Rigid Body Dynamics

rigid body

Rigid Body Dynamics

Equilibrium of Rigid Body

In this section, we will discuss the static equilibrium of rigid bodies.

A rigid body is said to be in static equilibrium when the resultant force acting on the body is zero and the resultant torsion at any point of the rigid body is zero.

Mathematically, the above statement is written as,

ΣF = 0 or  ΣFx = 0 and  ΣFy = 0    ...... 1

and

Σ 𝞃 = 0    ........................  2

Equation 1 is called the first condition for equilibrium and equation 2 is the second condition for equilibrium.

If the first condition for equilibrium, ΣF= 0, is fulfilled, the body is in translational equilibrium. A rigid body is said to be in a complete equilibrium when both conditions for equilibrium are fulfilled. Static equilibrium can be classified into three, i.e. stable equilibrium, unstable equilibrium, and indifferent (neutral) equilibrium.

1. Stable Equilibrium

A marble has the center of gravity at the center of the sphere. If the marble is put in a hemispherical (concave) container, it will stay at rest (be in an equilibrium state) at the bottom of the container. If the marble is given a disturbance by pushing it, its center of gravity will raise. It is indicated by the raise of the position of the marble in the container.If the disturbance is removed, the marble will return to its initial equilibrium. This kind of equilibrium is called stable equilibrium. A stable equilibrium is characterized by the raise of the center of gravity of the object whenever a disturbance is given. When the disturbance is removed, the object will return to its previous position. A rocking chair is another example of an object in a stable equilibrium.

2. Unstable Equilibrium

If a marble is so carefully put on top of a ball that it reaches equilibrium (still) and then given disturbance, the marble will never return to its initial position. This kind of equilibrium is known as the unstable equilibrium. It is characterized by the lowering of object's center of gravity whenever return to its initial position. An upright piece of wood is one example of unstable equilibrium

3. Indifferent (Neutral) Equilibrium

Now, consider a marble put on a flat and slippery floor. Whenever the marble is given a disturbance, its center of gravity will not undergo any change in height. When the disturbance is removed, the marble will soon be in its new equilibrium position. This is known as indifferent or neutral equilibrium.

It is characterized by the unaffected position of the center of gravity before and after a disturbance is given. Another example of indifferent equilibrium is cylinder put on a plane floor.

B. Center of Gravity of an Object

An object can be considered a composite of many particles. Each particle has weight. The weight of an object is the resultant of all gravitational forces undergone by the particles that make up the object. The direction of the gravitational force each particle is toward the center of the earth.

The resultant of all gravitational forces (weight forces) of the particles that make up an object is located at a certain point. The capture point of the gravitational force is called the center of gravity of the center of mass of an object. What about the center of gravity of two or three-dimensional objects? Objects that have symmetries such as a triangle plate, polygon, round plate, cylinder, and ball are shaped regularly and are homogeneous. Hence the weight center is at their respective equilibrium point.

Exercise

1. Three particles with masses of 2 kg, 4 kg, and 6 kg respectively are located at the three ends of a triangle of 1.5 m length on its sides.
2. On a rod AB, a block with masses of 8 kg is attached. At a distance of 1 m from point A, a load x with masses of 6 kg is placed. If the length of the rod AB is 5 m, calculate the tension of the rope T.
3. Why is the final velocity of an object sliding on an inclined plane faster than an object rolling on it?
4. In a wheel of 12 kg.m2 moments of inertia, a constant torsion of 50 N.m is applied. Determine the angular acceleration.
5. A solid sphere rolls from its original position down an inclined plane with the height of 1.4 m. Determine the linear velocity of the sphere at the bottom of the inclined plane. ( g = 10 m/s²)
6. A particle with masses of 1 g rotates around an axis at 90 rotations per minute. If the distance between the particle and the axis is 50 cm, calculate its angular momentum.

Multiple Choice

1) A particle revolves with angular velocity of 10 rad/s. If the mass of the particle is 2 g and the angular momentum is 8 x 10-6 kg m2/s, the trajectory radius will be ........
a. 2 cm
b. 4 cm
c. 6 cm
d. 8 cm
e. 10 cm

2) A solid cylinder made of iron is rolling down the floor at 10 m/s velocity. Its mass and diameter is 4 kg and 80 cm respectively. The total kinetic energy of the cylinder is ...........
a. 250 Joule
b. 300 Joule
c. 400 Joule
d. 500 Joule
e. 625 Joule

3) The rod AC has the mass of 40 kg and the length of 3 m. The footing distance of A and B is 2 m (the block can actually rotate at B). A boy with mass of 25 kg mass walks from A to C. The minimum distance from C for the boy so that the block would still be in balance (with the tip of A slightly up) is ........
a. zero
b. 0.1-meter
c. 0.2-meter
d. 0.3-meter
e. 0.4 meter

4) For an object in unstable equilibrium, after a disturbance is given, the position of the center of gravity will be ..........
a.. the same
b. higher
c. lower
d. up-side-down
e. higher and lower

5) A solid sphere with the diameter of 20 cm rotates with the axis at the center of the sphere. The sphere has the angular velocity of ω = (10 + 25t) rad/s, where t is in seconds. If the mass of the sphere is 4 kg, the magnitude of the torsion working on the sphere is ...........
a. 0.32 N.m
b. 0.4 N.m
c. 0.65 N.m
d. 0.8 N.m
e. 1.6 N.m

6)  A grindstone has 4 kg mass and 10 cm radius. If a fixed moment of force is applied, the grindstone will reach an angular velocity of 1200 rpm in 20 s. If the grindstone is initially at rest and having shapes of a solid cylinder, the magnitude of its moment of force is ..... N.m
a.   8 π x 10^-2

b.   4 π x 10^-2
c.   2 π x 10^-2
d.   8 π x 10^-4
e.   4 π x 10^-4

7)  A solid sphere undergoes translation and rotation with linear and angular velocities v and ω respectively. The total kinetic energy of the sphere is ..........

a.  6/10 mv²

b.  7/10 mv²

c.  8/10 mv²

d.  9/10 mv²

e.  11/10 mv²  

You Need To Know

Have your ever seen an acrobatic performance? Walking on a rope is one of them. How Could the man possibly keep his balance on a rope?

