WORK, POWER AND ENERGY TRANSFER IN DYNAMIC ENGINEERING SYSTEMS

(LINEAR MOTION)

EED2023 Mechanical Principles and Applications

Linear Motion

Speed, velocity and acceleration

Force and momentum- Newton’s Laws of Motions- Linear and angular motion- Momentum- Conservation of momeNtum

Speed, velocity and acceleration

Speed – distance moved per unit second

Velocity – speed and direction of travel

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(m/s in a particular direction – north for example)

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Increasing velocity Uniform velocity Decreasing velocity

Acceleration – the change in velocity per unit second

)/(change for this taken time

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graph time- velocity theofgradient the;OQ

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timeUniform acceleration Non-uniform acceleration

Speed, velocity and acceleration

Uniformly Accelerated Motion

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Linear MotionSo with reference to the statement below:

Uniform velocity is motion along a straight line at constant speed.

Uniform acceleration is motion in a straight line with constant acceleration.

We can deduce that in a straight line from stop we get uniform acceleration, then we get uniform velocity and then we get uniform deceleration as depicted in the velocity graph above.

Speed, v

Time, t

A train accelerates uniformly from rest to reach 54km/h in 200s after which the speed remains constant for 300s, at the end of this time the train decelerates to rest in 150s.Determine:

- The total distance travelled- The acceleration- The deceleration

Answer:Distance travelled = 7125 m / 7.125 kmAcceleration = 0.075 ms-2

Deceleration = 0.1 ms-2

Linear Motion – Example 1

A truck accelerates uniformly from rest to a speed of 100km/h taking 2 minutes, it then remains at a constant velocity for 32 minutes, coming up to a set of traffic lights the truck decelerates to rest in 30s.Determine:

- The constant velocity in m/s- The total distance travelled- The acceleration- The deceleration

Answer: 27.78 m/s; 55.42 km; 0.2315 ms-2; 0.926 ms-2

Linear Motion – Example 2

A train starting from rest leave a station with uniform acceleration for 20s, it then proceeds at a constant velocity before decelerating for 30s, coming to a total standstill. If the total distance travelled is 3 km and the total time taken was 5 minutes, sketch a velocity time graph and determine:

- The uniform speed- The acceleration- The deceleration- The distance travelled during the first minute

Answer: 10.91 m/s; 0.545 ms-2; 0.3637 ms-2; 545.5 m

Linear Motion – Example 3

Angular and Linear Motion

frequency is 1

where

22

speed,angular The

Tf

fT

R

s

Length of a circular arc, s = R

For a point object moving on a circular path at a uniform speed, its speed, v, may be calculated if the radius, R and time period, T are known. In one complete rotation, the object travels a distance of in time T. Therefore, its speed is given by,

R2

T

Rv

2

where is in radian

Rv Therefore,

Angular accelerationWhen a flywheel speeds up, every point of the flywheel moves at increasing speed. Consider a point on flywheel rim, at distance R from the axis of rotation. If the flywheel speeds up from initial angular speed ω1 to angular speed ω2 in time t, then the speed of the point increases from speed u = ω1 R to speed v = ω2 R in time t. By using acceleration, a = (v-u)/t, the linear acceleration of the point is

where α is the angular acceleration of a rotating object, defined as the change of angular velocity per second, and given by

Hence, linear acceleration of a point along its circular path is given by,

R

ta 12

t

12

Ra

Exercise1. An aircraft accelerates from 100m/s to 300m/s in

100s. What is its acceleration?2. A car joins a motorway travelling at 15m/s. It

accelerates at 2ms-2 for 8s. How far will it travel in this time?

3. A vehicle accelerates from rest, covering 500m in 20s. Calculate

a) the accelerationb) the velocity of the vehicle at the end of the run

4. A fly wheel is speeded up from 5 rpm to 11 rpm in 100s. The radius of the fly wheel is 0.08m. Calculate

a) the angular acceleration of the flywheelb) the acceleration of a point on the rim along its circular path

Newton’s Laws of Motion

1. A particle remains at rest or continues to move in a straight line with uniform velocity unless an unbalancing force act on it.

2. The acceleration of a particle is proportional to the resultant force acting on it and in the direction of the force.

3. The forces of action and reaction between interacting bodies are equal in magnitude, opposite in direction and collinear.

MomentumThe momentum of a moving object is defined as:

Momentum = mass x velocityThe principle of conservation of momentum states

that for a system of interaction objects, the total momentum remains constant, provided no external resultant force acts on the system.

A B

A B

A B

uA uB

vA vB

Before impart

During impart

After impart

mAvA + mBvB = mAuA + mBuB

Exercise

1. A trolley of mass 2kg is moving at 5m/s. It collides with a second, stationary, trolley of mass 8kg; it bounces back with a velocity of 3m/s. With what velocity does the second trolley move off?

2. Two billiard ball collide. Before the collision, ball A is travelling at 1.5m/s towards stationary ball B. After the collision, ball A travels in the same direction but at 0.6m/s. If ball A has a mass of 0.3kg and ball B has a mass of 0.35kg, determine the speed of ball B after impact.

Answer: 1. 2m/s 2. 0.77m/s

Exercise

3. A railway wagon of mass 1500kg travelling at a speed of 2m/s, collides with three identical wagon initially at rest. The wagons coupled together as a result of the impact. Calculate the

a) speed of the wagons after impact, b) loss of kinetic energy due to the

impact.Answer:a) 0.5m/s b) 2250J

Exercise

4. The 2000kg vehicle, travelling at 20m/s, collides with a vehicle of mass 2200kg, parked out of gear with the handbrake on, and the two vehicles lock together. Calculate the

a) momentum of the first vehicle before impact and the velocity of both vehicles immediately after impact.

b) The handbrake exerts a braking force of 10kN on the vehicles. Calculate the deceleration of the two vehicles and the distance travelled before they come to rest.

c) Calculate the angular deceleration of the road wheels of the first vehicle. The diameter of the vehicle’s road wheels is 72cm

Solution:

Exercise

5. A vehicle of mass 2000kg accelerates from rest, covering 400m in 28 seconds. Find the

a) average acceleration and the velocity of the vehicle at the end of the run.b) the momentum of the vehicle at the end of the run.

c) The diameter of the vehicle’s road wheels is 72cm. Calculate the angular velocity of the road wheels at the end of the run and the average angular acceleration.

Solution:

α = a/r = 1.02/0.36 = 2.83 rad/s2