exhaust is directed towards a tocket movement (that is, ve is positive), the increment of a rocket's speed is negative, a rocket slows down.



The formula above is the Tsiolkovsky's formula and is called the "ideal rocket equation".

Tuesday, July 17, 2018

Unizor - Physics4Teens - Mechanics - Dynamics - Spring Oscillation





Notes to a video lecture on http://www.unizor.com



Spring



Consider a point-object of mass m, hanging vertically at
the lower end of a weightless spring, that is fixed at the upper end.
Under the weight of this object a spring will stretch a little from its
neutral position.



The Hook's Law for a spring, which will be used to solve this problem, involves a spring's elasticity constant k, that we assume is given.



Let's stretch this spring even more, so that the distance between an object at its bottom and a spring's neutral level is L and let it go without any push.



Our task is to analyze the oscillation of the object as a function x(t) of its vertical deviation from a spring's neutral position.







The obvious initial conditions of the motion of our object are:

x(0) = L

x'(0) = 0



There are two forces acting on our object:

(a) its weight W, directed vertically down and equal in magnitude to m·g, where g is the acceleration of free falling

W = m·g

(b) the spring's elasticity force F, equal in magnitude to a coefficient of elasticity k
multiplied by a displacement of the spring's bottom end from a neutral
level; the direction of this force is always against the direction of
the displacement

F = −k·x(t)



The resultant of the superposition of these two forces can be equated to
mass times acceleration of the object, according to the Newton's Second
Law:

m·g − k·x(t) = m·x"(t)

This is the differential equation that defines the movement of our object.



We don't have to resort to modifying this differential equation with an
approximate one to be able to solve it. It is fully solvable and the
general solution of this linear differential equation of the second
order is

x(t) = C1·cos(t·√k/m) +

+ C2·sin(t·√k/m) + m·g/k




Now we can apply the initial conditions to determine constants C1 and C2.

Since x(0) = L,

C1 = L − m·g/k

Since x'(0) = 0,

C2 = 0



This produces the following expression for x(t):

x(t) = (L − m·g/k)·cos(t·√k/m) +

+ m·g/k




Interestingly, if

L − m·g/k = 0 or L·k = m·g

(which means that the weight W=m·g is balanced by the force of spring's elasticity F=−L·k in its initial position with our object at its end) then there are no oscillations, and the object will remain at distance L=m·g/k from a spring's neutral position.



If there are oscillations, their period is

T = 2π√m/k