By using his hands, the acrobatic player controls his steps in such a way that the resultant of the moment of force is zero, as required to reach equilibrium state of a rigid body. If the rope is considerably long, the players will spread his arms to reduce his weight. If the rope is somewhat short, the arms do not have to be spread.

Case Study

The design of racing cars is much more stable than the design for trucks. That is because the racing cars have lowered center of gravity and their base is also wider. That kind of design prevents the racing cars from rolling up-side-down while taking on curves with high speed. Whereas trucks or buses will easily roll over whenever they take on curves with high speed.

Why does the object with the higher center of gravity tend to be a more unstable tan object with the lower center of gravity, despite having the same base?

Thermodynamics Formulas

inernal energy Thermodynamics

Thermodynamics Formulas

Thermodynamics Formulas. The Steam engine is the first machine to be invented. It uses high-pressure steam to push a piston in a cylinder. The steam is produced by heating the water by using the flame from coal or gasoline. The Steam from the boiler is then delivered to a cylinder through a valve; this high-pressure steam will push the piston. When the piston is pushed to the edge of the cylinder through one edge and allows new steam pushes the piston back to its initial position. This cycle will keep on going as long as the steam is provided. The perpetual motion of the piston can be used to move another part of the engine.

Steam engine applies the concept of thermodynamics. And so do diesel engine, refrigerator, and air conditioner (AC); their working principles are the applications of thermodynamics concept.

Basic Concept Of Thermodynamics

Thermodynamics is a branch of physics that studies the relation of heat, mechanical energy, and its conversion from one into the other. Before studying thermodynamics, we first need to understand the definition of system and environment besides the terms of equilibrium, heat, work, and energy.

1. System and Environment

the system can be defined as a separated part which becomes the point of observation. A system can be a room, body, quantity of matter, etc. Environment or surroundings can be defined as anything outside the system which may influence the state of the system directly.

2. Equilibrium 

There is term "equilibrium" in thermodynamics. There are three types of equilibrium in thermodynamics, i.e. mechanical, chemical, and thermal equilibrium. when the conditions for the equilibrium to take place are fulfilled, then the system is said to be in a thermodynamic equilibrium. In that state, there is no tendency of state alteration in either system or its environment.

3. Definition of Heat

Heat is energy which moves due to the temperature difference between the system and its environment. The system can release heat to its environment, and on the contrary, the environment can also send out its heat to the system.

4. Definition of work thermodynamics

Figure 8.2 shows liquid in a cylinder equipped with a freely moving piston. If the cylinder's cross-section is A and the pressure given by the piston is P, then the force produces is F = PA. When the piston moves outward at a distance of dx, the work dW performed by the force F is:

dW = F dx = PA dx

However, A dx = dV or the volume change and so we have

dW = P dV

During the volume change from V₂ to V₁, the pressure P is constant as shown in Figure 8.3, and so the work performed by the system is

W = P(V₂ - V₁) 

Therefore, in every process with constant volume (V₁ = V₂), the work performed by the system in equal to zero. If V₂ > V₁ it means the work is performed by the system; its value is positive. Inversely, If V₂ > V₁, it means the work is given to the system; its value is negative.

Example :

A certain gas undergoes expansion at a constant pressure of 5 atm (1 atm = 10⁵ N/m²). Its initial volume was 300 L and the volume gets doubled soon after the expansion. Determine the work done by the gas.

Answer :

P = 5 atm = 5 x 10⁵ N/m²

V₁ = 300 L = 0,3 m³
V₂ = 2V₁ = (2) (0,3 m³ ) = 0,6 m³
W = P ΔV
 W =  P(V₂ - V₁)  = (5 x 10⁵ N/m²) ( 0,6 m³ - 0,3 m³) = 1,5 x 10⁵ J

Internal Energy

Every particle in a system always moves. Given that kinetic energy of each particle in a system is E_k, and the total number of particle in the system is N, the internal energy (U) of the system can then be expressed by:

U = NE_k

Internal energy is conservative; meaning that its value does not depend on the path or process was taken; it depends only on the initial and the final state of the system. Internal energy is a characterized state of gas which can not be measured directly. The thing that can be measured directly is the change of internal energy (ΔU); it is when the system changes from its initial state (U₁) to its final state (U₂). Internal energy is a state function which the change is formulated by 

ΔU = U₂ - U₁

 Thermodynamic Process

Process or the change of thermodynamic state can be classified into two, i.e irreversible and reversible processes. The irreversible process is a spontaneous process in one direction; it can not occur in the reverse direction. The reversible process is a process of which the direction can be reversed. The reversible process is an equilibrium process. The system that undergoes a reversible process will always be in a state of thermodynamic equilibrium.

The thermodynamic process can be in the form of expansion, compression, heating, or cooling processes. The thermodynamic process can take place in a state of isothermal, Isochoric, isobaric, or adiabatic processes.

1. Isothermal

An isothermal process is a process of state alteration of gas which takes place at constant temperature.

This process agrees with Boyle's law, which states that the value of PV is constant at a fixed temperature or more clearly stated by the general equation of ideal gas, PV= nRT. The work performed received by the system can be obtained by calculating the area under the P-V curve.

In fact, the work W in the equation above has the same meaning with the area below the curve on P-V graph. Suppose you have learned the integral concept as the area below the curve of a function, you certainly understand that formulation.

2. Isochoric

Icochoric process is a process of the state alteration of gas which takes place at constant volume.

 Isochoric process agrees with Charles' law, which states that the value of P/T is constant at a fixed volume. What is the value of work in the isochoric process? By remembering the work definition as the area below the curve of P-V graph, you can conclude that the work in isochoric process is zero (W=0). It can also be understood that there is no volume change, therefore we have ΔV = 0, W = ΔV = 0.

3. Isobaric

Isobaric process is a process of gas state alteration which takes places at a constant pressure.

Isobaric process agrees with Gay-Lussac's law, which states that the value of V/T is constant for a fixed pressure. How much work is involved in the isobaric process? By noticing the curve of P-V graph, you can determine the value of the work as an area below it.
W = P(V₂ – V₁) = P ΔV
The process of gas state alteration of gas state alteration in a steam engine's boiler is usually isobaric. The steam volume addition process in the boiler is carried out without changing the pressure inside it.

