Physics+ Newton Laws, Problem 2: UNIZOR.COM - Physics+ - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton - Problem 2

Problem

A chain of infinitesimally small links with overall length L and mass M is stretched on a table with its one end hanging off the edge of a table.
Half of a chain is lying on the table, while another half is hanging down.
The chain is stretched on a table along a straight line perpendicular to the edge of a table.

Initially, the chain is held at rest. Then at time t=0 it is let go, so the hanging part of a chain pulls the rest of it off the table.

The experiment ends at time t=T when the chain's endpoint B will be at the table's edge, where endpoint A was in the beginning of movement.

Assume that table legs are long enough, so chain, when it slides all the way down, does not reach the floor until it is fully off the edge of a table.
There is no friction between a chain and a table.
The acceleration of free falling is g.

What will be the speed of a chain at time t=T?
In other words, what will be the speed of the endpoint B, when this endpoint will be at the table's edge?

Solution

Let's try to solve this problem using only the Newton's Laws.

Assume that at a moment in time t the chain slides off the table by the length s(t).
Then the mass of the part of a chain that is not supported by a table will be
(½L+s(t))·M/L
It will pull down the part of a chain still on the table with the force
F = (½L+s(t))·(M/L)·g
This force pulls the chain of mass M with acceleration
a(t) = d²s(t)/dt² = s"(t).

Knowing the force, the mass and the acceleration, we can use Newton's Second Law getting a differential equation for s(t):
(½L+s(t))·(M/L)·g = M·s"(t)
or
s"(t) − s(t)·g/L − ½·g = 0

Initial conditions are
s(0)=0; s'(0)=0

To solve the above differential equation is not easy, which makes the whole approach less practical.

Let's try to solve it differently.
Assume, the length s(t) is an independent variable.
We know the force of gravity F(s(t)) in terms of s(t):
F(s(t)) = (½L+s(t))·(M/L)·g

Multiplied by infinitesimal increment of the length of the chain ds(t) that slides from the table during the infinitesimal increment of time dt, it will give an infinitesimal increment of work performed by this force
dW = F·ds

Integrating this from s(0)=0 to s(T)=½L, we will get the total work performed by the force of gravity
W_[0,T] = ∫_[0,½L]F(s)·ds =
= ∫_[0,½L](½L+s)·(M/L)·g·ds =
= ∫_[0,½L]½L·(M/L)·g·ds +
+ ∫_[0,½L]s·(M/L)·g·ds =
= ½M·g·s|_[0,½L] +
+ (M/L)·g·(s²/2)|_[0,½L] =
= ¼M·L·g + ⅛M·L·g =
= ⅜M·L·g

All this work was done by a gravitational field. The potential energy of a chain in a gravitational field has decreased by this amount, and this work was converted into kinetic energy of the chain.

Since the chain, when it completely slides from the table at time T has kinetic energy E=½M·v², where v its speed, ⅜M·L·g = ½M·v²
Therefore,
v² = ¾L·g
from which we derive the value of the chain's speed at the moment it completely slides off the table
v = √¾L·g
Notice that the speed of a chain at time t=T does not depend on the mass of a chain.

Yet another, even simpler approach to this problem is to use only a loss of potential energy as a reason for an increase in kinetic energy.

Let's take the surface of a table as the level zero of potential energy of a chain.

In the beginning, at time t=0 half a chain is on this surface, the potential energy P₁ of this half is zero:
P₁ = 0

Another half of a chain is below the level zero and, therefore, its potential energy is some negative value P₂.

The chain is at rest and, therefore, its total energy is
E = P₁ + P₂ = P₂

At the end of motion at time t=T the first half of a chain that had zero potential energy P₁ takes the place of the second one in the beginning of motion and its potential energy will be
P'₁ = P₂
The second half of a chain moves down by half the length of chain ½L, decreasing its potential energy to level P'₂.

Now the chain is moving and has some kinetic energy K.
Therefore, the total energy of a chain is
E = P'₁ + P'₂ + T =
= P₂ + P'₂ + T

From the Law of Energy Conservation
E = P₁ + P₂ = P'₁ + P'₂ + T

Knowing that
P₁ = 0 and P'₁ = P₂
we conclude
E = P₂ = P₂ + P'₂ + T

We see now that
P₂ = P₂ + P'₂ + T
Hence, canceling P₂,
0 = P'₂ + T
which allows to find kinetic energy T
T = −P'₂

Therefore, the loss of potential energy is the negative potential energy P'₂ of the second half of a chain at the end of motion.

It can be calculated by intuitively obvious method that considers all the mass concentrated in the middle of this half of a chain hanging below the surface of a table by ¾L, which leads to its potential energy
P = −½M·g·¾L = −⅜M·g·L.

