IBRA COLLEGE OF TECHNOLOGYENGINEERING DEPARTMENT

APPLIED MECHANICS IIMIME 3101 (Summer-2015)

Higher Diploma in Mechanical Engineering

Prepared by:Dr.K.S.Seetharama

Lecturer

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Contents

1 Kinetics and kinematics of particles 1-4

2 Motion with uniform acceleration 4-6

3 Linear motion of a particle with variable acceleration 7-10

4 Curvilinear motion of a particle. Motion of projectiles 11-16

5 Angular motion of a particle 17-19

6 Oscillatory motion of a particle, Simple Harmonic Motion 20-23

7 Absolute motion and relative motion .relative velocity 24-27

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9

Newton’s laws of motion . work power and energy. Law of conservation of momentum

Work Power And Energy

28-30

31-32

10 Elastic systems. Collision of elastic bodies 33

11 Kinematics of rigid bodies. D Alembert’s principle 34-37

12 References 38

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APPLIED MECHANICS

Statics Dynamics

Kinematics Kinetics

Chapter 1

Kinematics and Kinetics of a Particle

1.1 Introduction: Applied Mechanics or Engineering Mechanics is a foundation and framework for most of the branches of engineering. The primary purpose of study of engineering mechanics is to develop the capacity to predict the effects of force and motion while carrying out the creative design functions of engineering.

Statics provides analysis of stationary systems while Dynamics deals with systems that

change with time. The investigation of motion of a rigid body may be separated into two

parts, the geometrical part and the mechanical part. Within the geometrical part

Kinematics the transference of the body from one position to the other is investigated

without respect to the causes of the motion. The change is represented by analytical

formulae.

Statics : Study of a body at rest or a body with uniform motion.

Dynamics: Study of moving body.

Kinematics: Kinematics analyzes the positions and motions of objects as a function of time, without regard to the causes of motion. It involves the relationships between the quantities displacement (s), velocity (v), acceleration (a), and time (t). Thus

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O P Q

S1 S

S2

Kinematics : Study of Displacement, Velocity, Acceleration of a body.

Kinetics: Study of motion of body and the related forces.

Motion: Body changes position with reference to surroundings

Displacement :

The displacement, of a body between times t1 and t2 is defined to bethe change in position coordinate of the bodyDisplacement is a vector quantity.

Displacement (s) Change in position in a particular direction s= s2 – s1 (units: meter, m)

Velocity: (v): Rate of change of displacement.

Velocity= change∈displacementtime =

dsdt

Acceleration: Rate of change of velocity.

Acceleration ¿change∈velocity

time = dvdt

a = v−u

t (units: meter/second2, m/s2)

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where u = initial velocity, m/s v =final velocity, m/s t = time, sif v>u, acceleration is positiveif v<u , acceleration is negative (retardation)

1.2 Different Types of Motions

1.2.1 Motion with Uniform velocity.

Velocity of the body remains same irrespective of time.

For uniform velocity motion,

v ¿st

1.2.2 Motion with Uniform Acceleration.

Velocity of the body changes by equal amounts in equal intervals of time.Formulae:

(1) v = u + a t(2) s = u t +(1/2) at2

(3) v2 = u2 +2as

For a body moving under gravity, the above equations can be written as-

falling down moving up

v = u + g t v = u – g t

h = u t +(1/2) g t2 h = u t -(1/2) g t2

v2 = u2 +2 g h v2 = u2 -2 g h

Where g = acceleration due to gravity = 9.81 m/s2 h = height m

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1.3 Practice Problems:

(1) A car accelerates uniformly from a speed of 30 km/h to a speed of 75km/h in 5 seconds. Determine the acceleration of the car and distance travelled in 5 seconds

(2) An automobile is decelerating from a speed of 65km/h at the rate of 1.5 m/s2. How long will it take to come to rest and how far will it have gone

(3) An automobile travels 360 m in 30 seconds while being accelerated at a constant rate of 0.5m/s2 .Determine its initial velocity, final velocity and the distance travelled during first 10 seconds.

(4) A motorist travelling at 15m/s finds a child on the road, 40 m ahead. He instantly applies brakes so as to stop the car within 5m from the child. Calculate the retardation and the time required to stop the car.