Thermodynamics Eequation

1)  A thermodynamics process without any heat transfer between the system and its surroundings is called........
a.  Adiabatic
b.  Isochoric
c.  Isobaric
d.  Isothermal
e.  Isovolume

2) An amount of 2 x 10^-3  mol gas expands isothermally from V₁ = 20 cm³ to V₂ = 50 cm³ at the temperature of T = 300 K. If the universal gas constant is R = 8.3 J/mol.K, the work performed in the system is..........
a.  4.56 x 10³ Joule
b.  4.56 x 10² Joule
c.  4.56 x 10¹ Joule
d.  0.456 x 10^-1 Joule
e.  0.456  Joule

3) A kind of gas in a cylinder has an initial volume of  20 cm³. The gas experiences expansion at a constant pressure of 2 x 10⁴ N/m² in a way that its volume becomes 40 cm³; the work performed is.....
a.  0.2 Joule
b.  0.3 Joule
c.   0.4 Joule
d.  0.5 Joule
e.  0.6 Joule

4) An amount of 1.5 m³ helium gas at a temperature of 27^oC is heated isobarically to 87^oC. If the pressure of the helium gas is 2 x 10⁵ N/m², its external work will be .........
a.  60 kJ
b.  120 kJ
c.  280 kJ
d.  480 kJ
e.  660 kJ

5)  A mol of ideal gas at a temperature of T is in a cylinder equipped with the frictionless piston. The gas is then heated at constant pressure in a way that its volume becomes fourth of its initial. If R is the universal gas constant, the work performed by the gas is..........
a.   RT/4
b.   RT ln 4
c.   6RT
d.   4RT
e.   3RT

You Need To Know

Refrigerator in an electronic appliance we find regularly in our everyday life. The working principle of the refrigerator is making use of the second law of thermodynamics. The room inside the refrigerator acts as a cold reservoir whose temperature is  T_C, whereas the room outside the refrigerator act as the hot reservoir whose temperature is Th,

The heat in the cold reservoir is generated by the foods, drinks, or other things in the refrigerator. This heat will be removed and delivered to the outside air around the refrigerator. The electric energy is needed to flow the heat from the cold reservoir to the hot one. The electric energy used to operate the refrigerator is provider by our household electric installation. We buy this electric energy from a provider. In order to save the electric cost, it is necessary for us to chose the high effectiveness refrigerator.

Ideal Gas Constant

ideal gas Ideal Gas Constant

Ideal Gas Constant

Model Of Ideal Gas

When a can of soda water is poured out into a glass, bubbles of gas are visible in the glass. The bubbles of gas are moving up. During their motion, the size of the bubbles is increasing, their volumes even become almost twice of their initial volumes when they reach the surface.

The bubbles of gas in a soft drink contain carbon dioxide (CO₂ ), a kind of gas produced by the fermentation process. A bubble of gas develops its volume and moves quickly. Why does this happen that way?

1. The  Characteristic  of Ideal Gas

If a  container  of  volume  V contains  N particles   of  gas,  each  has molecular  mass  m, and  then the  total  mass  of the  gas  is Nm,  while  its mass density  is

Thereby, the mass density of gas can be reduced by decreasing the number of particles or by transferring the gas into a bigger container (increasing  V).

The result of the experiment shows that at a small enough density, all gasses tend to show a simple relationship between pressure, volume, and temperature. This fact guides  us  to a concept of ideal  gas, which  is gas with  a relatively small  inter-molecule  interaction  so it can be neglected

2.    The general  Equation  of Ideal  Gas

Before discussing ideal gas equation and the physics laws concerning it,  we need to discuss the mol concept and  Avogadro's number.

a)  Mol concept and Avogadro's Number

1) Mol Concept

The mol number of a  substance is usually denoted by n.  In  SI, mol (mole)  is defined as the most abundant carbon isotope in the universe, which is  carbon-12  isotope or simply written as  C-12.  The following is the definition. 

A mol of substance is the amount of substance containing elementary entities as much as the number of atoms in 12 grams of C-12 isotope.

Elementary entities in the definition above are molecules that contain either more than one atom such as water (H2O) or the single atom (such as carbon atom (C) and helium atom (He).

2) Avogadro's Number

The experiment has proven that the number of an atom in one mol C-12 isotope is:
6.022 x 1023

That number is called Avogadro's Number (NA), named after the Italian physicist and chemist, Amedeo Avogadro.

The relative mass of a molecule (Mr) of a substance is the molecule mass in 1 mol substance. The relative mass of an atom (Ar) of a substance is the mass of an atom in 1 mol substance.

Mass of a particle in atom is the ratio between the atomic mass and the Avogadro' number,

M₀ = M/N_A

where

M = atomic mass (molecular mass or mole) in kg/mol
N_A = Avogadro's number = 6,022 x 10²³ atoms/mol or molecules/mol
= mass of a particle in the atom

The number of mole of gas (n) could be calculated by using the following formula:
n = m/M_r
n = N/N_A                                                                          

where
m  = mass of a substance
M_r = relative mass of molecules
N_A = Avogadro's number

b.  Boyle's Law

Boyle investigated the relationship between pressure (P) and volume (V) variables when the gas is at a constant temperature. If the temperature of gas in a container is maintained constant, the gas pressure is inversely proportional to its volume. The above statement was coined by Rober Boyle and called Boyle's law.

In general, Boyle's law is written as follows

PV = constant or

P₁V₁ = P₂ V_{2    ....... 7.4}

c. Charles Gay Lussac's Law

If the pressure of gas in an enclosed container is maintained constant, the volume (V) of gas is proportional to its absolute temperature (T). The above statement was made by Charles and Gay-Lussac and called Charles-Gay Lussac' law.
In general, the law of Charles-Gay Lussac is written as follows:
V/T = constant or

V₁/T₁ = V₂ /T₂                 ....... 7.5  

P₁V₁/T₁= P₂ V₂ /T₂      ...... 7.6

d.  The General Equation of ideal gas

Equations [7.4] - [7.6] hold only on gas in an enclosed container (with no leakage), in a way that mass of the gas is constant. What about gas that leaks or gas with changing mass? To answer this question, we need a gas equation the gives solution when conditions of the three variables (pressure, volume, and temperature) are changing.

The laws of Boyle, Charles, and Gay-Lussac were obtained by maintaining one variable constant. These laws could be combined in a way that we get the relation of the three variables P, V, and T.