Alternatively, we can calculate the potential energy of this half of a chain directly integrating potential energy of each infinitesimal part of it.
Let x be a distance of the point of the chain's second half from the table (level zero).
Then
P'₂ = −∫_[½L,L](M/L)·g·x·dx =
= −(M/L)·g·(x²/2)|_[½L,L] =
= −(½−⅛)M·g·L =
= −⅜M·g·L

In all cases the loss of potential energy is ⅜M·g·L.
From the Law of Conservation of Energy follows that kinetic energy of a chain must increase by the same amount, which leads to same value of a speed of a chain at time t=T
v = √¾L·g

Physics+ Newton Laws, Problem 3: UNIZOR.COM - Physics+ - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton - Problem 3

Problem

A chain of infinitesimally small links with overall length L and mass M is hung vertically by its top end A.
Its bottom end B touches a flat platform with a spring under it.
Assume that a platform and a spring are weightless, so the spring is in neutral position.
The elasticity coefficient of a spring is k.
Gravity acceleration is g.


If the chain is let go from point A, it will fall down on a platform, which will squeeze the spring.
What is the maximum distance the platform will go down, shortening the spring?

Solution

We will compare the potential energy of a chain in its initial position to potential energy of a squeezed spring.
As the chain falls down to a platform, its potential energy is gradually transformed into kinetic energy, which, in turn, is transformed into an energy of a spring.

When chain is fully on a platform, the energy of an oscillating spring (kinetic plus potential) should be equal to an initial potential energy of a chain.
At that time a spring will oscillate between maximum squeeze and maximum elongation with its total energy at the end points be only potential and equal to initial potential energy of a chain before it started to fall down.

Let's take the level of a platform in its neutral position as level zero of potential energy.
The mass density of a chain per unit of length is
μ = M/L
An infinitesimal piece of a chain of length dy has mass μ·dy.
Its potential energy on a height y above the surface of a platform in its neutral position, which is level zero, is
dP = μ·g·dy·y = (M/L)·g·y·dy
Potential energy of the whole chain is
P = ∫_[0,L](M/L)·g·y·dy =
= (M/L)·g·y²/2|_[0,L] = M·g·L/2
Check units:
(kg/m)·(m/sec²)·m² =
= (kg·m/sec²)·m = watt

According to the Hook's Law, the force of resistance of a spring F is proportional to the length of its shortening x.
F = k·x
When a spring is squeezed from the length x to x−dx, the work performed by an external force that caused it is
dW = F·dx = k·x·dx
To shorten a spring by the length S we have to do the work
W = ∫_[0,S]k·x·dx = k· S²/2
Check units:
[(kg·m/sec²)/m]·m² =
= (kg·m/sec²)·m = watt

From the Law of Energy Conservation follows that the potential energy of a chain in its initial position should be equal to the work performed by the force of a falling chain to squeeze a spring under a platform, that is
P = W
from which we derive
M·g·L/2 = k·S²/2
and
S = √M·g·L/k

Answer:
A spring will shorten its length by
S = √M·g·L/k
Check units:
√kg·(m/sec²)·m/(kg/sec²) = m

Physics+ Newton Laws, Central Field: UNIZOR.COM - Physics+ - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Central Force Field

The subject of this lecture is a theory behind practical problems of space travel around some gravitating mass or a movement of an electron around an electrically charged nucleus, or other types of movement in a central force field, where the force at any point A is always directed towards or away from some central point O and its magnitude is the same at all points that lie on the same distance from the center O.

Consider an inertial reference frame in our three-dimensional space with origin at point O and a force field F defined in all points with the following conditions satisfied:
(a) for any point A in our space the vector of force F(A) at this point is always colinear with position vector r=OA;
(b) for any two points A and B lying on the same distance from the center O vectors F(A) and F(B) have the same magnitude and both directed either towards or away from the center O.

Obviously, from condition (b) follows that all force vectors at all points lying on the same distance around center O have the same magnitude and all of them are directed the same way relatively to center O - either towards it or away from it.

Defined in such way, the force field is called central force field.

The vector of central force at any point A is defined by position vector r=OA and can be expressed as a scalar f(r) that depends only on the distance r=|r| of a point A from the origin of coordinates (positive or negative to differentiate between the directions towards or away from center O) multiplied by the vector of position r.
F(A) = F(r) = f(r)·r.

Imagine that at time t=0 some object of point mass m is at position defined by a vector r₀ and its initial velocity is v₀.

Velocity vector, by definition, is the first derivative of position
v(t) = r'(t)

Vector of acceleration is the first derivative of velocity by time or the second derivative of position
a(t) = v'(t) = r"(t).

