Topic 2: Mechanics (Part 2)

1. Introducing momentum

• Momentum is a measure of an object’s mass and velocity.
• It’s a vector quantity, meaning it has both magnitude (amount of movement) and direction.
• The momentum of an object is calculated as:
• [math] \text{Momentum} \, (\vec{p}) = \text{Mass} \, (m) \times \text{Velocity} \, (\vec{v}) \\ \vec{p} = m \vec{v} [/math]
• The unit of momentum is typically measured in kilogram-meters per second (kg · m/s).
• Figure 1 Momentum is transferred from one object to another
• Here are some key aspects of momentum in physics:
  1. Newton’s laws: Momentum is closely related to Newton’s laws of motion.
  2. The second law states that force ([math]\vec{F}[/math]) is equal to the rate of change of momentum ([math] \Delta \vec{p} / \Delta t [/math]).
  3. Momentum transfer: When objects interact, momentum is transferred from one object to another. This is why a baseball bat can transfer momentum to a ball, making it fly!
  4. Relativity: In special relativity, momentum is affected by an object’s speed and mass. As an object approaches the speed of light, its momentum increases exponentially.
• In the example of the exploding firework, chemical potential energy is transferred to thermal energy, light energy and kinetic energy of the exploding fragments.
• However, during the explosion the momentum remains the same.

Example

A ball of mass 0.1kg hits the ground with a velocity of 6m/s and sticks to the ground. Calculate its change of momentum.
Given data:
Mass of a ball = m = 0.1 kg
Change in velocity = [math]\Delta \vec{v}[/math] = 6m/s
Find data:
Change in momentum=[math]\Delta \vec{p}[/math]  = ?
Formula:

[math] \Delta \vec{p} = m \Delta \vec{v} [/math]

Solution:

[math] \Delta \vec{p} = m \Delta \vec{v} [/math]

Put values

[math] \Delta \vec{p} = 0.1(6) [/math]

[math] \Delta \vec{p} = 0.6 kg .\frac{m}{s} [/math]

2. Momentum and impulse

• Momentum and impulse are closely related concepts in physics.
• Impulse is a measure of the change in momentum of an object.
• In other words, impulse is the product of the force applied to an object and the time over which it’s applied.
• The unit of impulse is typically measured in newton-seconds (N·s).
• Here are some key points about the relationship between momentum and impulse:
  1. Impulse changes momentum: When an impulse is applied to an object, it changes the object’s momentum.
  2. Momentum conservation: If the total impulse is zero, the total momentum remains constant.
  3. Force and time tradeoff: A smaller force applied over a longer time can produce the same impulse as a larger force applied over a shorter time.
• Impulse- momentum theorem: The impulse applied to an object is equal to the change in momentum of the object.
• Newton’s second law can be used to link an applied force to a change of momentum:
• [math] \vec{F} = m \vec{a} [/math]
• Substituting
• [math] \vec{a} = \frac{\Delta \vec{v}}{\Delta t} [/math]
• Gives
•  [math] \vec{F} = \frac{ m\Delta \vec{v}}{\Delta t} [/math]

• Or

  [math] \vec{F} = \frac{\Delta (m\vec{v})}{\Delta t} [/math]

• So,

• Force equals the rate of change of momentum.
• This is a more general statement of Newton’s second law of motion.
• The last equation may also be written in the form:
• [math] \vec{F} \Delta t = \Delta (m \vec{v}) [/math]
• or
• [math] \vec{F} \Delta t = m \vec{v}_2 – m \vec{v}_1 [/math]
• where [math]\vec{v_2}[/math] is the velocity after a force has been applied and [math]\vec{v_1}[/math] the velocity before the force was applied.
• The quantity [math] \vec{F} \Delta t = I [/math] is called the impulse. Where [math] \vec{F} [/math]  is constant act.
• 
  Figure 2 Impulse and momentum