Based on research, the relation of pressure (P), volume (V), and temperature (T) of gas is

PV = n RT .... 7.7

where n = the number of mole of gas, and R = the universal gas constant whose value is 8.314 J/mol.K or 0.0821 L. atm/mol. K

Equation [7.7] is called general equation of ideal gas.

The general equation of ideal gas could be expressed in term of the number of gas particles (N). The number of gas particles (N) is the product of the number of gas mole (n) and Avogadro's number (N_A) following the equation

N = nN_A

In a form that includes quantity of the number of gas particles (N), the general equation of ideal gas can be written as

PV = NkT

where k = Boltzmann's constant = 1.38 x 10^-23 J/K

Exercise 7.1

1. A cylindrical pump contains oxygen gas at a temperature of 273 K and pressure of 20 atm. At the certain time, the piston is pushed down in a way that its volume becomes 50 liters. As a result, the temperature of the gas increases to 300 K. what is the current pressure given that the volume of the cylinder is 100 liters?
2. An amount of gas with the volume of 600 liters, temperature of 27 ^oC and pressure 5 atm possesses masses of 1.95 kg. Determine the relative mass of the gas.
3. Container A which contains ideal gas at pressure 5 x 10⁵ Pa and at temperature 300 K is connected to container B which whose volume is four times larger than A's. Container B contains the same ideal gas at pressure 1.0 x 10⁵ Pa and at temperature 4,000 K. Valve X is then opened in a way that both containers reach a new equilibrium state.If the temperature of each container is maintained at its initial value, determine the new equilibrium pressure
4. A rubber with a volume of 20 liters is filled with oxygen at the pressure of 135 atm and temperature of 27 ^oC. Determine the oxygen mass if given that R = 8.314 J/mol.K. and M_r(O₂) = 32 g/mol.
5. Oxygen gas at the temperature of 27 ^oC and pressure of  10⁵ Pa has the volume of 30 liters. Determine the volume of the oxygen given that the pressure becomes 2.5 x  10⁵ Pa and the temperature becomes 127 ^oC.

You Need To Know

Have you ever seen an airbag in a car? The airbag has the almost similar function with a seat belt, which is preventing a direct collision between the driver and the car's dashboard. If there is hard enough collision or crash, the airbag acts as an air cushion for the driver's obverse body. The airbag is made based on the principle of the ideal gas equation, which is: PV = NkT. The airbag is made in a way that a collision which occurs at the car will be detected as a change of thermodynamic variable and it triggers the flow of gas into the bag causing it to expand all at once.

One Dimensional Elastic Collision

Collision Momentum

One Dimensional Elastic Collision

Now, we are going to discuss the application of the law of momentum conservation on the collision of two bodies. Collision process may happen very shortly, for example on the collision between two billiard balls, or it may happen in a long time, for example on the collision among stars in heaven. On all collision processes, the colliding bodies will interact with each other strongly only during the collision. If there are external forces, these forces are much smaller compared to the interacting forces and therefore are neglected.

If the total kinetic energy of the colliding bodies after the collision is equal to the total kinetic energy before the collision, the collision process is called completely elastic collision. However, if the total kinetic energy of the two bodies before the collision is not the same with the total kinetic energy after the collision, then the collision process is called inelastic collision. In an inelastic collision, part the of the kinetic energy is changed into another kind of energy, for example, thermal energy. Thus, the total kinetic energy after the collision the two bodies are joined together, the collision process is called completely inelastic collision, as studied in Example 4.2. It should be underlined here that in every collision process where the influence of external forces is neglected, the law of momentum conservation always holds.

1. One-Dimensional Elastic Collision

I have been mentioned that if in a collision there is no lost of kinetic energy, then a collision is elastic. We will discuss a one-dimensional collision, in which the velocities of colliding bodies is on the same straight line, for example along the x-axis.

Figure 4.6 (a) shows two bodies of masses m1 and m2, which are moving with velocities of v1 and v2 along the same straight line. Having collided elastically, their velocities become v1' and v2' (Figure 4.6 (b)). Based on the law of momentum conservation, it is obtained:

2. Two-Dimensional Elastic Collision

If often happens that the motion of colliding bodies is not in a straight line but in the two-dimensional plane. One example of such collisions in the collision between two balls in a billiards game.

Figure 4.7 shows a ball of mass m1 moving along the x-axis is colliding with another ball of mass m2, which in initially at rest. Having colliding, the two ball move into directions that make the angles of Ө₁ and Ө₂ respectively toward the x-axis.

By applying the law of momentum conservation in the x-axis direction, it is obtained

m₁v₁ = m₁v₁’ cos Ө₁ + m₂v₂’ cos Ө₂

 Since initially both bodies are not moving in the direction of the y-axis then the component of momentum in the y-axis directions is zero.

m₁v₁’ sin Ө₁ + m₂v₂’ sin Ө₂ = 0

3. Completely Inelastic Collision

In a completely inelastic collision, after the collision, the two bodies will move together so that it holds v₁’ = v₂’= v'. Thus, the law of momentum conservation becomes:

m₁v₁ + m₂v_{2 = (} m_{1 +} m₂) v’

Therefore, the velocities of the two bodies after the collision can be calculated with the formula. Thus, by measuring the masses and velocities of the bodies before the collision, the velocity after collision could be predicted. In a complete collision, kinetic energy after collision is always smaller than kinetic energy before collision. If kinetic energy before the collision is E_k = ½ m₁V₁'². The kinetic energy after collision is

E_k' = ½ (m₁ + m₂) v’²

4. Inelastic collision

Most collisions between two bodies found in nature are inelastic collisions, for example, a tennis ball colliding with racket or the hit ball in a baseball game. The analysis of inelastic collision involves a quantity called restitution coefficient (e).

Restitution coefficient is defined as the negative value ofthe ratio of relative velocities before and after collision. 

For inelastic collision, the value of restitution coefficient is between zero and one, 0 < e < 1.