(5) A stone is dropped from the top a tower 100m high. Another stone is projected upwards at the same time from the foot of the tower and meets the first stone at a height of 40 m. Find the velocity with which the second stone is projected upwards

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O P Q

s ∂s

(6) A body falling freely under the action of gravity passes two points 15 m apart vertically in 0.3 second. From what height above the higher point, did it start to fall

ASSIGNMENTA motorist is travelling at 72km/h . He observes traffic light which is 320 m away

turns red. The traffic light is timed to stay red for 22 seconds. If the motorist wishes to pass the light without stopping, just as it turns green again, determine (i) the required uniform deceleration of the car and (ii) the speed of the car

1.4 Linear Motion of a particle with variable acceleration.

Consider a particle moving in a straight path starting from O with variable acceleration as shown in figure. S is the distance covered in time t when it moves from O to P. ∂s is the distance covered from P to Q in a small interval of time ∂t.

Displacement = s mVelocity v = ds/dt m/sAcceleration a = dv/dt. m/s2

= d(ds/dt) dt = v (dv/ds) m/s2

Similarly,

Velocity, v= a.dt

Displacement s= v.dt

Example 1

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A particle has a displacement from point O given by s = t3 – t2 + 4 t.Where s is in metres and t is time in seconds.Find the displacement from point O, velocity and acceleration of the particle at time t = 2 s.Solution

s = t3 – t2 + 4 t.

for t =2

s = 23 – 22 + 4 *2

= 8- 4 + 8

= 12 m.

Differentiating this equation with respect to t, we get the velocity

v = 3 t2 – 2 t +4

put t =2

v = 3 *22 – 2 *2 +4

= 12 -4 +4

= 12 m/s

Again differentiating the velocity equation w.r.t. time we get acceleration.

v = 3 t2 – 2 t +4

a = 3 *2t – 2 + 0

put t =2

a = 3*2*2 -2

= 10 m/ s2

Example 2

A body starting from rest moves with a variable acceleration such that the acceleration, a , of the particle, at time t seconds is given by:

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a=4− t2 0 m/ s2 . Find the distance travelled by the body after 80 seconds.

Solution

First integrate the acceleration to obtain the velocity

Velocity, v = ∫ a dt

= ∫ (4 – t/20) dt

= 4t – t2/ 40 + C1

To find the value of the constant C1  , note that the particle is initially at rest, so that

v = 0 when t =0. Substituting these values shows that C1 =0. Hence the velocity is:

v = 4t – t2/ 40 m/ s2

The displacement of the particle can be found by integrating the velocity:

s = ∫ (4t – t2/ 40 ) dt

= 4 t2/ 2 – t3/ 3 *40 + C2

To find the constant C2 , assume that the particle starts at the origin, so that s =0 for t =0 Hence C2 =0and the displacement at time t is given by:

s = 4 t2/ 2 – t3/ 120

To find the distance travelled in 80 seconds, substitute s = 80 in the equation.

s = (4 * 802 /2) - (803 / 120)

=4* 6400 /2 -512000 /120

= 4* 3200 – 4266.66 m

= 8533.33 m.

1.5 Practice Problems

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1 The equation of motion of a particle is s = - 6 - 5t2 + t3

where s is in meters and t is in seconds.Calculate (i) The displacement and the acceleration when the velocity is zero. (ii) The displacement and the velocity when the acceleration is zero.

2 A body moving in a straight line has the equation of motion given by s = 2 t3 -4t +10where s is measured in meters and t in seconds.Determine

(i) Time required for the body to reach a velocity of 68 m/s starting from rest.(ii) The acceleration of the body when the velocity is equal to 32 m/s.(iii) The net displacement of the body between the time interval of t= 0 to t = 4 s.

3 A car travels along a straight path such that the distance travelled is directly proportional to the cube of time of travel .If it travels 216 m during the first minute, find the displacement, velocity and acceleration just at the start and 10 s after starting.

4 A train starting from rest is accelerated and the acceleration at any instant is given by, 3/(v+1) m/s. where v is the velocity of the body in m/s. find the distance in which the train attains a velocity of 48 km/ hr.

5 The acceleration of a particle moving in a straight line is given by a = 30t – 10.where a is in m/s2 and t is s. If it is known that its displacement at t = 0 and t = 5 s are, -25 m and + 25 m respectively, calculate displacement and velocity at t = 6 s.

6 The equation for acceleration of a particle starting from rest and moving in a straight line is given by a = 10 – 0.006 s2 , where a is in m/ s2 , s in meters, determineThe velocity of the particle when it has travelled 20 meters.The distance travelled by the particle when it comes to rest.

ASSIGNMENTA particle moves along a straight line in a viscous medium with acceleration a = -2/x2

where a and x are in m/s2 and m respectively. Knowing that x = 1 m and v = 2 m/s at t = 1 s , determine the position and velocity of the particle at t = 4 s.