3. Conservation of linear momentum

• Conservation of linear momentum states that in a closed system, the total linear momentum remains constant over time.
• This means that the total momentum before an event (like a collision) is equal to the total momentum after the event.
• Mathematically, this is expressed as:
  Before collusion (initial) total momentum ([math]P_i[/math]) = After collusion (final) total momentum ([math]P_f[/math])
• [math]P_i = P_f \\ m\vec{v-i} = m\vec{v-f}[/math]
• For two bodies
• [math] m_1 \vec{v}_1 + m_2 \vec{v}_2 = m_1 \vec{v}’_1 + m_2 \vec{v}’_2 [/math]
• The total momentum of two bodies in a collision (or explosion) is the same after the collision (or explosion) as it was before.
• 
  Figure 3 Law of conservation of momentum
• This law applies to all closed systems, where no external forces are acting. In other words, the total momentum is conserved if there are no external influences that can change the momentum.
• Momentum is always conserved in bodies involved in collisions and explosions provided no external forces act. This is the principle of conservation of momentum.
• Here are some key aspects of conservation of linear momentum:
  1. Closed system: The system must be closed, meaning no objects can enter or leave, and no external forces can act upon it.
  2. Total momentum: The total momentum is the vector sum of all individual momenta in the system.
  3. Conservation: The total momentum remains constant over time, unless acted upon by an external force.
  4. Collisions: In collisions, momentum is transferred between objects, but the total momentum remains conserved.
  5. Explosions: In explosions, momentum is conserved, but the total momentum is distributed among the products.
• 
  Figure 4 A simple experiment that demonstrates the conservation of linear momentum
• In all collisions and explosions, both total energy and momentum are conserved, but kinetic energy is not always conserved.
• In this case, the chemical potential energy in the match head is transferred to the kinetic energy of the foil and matchstick, and also into thermal, light and sound energy.
• As the match head explodes, Newton’s third law of motion tells us that both the matchstick and foil experience equal and opposite forces, [math] \vec{F} [/math].
• Since the forces act for the same interval of time, Δt, both the matchstick and foil experience equal and opposite impulses, [math] \vec{F} \Delta t[/math].
• Since [math] \vec{F} \Delta t = \Delta(m\vec{v})[/math], it follows that the foil gains exactly the same positive momentum as the matchstick gain’s negative momentum.
• We can now do a vector sum to find the total momentum after the explosion (see Figure 5c):
• [math] + \Delta (m \vec{v}) + \left[ – \Delta (m \vec{v}) \right] = 0[/math]
• So, there is conservation of momentum: the total momentum of the foil and matchstick was zero before the explosion, and the combined momentum of the foil and matchstick is zero after the explosion.

4.Turning moments

⇒Introducing moments

• Turning moment, also known as torque, is a measure of the rotational force that causes an object to rotate or turn. (Also known as moment of a force about a point)
• It is a vector quantity.
• It’s unit is typically measured in newton-meters (N·m)
• Formula:
  Turning momentum = Forced appiled * distance prependicular from the piovt
  

   

  • [math] \vec{\tau}[/math] is the turning moment (torque)
  • [math] \vec{r}[/math]  is the distance from the axis of rotation to the point where the force is applied
  • [math] \vec{F}[/math] is the force applied
• A pivot is a point or axis around which something rotates or turns.
• It is a fixed point that remains stationary while other parts move or rotate around it.
• An object is in equilibrium, when the sum of the forces = 0 and the sum of the turning moments = 0.
• Newton’s first law of motion that an object will remain at rest if the forces on it balance.
• However, if the body is to remain at rest without translational movement, or rotation, then the sum of the forces on it must balance, and the sum of the moments on the object must also balance.
• Another way of expressing Newton’s first law is to say that when the vector sum of the forces adds to zero, a body will remain at rest or move at a constant velocity.

5. Centre of mass

• The center of mass (COM) is the point where the total mass of an object or system is concentrated.
• It is the point where the object or system would balance if it were suspended from that point.
• Properties of Center of Mass:
  • Unique point: The COM is a unique point for any object or system.
  • Mass concentration: The COM is where the total mass of the object or system is concentrated.
  • Balance point: The object or system would balance if suspended from the COM.
  • Invariance: The COM remains unchanged under rigid body transformations (translation and rotation).
• Importance of Center of Mass:
  Motion analysis: COM is used to describe the motion of objects and systems.
  Stability analysis: COM is used to determine the stability of objects and systems.
  Collision analysis: COM is used to determine the outcome of collisions.
  Robotics and computer graphics: COM is used to simulate and animate objects.
• Figure 5a shows a rod in equilibrium on top of a pivot.
• Gravity acts equally on both sides of the rod, so that the clockwise and anticlockwise turning moments balance.
• This rod is equivalent (mathematically) to another rod that has all of the mass concentrated into the midpoint, this is known as the centre of mass.
• In Figure 5b the weight acts down through the pivot, and the turning moment is zero.
• The centre of mass is the point in a body around which the resultant torque due to the pull of gravity is zero.
• This means that you can always balance a body by supporting it under its centre of mass. In figure 10c the tapered block of wood lies nearer to the thicker end.