Exercise 4.3 

1. A body with the mass of 1 kg is moving to the direction of positive x-axis with a velocity of 2 m/s. The other body which mass is 3 kg, is moving in the opposite direction with a velocity of 2 m/s. After collided, both bodies are moving together as one. Calculate the velocity of both bodies after the collision.
2. Two balls with masses of 3 kg and 6 kg are moving with speeds of 4 m/s and 1 m/s respectively. The two balls are approaching each other along a straight line connecting the center of masses of the balls. After the collision, the ball with a mass of 3 kg stops moving. Determine the value of the restitution coefficient of the two balls.
3. Two bodies A and B have the same masses. Initially, A is moving to the right with an initial velocity of 5 m/s, and after 2 seconds A covers a distance of 14 m. At that time, A and B undergo a completely inelastic collision. If B is initially moving to the left with a velocity of 15 m/s, determine the velocity of both after the collision.

Multiple Choice

1) Andre Agassi is playing tennis using a ball with the mass of 100 g. The ball is approaching him with a velocity of 100 m/s. Andre hits back the ball with a force of 120 N. If the racket touches the ball for 0.02 s, the velocity of the ball is .....
a. 46 kg.m/s
b. 56 kg.m/s
c. 66 kg.m/s
d. 76 kg.m/s
e 86 kg.m/s

2) Among the following moving bodies, the one that experiences the greatest force when the body hits the wall and stops in the same interval of time is ...........
a. 40 kg with a speed of 25 m/s
b. 50 kg with a speed of 15 m/s
c. 100 kg with a speed of 10 m/s
d. 150 kg with a speed of 7 m/s
e. 200 kg with a speed of 5 m/s

3) Someone with the mass of 40 kg is standing on a skateboard, which mass is 2 kg and is moving at a speed of 10 m/s. If he is suddenly jumping forward with the velocity of 4 m/s while the skate board keeps running, the magnitude of the skate board's velocity is ............
a. 120 kg.m/s
b. 130 kg.m/s
c. 135 kg.m/s
d. 140 kg.m/s
e 150 kg.m/s

4) Ball A is moving with momentum mv and colliding with ball B which is moving on the same straight line. If after the collision the momentum of ball A become -3 mv, the increase of the momentum of ball B is ..............
a.  2 mv
b.  -2 mv
c.  3 mv
d.  -4 mv
e.  4 mv

5) A bullet with the mass of 10 g is fired and hits a block which masses 1.49 kg. The block hanging freely on a piece of rope which length is 0.2 m. If the acceleration due to the earth's gravity is 10 m/s² and the rope is displaced by 60 cm measured from its initial position/vertical position, the velocity of the bullet when fired is ........... m/s
a.  300 √3
b.  300 √2
c.  150 √3
d.  50 √2
e.  150

6) A grenade that is initially at rest is suddenly exploding and breaking up into two pieces, which are moving in the opposite directions. The ratio of masses of the two pieces m₁ : m₂ = 1 : 2. If the released energy is 3 x 10⁵ joules, the ratio of the kinetic energy of the first piece and the second one is ..........
a. 1 : 1
b. 2 : 1
c. 1 : 3
d. 5 : 1
e. 7 : 5

7)  A ping pong ball is falling freely from the height of h₁ and reflected back with the lower height of h₂. The restitution coefficient is ..........
a.   e = ( h₂/ h₁ )^1/2
b.   e = ( h₁/ h₂ )^1/2
c.   e = ( h₁/ h₂ )
d.   e = ( h₂/ h₁ )
e.   e = h₁ h₂

Conservation of Momentum

Momentum

Conservation of Momentum

Conservation of Momentum

Show two balls with masses of m₁ and m₂ are moving in opposite directions in a straight line with masses of m1 and m2 are moving in opposite directions in straight line with velocities of v₁ and v₂, respectively. Having collided, their velocities become v₁' a v₂'.

The sum of momentum of both balls before collision is

p = p₁ + p₂  = m₁v₁ + m₂v₂ 

The sum of momentum of both balls after collision is

p' = p₁' + p₂'  = m₁v₁' + m₂v₂' 

Based on the law of momentum conservation p=p'. thus

m₁v₁ + m₂v₂  = m₁v₁' + m₂v₂' 

The law of momentum conservation holds only if the sum of external forces working on the colliding bodies is equal to zero.

Example 4.2

A trained coach with a mass of 10.000 kg is running with the speed of 24 m/s to the right. The train coach collides with a similar coach which is initially at rest. If the two coaches are then connected as the result of the collision, what is their velocity after the collision?

Answer :
The sum of momentum before collision is
 p₂  = m₁v₁ + m₂v₂ = (10.000 kg) (24 m/s) + (10.000 kg) ( 0 m/s)

       = 2,4 x 10⁵ kg.m/s   (the direction of the initial total momentum is to the right)

The law of momentum conservation

Having collided, the two coaches merge and move together with velocity v'. the sum of momentum before the collision is equal to the sum of momentum after the collision and the momentum becomes:

 p₂  = (m₁ + m₂)v'

Based on the law of momentum conservation, the sum of momentum before collision is equal to the sum of momentum after collision;  p₂  = p₁    
(m₁ + m₂)v' = p₁  ⇒   v'  =  p₁ /(m₁ + m₂)
                     = 2,4 x 10⁵ kg.m/s (10.000 kg + 10.000 kg) = 12m/s.
After a collision, both coaches are moving together to the right with a velocity of 12 m/s.

Exercise 4.2

1) The velocity of a bullet when it is fired is 200 m/s. If the masses of the bullet and the rifle are 10 g and 5 kg respectively, then what is the recoil velocity of the rifle toward the shoulder of the man when the bullet is fired?2. Bodies A and B have masses of 3 kg and 2) kg respectively. Body A is moving to the right with velocity of 5 m/s and B is moving to the left with velocity of 10 m/s in such a way that they are colliding with each other. If after the collision the two bodies are joined together then what is the velocity of the two bodies after collision.

Working Principles of a Rocket

There is wrong assumption that the forward movement of a rocket is due to the reaction of the thrust face done by the rocket to the air. However, this concept does not satisfy when the rocket is in outer space since there is no air to push the rocket.

The working principle of a rocket is similar to the working principle of a rifle firing a bullet. Consider that the rifle in a position to move freely without friction as illustrated in Figure 4.8. When the bullet is fired, the rifle will move backward with a certain velocity. In this situation, the law of momentum conservation holds.

The acceleration experienced by the rifle is similar to the acceleration acquired by the rocket. The acceleration of the is obtained from the thrust done by the rocket. Every molecule of gas can be assumed as on small bullet fired by the rocket. In that system, the total momentum of the rocket is always equal to the total momentum of the gas; as long as there is no external force that influences the rocket (the external forces that influence the rocket are neglected).