Now we can express the Newton's Second Law as
F = f(r)·r = m·a = m·r"
and express an acceleration in terms of a central force and mass
a = f(r)·r/m

Consider a vector of angular (rotational) momentum of motion of our object L defined as the vector (cross) product of the position vector r(t) by the momentum of motion p=m·v(t):
L = r⨯(m·v) = m·r⨯v
and analyze how this angular momentum changes with the time.
More precisely, let's prove that the angular momentum of an object in a central field is a constant (the Law of Conservation of Angular Momentum in Central Field).

The first derivative of an angular momentum L is
L'(t) =
= m·[r(t)⨯v'(t) + r'(t)⨯v(t)]
Take into account that the derivative of a position is a velocity and the derivative of a velocity is an acceleration.
Then L'(t) =
= m·[r(t)⨯a(t) + v(t)⨯v(t)]

The second component in the above expression is a zero-vector because the vector product of collinear vectors is always a zero-vector.
Substituting acceleration for its expression in terms of force, position and mass a=f(r)·r/m, we obtain
L'(t) = f(r)·r(t)⨯r(t) = 0
because, again, a vector (cross) product of two collinear vectors is a zero-vector.

As we see, the derivative by time of the vector of angular momentum of an object in a central force field equals to zero-vector, which means that the vector of angular momentum in a central force field is a constant.

Notice that in our proof of the Angular Momentum Conservation Law we relied only on the central character of the force field, not its specific form for gravitational or electrostatic fields.

Physics+ Newton Laws, Problem 4: UNIZOR.COM - Physics+ - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton - Problem 4

Problem

This problem is related to an effect on rotation of a figure skater when she pulls the hands closer to her body, thereby increasing the angular speed of rotation.

The physical problem that models this situation is as follows.
A point mass m initially rotates on a thread of length r₀ with initial angular speed ω₀.
We would like to shorten the radius of rotation and see how this is related to a speed of rotation.

To accomplish this, let's put the thread that holds a rotating mass through a thin tube positioned perpendicularly to a plane of rotation with its one end used as a center of rotation, so we can shorten the radius of rotation by pulling a thread from the other end of a tube.


Assuming the radius of rotation is changing with time and, finally, equals to r, what then will be the angular speed of rotation ω?

Solution

The solution is based on the Rotational (Angular) Momentum Conservation Law.
This law states that vector
L = r⨯ p
where r is a position vector relative to a center of rotation
and p is a momentum of motion of an object,
is preserved and does not change in direction or magnitude.

Consider the beginning and the end of the process of reducing the radius of rotation. In both these cases the motion of our object is rotational, and vectors of position r and velocity v are perpendicular to each other.
Therefore, their vector product has a magnitude
L = m·r·v
Since v=r·ω, L = m·r²·ω
This value is the same at the beginning and at the end of the process.
Based on the Law of Conservation of Angular Momentum, dependency of angular speed ω from radius r can be easily derived as
L = m·r²·ω = L₀ = m·r₀²·ω₀
r²·ω = r₀²·ω₀
ω = ω₀·r₀²/r²

Incidentally, the expression I=m·r² is called moment of inertia, so the Law of Conservation of Angular Momentum can be expressed as
L = I·ω = L₀ = I₀·ω₀

For example, if the radius is reduced by half, the angular speed will increase by a factor of four.

Let's consider a more practical case of a rotating figure skater, when she, to increase the speed of rotation, brings her arms close to a body.
This is a rotation of a solid object of variable geometry, and we will simplify our job by considering the following model.

Assume, there is a rotating object that has two parts.
One part, modeling arms initially stretched, but later brought tightly to the body, has mass m and rotates on initial radius r₀, which will be changed to a smaller radius r.
Another part, modeling the rest of a figure skater's body, has mass M and a constant radius of rotation R.

Assume, initial angular speed of rotation of a figure skater with arms stretched is ω₀.

The total angular momentum with arms stretched is
L₀ = m·r²₀·ω₀ + M·R²·ω₀.

Bringing the arms close to the body of a figure skater in our model will be reducing the radius of rotation of mass m from initial r₀ to final R which is smaller.
This will change the angular speed of rotation to ω.

The final angular momentum with arms close to the body and a new angular speed of rotation ω is.
L = (M+m)·R²·ω.

According to the Law of Conservation of Angular Momentum, the final angular momentum L is equal to the initial one L₀
L = L₀
Therefore,
m·r²₀·ω₀ + M·R²·ω₀ =
= (M+m)·R²·ω.
from which follows that the final angular momentum is
ω =
= ω₀·(M·R²+m·r²₀)/[(M+m)·R²]

Since r₀ (radius of rotation of stretched arms) is greater than R (radius of rotation of the rest of the body), the numerator of the above fraction is greater than denominator, which makes final angular momentum ω greater than the initial one ω₀.