Figure 5 Shows centre point of a rod where total weight of the body acting downward

6. Moments in action

• Moments in action refer to the turning effect of a force around a pivot or fulcrum.
• Here are some key aspects of moments in action.
  Clockwise and counterclockwise moments: Depending on the direction of rotation.
  Moment arm: The distance from the pivot to the point where the force is applied.
  Force and moment: The product of the force and moment arm.
  Torque and rotation: The relationship between moments and rotation.
• Moments in action are used in various applications, such as:
  Mechanical advantage: Moments are used to gain a mechanical advantage in tools and machines.
  Rotation and torque: Moments are used to understand and calculate rotation and torque in mechanical systems.
  Stability and balance: Moments are used to analyze and determine the stability and balance of objects and systems.
• Figure 6 shows the action of a force to turn a spanner, but the line of action lies at an angle of 45° to the spanner.
• How do we determine the turning moment now?
• This can be done in two ways.
  • First, a scale drawing shows that the perpendicular distance between the line of the force and the pivot is 0.21m.

    
    Figure 6 The action of a force to turn a spanner

  • So, the turning moment is:
  • 100N * 0.21m = 21N.m
  • Secondly, the turning moment can be calculated using trigonometry, because the perpendicular distance is (the length of the spanner) × sin θ.
  • Moment=F * l * sin θ
    Moment = 100N * 0.3m * sin 45 = 21Nm

7. Work:

• Work is defined as the transfer of energy from one object to another through a force applied over a displacement.
• In other words, work is done when a force is applied to an object and the object moves in the direction of the force.

  Figure 7 Displacement covers in the direction of applied force

• Formula:
• Work is a scalar quantity, meaning it has only magnitude (amount of work) but no direction.
• The unit of work is typically measured in Joules (J).
• Conditions:
  • Positive work: when the force and displacement are in the same direction.
  • W = Fdcosθ         θ=0°
    W = Fd(1)
    W = Fd

• 
  Figure 8 negative work

• Negative work: when the force and displacement are in opposite directions (e.g., lowering a book)

  W = Fdcosθ  θ = 180°
  W = Fd(-1)
  W = – Fd

• Zero work: when the force and displacement are perpendicular (e.g., pushing a wall)

  
  Figure 9 Zero work done

• W=Fdcosθ θ = 90
  W=Fd(0)
  W=0

8. Calculating the energy

• Work is converted to heat energy when it is done against resistive forces, like moving a luggage through an airport.
• As a result, the suitcase’s wheels and the area around them somewhat heated.
• Energy may also be transformed into many forms via work.
• This concept allows us to obtain the formulae for kinetic energy, elastic potential energy, and gravitational potential energy.

⇒ Gravitational potential energy

• Gravitational potential energy is the energy an object possesses due to its height or position in a gravitational field.

  
  Figure 10 gravitational potential energy converted into the kinetic energy

• It’s the energy an object has because of its potential to fall or move downward.
• Gravitational potential energy is a type of potential energy, which means it has the potential to become kinetic energy (the energy of motion).
• The higher an object is, the more gravitational potential energy it has.
• Gravitational potential energy is always relative to a reference level, such as the ground or a table.
• When an object falls or moves downward, its gravitational potential energy is converted into kinetic energy.
• A ball at the top of a point has gravitational potential energy due to its height.
• We can calculate the gravitational potential energy by using formula
• ∆E_p = mg∆h
• Where
  • ΔE_p is gravitational potential energy
  • m is mass of an object
  • g is gravitational acceleration
  • ∆h is height of an object in which potential energy store

• 
  Figure 11 A load with a weight W lifted through a height Δh.