The force of gravity, which is acting as the external force, will decrease the momentum of the rocket (remember, a force can change the momentum of a body). Consider that the initial velocity of the rocket is v and its mass is m. The rocket ejects gas with the mass of Δm and velocity v_g so that the velocity of the rocket eject increases to v + Δv. The velocities of rocket and gas are measured relatively toward the same frame of reference, for example; toward the earth. Based on the values of velocities, the initial and final momentum of the rocket could be calculated as follows ( p₁ and p₂, respectively);

P₁ = mvP₂ = (m – Δm) (v + Δv)

The final momentum of the gas is p_g = - Δmv_g, where the negative sign shows that the gas velocity is opposite the rocket velocity.
Base on the law of momentum conservation, it is obtained

 mv = (m – Δm) (v + Δv) – Δmv_g

mv = mv – Δmv­ + m Δv – ΔmΔv - Δmv_g  

0   = – Δmv­ + m Δv - Δmv_g
The value Δm.Δv is neglected since it is much smaller compared to the other terms, thus we obtain the change of the rocket's velocity as follows

Exercise 4.4

1. Determine the escape velocity needed by a rocket to leave the earth's surface. (g = 9.8 m/s², earth's radius is 6,400 km).
2. A rocket is flying with the rate of gas ejection of 300 kg/s. If the mass of the rocket is 450 kg and its relative velocity toward the gas is 300 km/hours, determine the average acceleration of the rocket.
3. A block with a mass of 1.5 kg is placed on a horizontal plane. The friction coefficient of the block with the horizontal plane is 0.2 A bullet with a mass of 10 g is fired horizontally, hits the block and buried in it. The block is displaced by 1 m. If g = 10 m/s², then determine the speed of the bullet when it collides with the block.
4. Block A which mass is 3 kg, is at rest on a floor. The kinetic friction coefficient of the floor is 0.2. Block A is collided by block B, which mass is 2 kg and moving at the speed of 20 m/s. After the collision, block A and B are moving together. Determine the distance covered before both of them stop.
5. A ball with the mass of 60 g is colliding with the wall perpendicularly. The ball is then turning back with the speed of 40 m/s, which is the same speed as the ball's speed before it collided with the wall. What is the momentum change experienced by the ball during the collision?
6. A bullet with 10 gram of mass hits the block with 1.5 kg of mass on a surface of a horizontal plane, with the friction coefficient of 0.2. Due to the shot, the block is displaced by 1-meter. Determine the speed of the bullet just before it hits the block. 

Multiple Choice

1)  A bullet with 10-gram mass is fired at the velocity of 300 m/s to 1.49 kg mass of block which is freely hung on a 1.6-meter long rope. If the earth's gravitational acceleration is 10 m/s², the rope will deflect from the vertical position with an angle of...

a.  30^o
b.  37^o
c.  45^o
d.  60^o
e.  67^o

2)  A 200 gram mass of marble is rolling on a frictionless floor so that its momentum is 4 kg.m/s. The marble's speed is....

a.  5 m/s

b.  10 m/s

c.  15 m/s
d.  20 m/s
e.  25 m/s

3)  A 100 gram mass of the ball, which slides at speed of  100 m/s is struck back so that its speed becomes 76 m/s. If the duration of the contact between the ball and the stick is 0.02 second, the force done by the stick is ...........
a.  100 N
b.  120 N
c.  176 N
d.  24 N
e.  2 N

4)  A car with 800 kg mass, which moves at speed 2 m/s, is braked by friction force of 32 N. The distance covered by the braking car, before it finally stops, is....
a.  50 m
b.  25 m
c.  15 m
d. 5 m
e. 10 m

5)  A 2 kg particle moves in a straight path at positive x-axis direction with the velocity of 3 m/s. If a 6 N force has been working on the particle for 3 s, after 3 s ...........
a.  its speed will be 3 m/s
b.  its velocity direction is toward the negative x-axis
c.  the particle never stops
d.  its velocity keeps on increasing
e.  the direction of the particles is unchanged

You Need to Know

The Function of Helmet

One of the equipment of a motor rider is the helmet. The inner layer of a helmet is made of a soft material such as foam rubber sponge. This layer has a function to lengthen the time of impact to the head so that it reduces the contact force to the head during the shock. What happens if the helmet does not have soft layer? When shock happens, the interaction force with the head will take place shortly and it can cause a headache.

Impulse and Momentum Theorem

Momentum momentum and impluse physics

Impulse and Momentum Theorem

The Meaning of linear momentum and Impulse. Rocket launching is a spectacular example of the application of the law of momentum conservation. The rocket gains propulsion by burning and ejecting its fuel's mass. At a certain hight, the rocket also gains velocity increase by releasing its fuel tanks. What is momentum? what is its relationship with mass and velocity

When a billiards game is started, the white ball is shot hard so that is moves and collides with the other set of balls. The other balls the move to several directions with different velocities. Some are moving to the right, left, slanting to the right, to the left, etc. some are moving fast, some are moving slowly, some even stay at rest. Why?

What are the factors made the balls to move in different directions and at different speeds? To answer such question, we should learn about linear momentum, Impulse, and the law of momentum conservation.

Impulse and Momentum Theorem

Linear momentum (often simply called momentum) of a body is defined as the product of the body's mass and its velocity. momentum, symbolized by p, is a vector quantity. Mathematically, it is written

P = mv,

where m is the mass of the body and v is the velocity.

The SI unit of momentum is kg.m/s. the direction of momentum is the same as the direction of its velocity.

In daily lives, according to equation (4.1), a body with a certain mass, which is moving at high velocity, will have greater momentum than another body with the same mass but is moving at a lower speed. On the other hand, two bodies that are moving with the same velocity but having different masses will also have different momentums. For example, a container truck will have a greater momentum than a sedan, even though they both are moving with the same velocity.

If force F is applied to a body with mass m, then Newton's second law holds F = ma. However,
a = Δv/Δt  so that we get;
Where Δp represents a change in momentum.
Equation (4.2) is another statement of the Newton's second law are have discussed before. It states that.

The force working on a body equal to the change of momentum per unit interval of time.