  In Figure 11 a load with a weight W, has been lifted through a height ∆h.
  The work done = increase in gravitational potential energy.

• Work = W * Δh

Weight W=mg

• So, the increase in potential energy,  , is given by
• ∆E_p = mg∆h

⇒ Elastic potential energy

• Elastic potential energy is the energy stored in an object that has been stretched, compressed, or deformed in some way.
• It’s the energy an object has due to its elastic properties, like a spring or a rubber band.
• When an object is stretched or compressed, its molecules are moved apart or pushed together, creating a force that wants to return the object to its original shape. This force is called the restoring force, and the energy associated with it is the elastic potential energy.

  
  Figure 12 a spring may be stretched

• The more an object is stretched or compressed, the more elastic potential energy it has.
• When an object is released from its stretched or compressed state, the elastic potential energy is converted into kinetic energy.
• Elastic potential energy (E_ep) is stored in a stretched spring.
• It has the ability to transform its stored energy into the kinetic energy when it is released.
• The relationship between the force needed to stretch the band and its pullback distance is seen in Figure 10. We must compute the stored energy using the average force since the force varies.
• [math]\Delta E_{\text{ep}} = \vec{F}_{\text{av}} * \vec{s}[/math]

• 
  Figure 13 How the force required to stretch the band depends on the distance it is pulled back.

  In the case of Figure 10, where the force to stretch the band is proportional to the distance moved.

• [math]\text{Average force} = \frac{1}{2} \, \text{final force}[/math]
• So,         [math]\Delta E_{\text{ep}} = \frac{1}{2} F s[/math]
• The stored elastic potential energy can also be calculated more generally using the area under the force-extension graph.

⇒ Kinetic energy

• Kinetic energy is the energy of motion. It’s the energy an object possesses due to its motion, and it’s defined as the work required to bring the object to rest.
• In other words, it’s the energy an object has because it’s moving.
• Kinetic energy is a scalar quantity, meaning it has only magnitude (amount of energy) but no direction.
• The more massive an object is and the faster it moves, the more kinetic energy it has.
• Kinetic energy can be converted into other forms of energy, such as potential energy, and vice versa.
• 
  Figure 14 A constant force accelerates a car
• In Figure 14, a constant force F accelerates a car, starting at rest, over a distance s.
• Work is done to increase the kinetic energy of the car.
• We can use this idea to find a formula for kinetic energy, E_k, in terms of the car’s speed and mass.
• [math]\Delta E_k = F s \quad (1)[/math]
• but from Newton’s Second Law:
• F = ma
• and from the equations of motion:
• [math] s = \frac{1}{2} a t^2[/math]
• Put in equation 1
• [math]\Delta E_k = ma * \frac{1}{2} at^2 \\ \Delta E_k = \frac{1}{2}ma^2t^2 [/math]
• Since
• 
• So the kinetic energy of a body of a mass m, moving at a velocity v, is given by:

   [math] E_k = \frac{1}{2} mv^2 [/math]

  Note this is a scalar quantity because  has no direction

9. The principle of conservation of energy:

• The principle of conservation of energy states that energy cannot be created or destroyed, only converted from one form to another.
• In other words, the total energy of an isolated system remains constant over time.
• This means that the sum of all the different types of energy within a system (kinetic, potential, thermal, electrical, etc.) remains the same at all times.
• Energy can be converted from one form to another, but the total energy remains constant.
• [math]Total kinetic energy = total potential energy[/math]

⇒ Examples:

(1)

• A boy throws a ball upwards with a speed of 16m/s.
• It leaves his hand at a height of 1.5m above the ground.
• Calculate the maximum height to which it rises
  Solution:
• • The ball gains gravitational potential energy as it rises and loses kinetic energy when it leaves the boy’s grasp.
  • The ball stops travelling at its highest point, converting all of its kinetic energy to potential energy.
  • Given data:
    Velocity = 16m/s
  • Find data:
    Maximum height = ?
  • Formula:
    
  • Solution:
    
  • but the total height gained is 13m + 1.5m = 14.5m.

10. Power

• Power is the rate at which work is done or energy is transferred.
• It’s the amount of energy transferred per unit time.
• In other words, power is the measure of how quickly energy is used or produced.
• Formula:
• [math] \text{Power} = \frac{\text{energy transferred}}{\text{time}} \quad \text{or} \quad \frac{\text{work done}}{\text{time}} [/math]
• The unit of power is the watt (W), or J/s.
• This definition can be used to produce a useful formula to calculate the power transferred by moving vehicles.
• [math] \text{Power} = \frac{\text{work done}}{\text{time}} \quad \text{or} \quad \frac{\text{W}}{\text{t}} \\ \text{Power} = \frac{\vec{F} \cdot \vec{s}}{t}[/math]

• or

• [math]\text{Power} = \vec{F} \cdot \vec{v} [/math]
• Power (P) is the rate of energy transfer (measured in watts, W)
• Work (W) is the energy transferred (measured in joules, J)
• Time (t) is the time over which the energy is transferred (measured in seconds, s)

⇒ Example:

• A car is moving at a constant speed of 18m/s. The frictional forces acting against the car are 800N in total. The car has a mass of 1200kg.
• a) Calculate the power transferred by the car on a level road.
• b) The car maintains its constant speed while climbing a hill of vertical height 30m in 16s. Calculate the power transferred by the car now.

Given data:

Speed = 18m/s

Force = 800N

Mass = 1200 kg

Find data:
(a) Power transferred by the car on a level road = ?
(b) Power transferred by the car when car climbing a hill = ?

Vertical height = 30m

Time taken = 16s

Formula:

[math]\text{(a)} \quad \text{Power} = \vec{F} \cdot \vec{v}\\
\text{(b)} \quad \text{Power} = \text{previous power} + \frac{W}{t}\\
\text{Power} = \text{previous power} + \frac{mg \Delta h}{t} [/math]

Solution:

[math] \text{(a)} \quad \text{Power} = \vec{F} \cdot \vec{v}\\
\text {Power} = 800 * 18 [/math]

(a) Power transferred by the car on a level road

[math] \text{Power} = 14 \, \text{KW} [/math]

(b) Power transferred by the car when car climbing a hill

[math]\text{Power} = \text{previous power} + \frac{mg \Delta h}{t} \\
\text{Power} = 14000 + \frac{1200(9.8)(30)}{16} \\
\text{Power} = 14000 + 22050\\[/math]
[math]\text{Power} = 36 \text{KW}[/math]

11. Efficiency

• Efficiency is a measure of how well a system or process uses energy to produce a desired output.
• It’s defined as the ratio of useful output energy to input energy.
• Formula:
• [math]\text{Efficiency} = \frac{\text{useful output energy}}{\text{input energy}} \\ \text{Efficiency} = \frac{\text{useful output energy}}{\text{input energy}} \times 100\% [/math]
• Efficiency is usually expressed as a percentage, with 100% being the maximum possible efficiency.
• In reality, no system can achieve 100% efficiency, as some energy is always lost as waste heat, friction, or other forms of energy.
• Types of efficiency:
  1. Mechanical efficiency: measures the efficiency of machines or mechanisms.
  2. Thermal efficiency: measures the efficiency of energy conversion from heat to work.
  3. Electrical efficiency: measures the efficiency of electrical systems or devices.
  4. Energy efficiency: measures the efficiency of energy use in buildings, industries, or processes.
• Efficiency is important because it:
  1. Saves energy and resources
  2. Reduces waste and pollution
  3. Increases productivity and performance
  4. Saves money and reduces costs
• Sankey diagram: A particular flow diagram where the width of the indicated arrows corresponds to the flow amount. This chapter uses Sankey diagrams to illustrate the flow of energy through different processes.
• 

• Figure 15 Sankey diagram

⇒ Example

• A Sankey diagram illustrating the conversion of 100J of electrical energy into different forms of energy may be seen in Figure 12.
• Just 30J of the energy is converted into useable gravitational potential energy.
• The rest is converted to heat and sound energy.
• [math]\text{Efficiency} = \frac{\text{useful output energy}}{\text{input energy}} \\
  \text{Efficiency} = \frac{30 \, \text{J}}{100 \, \text{J}} \\
  \text{Efficiency} = 0.3 \text{ or } 30\% [/math]