Equation [4.2] can be applied in the case where the mass of the body is changing toward the time, for example; in a motion of a rocket, which loses it mass when the fuel is burning out.
We could apply the concept of momentum change to a body where the applied force is not constant. For example is momentum in a ball, which is rolling at a certain velocity, and then kicked off in such a way that its velocity changing from to. In this situation, Equation (4.2) could be used to determine the average force working on the ball, according to the formula:

The equation above can be expressed as

P Δt = ΔP = P_t - P₀ = mv_t - mv

The product between F  and Δt is called impulse, denoted by I. Thus,

I = ΔP = P₁ - P₀ 

From its definition, the impulse has a unit of N.s, however, based on Equation (4.4), the unit of impulse is the same with the unit of momentum, since impulse is a momentum change. Thus, N.s = kg.m/s

Example 4.1

1. A wagon with 15 kg of mass moves at velocity of 0,25 m/s toward a concrete wall. At the same time, a bullet with 10 g mass is fired at velocity 1.000 m/s toward the wall. Calculate : (a)  the momentum of the wagon, and (b)  the momentum of bullet.

Answer :
m_wagon = 15 kg;  v_wagon = 0,25 m/s
m_bullet = 10 g = 10 x 10^-3 kg;   v_bullet = 1.000 m/s

(a) the momentum of the wagon

P_wagon = m_wagon  v_wagon
           = (15 kg) (0,25 m/s)
           = 3,75 kg.m/s

(b) the momentum of the bullet

P_wagon = m_wagon  v_wagon
           = (10^-2 kg) (1,000 m/s)
           = 10 kg.m/s


2.  A circus player who has 70 kg of mass is falling freely from the height of 3 m. (a) Calculate the impulse experienced by the circus player. (b) Determine the average force given by his leg when he is landing with the bending knee so that distance covered, which is measured from the first time he is touching the ground until completely stop, is 50 cm.

Exercise 4.1

1. A ball is initially at rest. The ball is then hit by a stick so that it is moving with the velocity of 10 m/s. If the mass of the ball is 250 g; calculate the impulse applied to the ball
2. A force of 70 N is applied to a body for 0.8 seconds. Determine the impulse experienced by the body.
3. A horizontal force of 200 N is working on a body, which mass is 2 kg. If the body is initially at rest and the force is working for 0.4 s, what is the final velocity of the body?
4. A stone with a mass of 500 g is thrown up vertically upward (g = 10 m/s² ). At the highest point, the stone is breaking up into two pieces with the mass ratio 2:1. If the two pieces are moving n opposite directions, determine the ratio of their kinetic energies.

Multiple Choice

1)  Impulse is classified into.......... quantity

a.  vector

b.  scalar

c.  derived and scalar

d.  fundamental

e.  vector and derived

2) The dimension of impulse is ......

a.   MLT

b.   MLT^-1

c.   ML²T

d.   MLT^-2

e.   ML²T^-2

3) A body that has a mass of 20 kg is hit by a force of 60 N. If the force is working for 0.2 s, the impulse working on the body is .............
a. 9 Ns
b. 10 Ns
c. 11 Ns
d. 12 Ns
e. 13 NS

4) A marble with mass of 200 g is sliding on a slipery floor with velocity of 20 m/s. The magnitude of momentum of the marble is ............
a. 2 kg.m/s
b. 3 kg.m/s
c. 4 kg.m/s
d. 5 kg.m/s
e 6 kg.m/s

5) An apple with mass of 200 g is falling down from a 5 m high tree, (g = 10 m/s² ). The momentum of the apple when it hits the ground is ............
a. 12 kg.m/s
b. 10 kg.m/s
c. 9 kg.m/s
d. 8 kg.m/s
e 7 kg.m/s

6)   A bullet with a mass of 8 gram is fired to a block with a mass of 250 gram, which is initially at rest on the edge of a table with the height of 1 meter. The bullet is buried into the block and after the collision, the block landed on the floor at a distance of 10-meter measure form the legs of the table. The initial velocity of the bullet is......
A)  224 m/s
B)  184 m/s
C)  145 m/s
D)  42 m/s
E)  86 m/s

7)  In a collision process, the following statement that always holds is ...........
A)  the law of momentum conservation
B)  The law of kinetic energy conservation
C)  the law of mechanical energy conservation 

D)  the law of momentum conservation and the law kinetic energy conservationE)  the law of momentum conservation and the low of mechanical energy conservation.

8)  A body with the mass of 2.5 kg is falling freely from the height of 3 meters above the floor (g = 10 m/s² ). If the body undergoes a completely inelastic collision with the floor, then the heat produced is ...
A)  7.5 cal
B)  18 cal
C)  30 kcal
D)  75 cal
E)  300 cal

9)  A body with mass of 0.1 kg is moving to the west with velocity of 0.2 m/s, and having a completely elastic collision with another body with mass of 0.15 kg, which is initially at rest. The velocity of the 0.10 kg body after collision is ......
A)  0.16 m/s to the east
B)  0.16 m/s to the west
C) 0.08 m/s to the west
B) 0.04 m/s to the east
E) 0.04 m/s to the west

10)  Two balls with masses of m1 = 20 kg and m2 = 25 kg respectively are moving into the same direction on a straight line. Ball 1 is moving with velocity of 8 m/s situated behind ball 2, which is moving with velocity of 4 m/s. If both balls are colliding inelastically with e = 0.3, the magnitude of velocities of the two balls respectively are....
a.  6.31 m/s and 5.11 m/s
b.  6.41 m/s and 5.11 m/s
c.  6.31 m/s and 5.31 m/s
d.  6.51 m/s and 5.11 m/s
e.  6.31 m/s and 5.41 m/s

Rectilinear motion with variable acceleration

physics rectilinear motion rigid body vector

Rectilinear motion with variable acceleration

The Analysis of Rectilinear Motion

If we pay close attention to the motions in a bicycle racing, we many notices that there are two kinds of motion occurred,i.e.: rotational and translational motions. The examples of the rotational motions are the motion of the racer's leg pedaling the pedal, the motion of the bicycle's wheels and the motion of the pedaled pedal. The examples of the translational motion are the motion of the bicycle and the racer when they are moving away from the start line.Base on the illustration of bicycle racing above, we have some questions related to physics. How about the velocity and accelerations of the bicycle? How about the motion of the bicycle's wheel? For answering the questions correctly, you need to study the following discussion.

During the motion, the position of a body always changes. The moving body always has velocity and probably acceleration as well. How about the relationship of the displacement, velocity, and acceleration of body?

1. Position of a  body in a Plane or in space

by using a coordinate system we can determine the position of a body. there are several types of coordinate system; some of them are Cartesian and polar coordinate systems. In discussing the rectilinear motion, the Cartesian coordinate is preferred, both the two or three dimensional systems. On the other hand, the polar coordinate system is preferred to be used in discussing the circular motion.

In a coordinate system, the position of a body is represented by a position vector. Before discussing position vector further, we need to discuss the unit vector.

a. Unit Vector

A unit vector is a vector, which magnitude is one. It has no unit and its direction is along the axes of the coordinate. For the Cartesian coordinate system, the directions of the unit vectors are along the x, y, and z-axes.

In a Cartesian coordinate system, unit vectors are usually denoted as i, j and k for the unit vectors along the x, y, and z-axes respectively, are:

Ax = A Cos α and Ay = A Sin α

In the form of unit vectors, vector A could be written as follows:

A= A_x i + A_yj

If vector A is spatial vector then

A= A_x i + A_yj +A_zk

b. Position Vector

A position vector is a vector that represents the position of a particle in a plane or in space.

The position of a  body in a  planar surface could be represented by a  position vector of r,  that is a  vector drawn from the origin to the position of the body.  Look at Figure 1.5.  The  position  vector of  point P(x, y) could  be  written as fo1lows: 

r = xi +yj

Whereas,  a  position vector in  a  three-dimensional space  could  be written as follows:

r = xi +yj +zk

In a rectilinear motion,  displacement is defined as the change of position of a body.  For example,  at time t₁,   a body is at point P(x, y);  this position is represented by position vector r₁. A few seconds later, at time r₂, the body is at point Q (x_1,y₂), which position is represented by position vector r₂. A vector. A vector (say Δr), which is drawn from P to Q can be written as follows.
Δr =  r₂-  r₁
     = (x₂i - y₂j) - (x₁i - y₁j)
     = (x₂ - x₁)i  + (y₂ - y₁)j
     = Δxi + Δyj

Example

A body is initially at  r₁ = 3i - 2j . then it moves to  r₂ = -i + 4j. Determine the magnitude of displacement of the body.

Answer :
The initial position of the body is  r₁ = 3i - 2j; the final position of the body is  r₂ = i - 4j. The displacement of the body is

 Δr =  r₂-  r_{1 
       =} (-1 - 3) i + 4 -(-2) j = -4i -2j

Exercise 1.1

1. A body move from point P (1, 0, 1) to point Q (5, 4, 3). Determine the displacement vector of the body and its magnitude.
2. A rabbit moves according to the equation x = t² and y = 5t. Determine: (a) the components of velocity at t = 0 second and at t = 4 seconds, (b) the average velocity vector, and (c) the magnitude and the direction of the average velocity.
3. The position of a moving particle is represented by r = 4t -3t² + t³, where r is n meter and t is in second. Determine the average velocity from t₁ = 0 to t₂ = 2 seconds.
4. Position vector of is represented by r = {(cos 2t) i + (4 sin2t) j)} m. Determine the particle velocity at t = 2 seconds.
5. The formula of displacement of a body, which is moving in a straight line, is x = -t² + 4t – 20, where r is in meter an t is in second. Determine the initial velocity and the acceleration of the body.
6. A particle is moving in a certain direction in such a way that its displacement components are given by x = 4t² + 2 and y = 4t + 2t², respectively. Determine the magnitude of the particle's acceleration at t = 2 seconds.
7. A ball is moving with the initial velocity of 20 m/s. If the acceleration of the ball is a = (4t + 2) m/s2, what are the equations of velocity and position of the ball? (its initial position is (0, 0)).

B. Analysis of circular Motion

A circular motion is a motion path is in the form of a circle. For examples, the motions of bicycle wheels, phonograph record, and helicopter's propeller. In the circular motion, the path's radius r is always constant, so that the component of the polar coordinate that determines the position of a circularly moving point is the traveling angle, which is Ө. A bicycle wheel rotating with respect to the fixed axes through point O. The line OP is a fixed line on the wheel that rotates with the same speed as the wheel. The position of point P is determined by the angle Ө formed by the line OP and the positive x-axis.

If the angle Ө is measured from the positive x-axis counter clockwise, then the Ө is negative. By definition, the angle Ө (in radian) is calculated using the formula:

Ө = s/r  rad

Ө = s/r = 2πr/r = 2π rad

However, one complete rotation is equal to 360^o. Then,

1 rotation = 360^o =  2π rad

1 rad = 360^o /2π = 57.3^o or 1^o = 0.01745 rad

The velocity possessed by a body undergoing circular motion is called angular velocity. Angular velocity is defined by the rotation angle covered in a unit interval of time. The unit of angular velocity is radian per second (rad/s).

Example:

A phonograph record with the diameter of 20 cm rotates with the angular speed of 30 revolutions per minute. Calculate the magnitude of the angular velocity and linear velocity of a point situated on the perimeter of the phonograph record.

Answer:

The magnitude of the angular velocity of the record:

ω = 30 rpm = 30 rotations per minute = 30 x 2π rad/ 60 s

     = π rad/ s

The magnitude of the linear velocity v = ω r = ½ ωd

v = π [ ½ (π rad/s) ( 20 cm)]

   = 31.4 cm/s

You Need to Know

Roller coaster is a high-speed train that passes through rectilinear and circular paths. When the train passes circular path it will undergo centripetal acceleration toward the center of the circle. To keep the train stay on track, it needs a big centripetal force obtained by accelerating the train. Thus, the passengers of roller coaster have to be safety strapped to the chair, to avoid them from being thrown away due to the high speed.

Case study

Between Soccer and Physics

What the soccer players do is very closely related to physics. For example, when he kicks the ball to the goal post, he should predict correctly the velocity as well as the elevation angle of the ball. If the elevation angle and the velocity are too great, the ball will miss the goal post. On the contrary, if the elevation angle and the velocity are too small or low, the ball will not reach the goal post.

Now is your turn to analyze and explain the following problems by using physical concepts.

1. Why does the path of the ball from a parabolic pattern?
2. Why is it difficult for the goalkeeper to block penalty kick?