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Chapter 2

Curvilinear motion of Particles.

When a particle moves along a curve, other than straight line, it is in curvilinear motionTo define the position of such a particle, we use x and y axes with O as origin.

Consider a particle moving along a curved path from position P(x,y) to P’ (x + ∆x, y + ∆y) , as shown in figure.Displacement of the particle ∆r = ∆x, + ∆y Magnitude dr = √ (∆x2 + ∆y2) Direction θ = tan-1 ( ∆y/∆x)Velocity of the particle at any instant is given by the x and y components. vx = x, where x = dx/dt vy = y where y = dy/dt.

Similarly, horizontal component of acceleration ax = d vx / dt = xVertical component of acceleration ax = d vy / dt = y

Example for a body having curvilinear motion Projectile, a car moving on a road which is not straight. Fish moving in water.

2.1Projectiles

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A particle thrown into space at an angle traces a parabolic path under the action of gravity and is called projectile. Consider a ball simply dropped from a height. It traces a straight-line path and falls just below the point of dropping.

Similarly if a ball is thrown exactly vertically up will move up in a straight path and falls back to the same point from where it was projected.

On the other hand if we throw the ball at an angle, it will follow a parabolic path before reaching the ground at another point. The ball travels in a curvilinear motion.

A shot fired from a gun from the top of a hill will trace a parabolic path as shown below.

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A Foot ball kicked from ground will have a curvilinear motion as shown below.

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Terminology:

Trajectory: path of projectile- a parabola

Velocity of projection u: Velocity with which projectile was thrown.

Angle of projection α : The angle made by initial velocity vector with horizontal

Range r : Horizontal distance travelled by the particle.

Time of flight: total time taken by the projectile during flight.

Maximum height : Maximum vertical distance covered from the point of projection.

Consider a particle thrown into space at an angle α as shown in figure.

if u = initial velocity of projection.

ux = horizontal component of velocity. = u cos α remains constantuy = vertical component of velocity. = u sin α

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Time of flight = (2usin α)/ g

Maximum height h = (u2sin 2 α)/ 2g

Range r = (u2sin 2 α)/ gEquation of motion of projectile: y = x tan α – (gx2/ 2u2cos 2 α)

An object projected with different angles of projection will trace different paths for the same initial velocity of projection

2.1,1 Practice problems:

1 A bullet is fired into space at an inclination of 60o with a velocity of 60m/s. Determine (i) Time of flight. (ii) Maximum height attained (iii) At what angle should it be fired to have maximum range and what would be its magnitude?\

2 A rifle bullet is fired and it hits a target at 100 m. If the bullet rises to a maximum height of 3 cm above the horizontal line between the muzzle of rifle and the point of impact on the target, find the muzzle velocity.

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3 A fighter jet moving horizontally at 108 km/hour, towards a target is at an altitude of 1000m,. It releases a bomb, which hits the target. Estimate the horizontal distance of the aircraft from the target when it released the bomb .Calculate also the direction and velocity with which the bomb hits the target.

4 An aircraft is flying at a height of 300 m with a horizontal velocity of 360 km/ hour. A shell is fired from the ground exactly when the aeroplane is above the gun. What should be the minimum initial velocity of the shell and the angle of inclination in order to hit the aeroplane?

5 A shot is fired from the edge of a 150 m high cliff with an initial velocity of 180 m/s at an inclination of 30o to the horizontal (upwards). Find the horizontal distance from the gun to the point where the shot strikes the ground. Also find the greatest elevation above the ground reached by the projectile.

6 A shot is fired with a velocity of 30 m/s from a point 15 m in front of a wall 6 m high. Find the angle of projection to the horizontal to enable the shot to just clear the top of the wall.

7 a bullet is fired upwards at an angle of 30o to the horizontal from a point P on a hill and it strikes a target which is 80 m lower than P. The initial velocity of the bullet is 100 m/s. Calculate:(i) The maximum height to which the bullet will rise above horizontal.(ii) The actual velocity with which it will strike the target.(iii) The total time required for the flight of the bullet

(iv) The horizontal distance between the hill position and the targe

ASSIGNMENTA man standing at the 18 m level of a tower throws a stone in the horizontal direction. Knowing that the stone hits the ground 25 m from the bottom of the tower, determine (i) the initial velocity of the stone (ii) the distance at which a stone would hit the ground if it were thrown horizontally with the same velocity from a 22m level of the tower.

2.2 ANGULAR MOTION OF A PARTICLEMotion of rotation

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θ

v, at

an

x

yFig 2.21

Fig 2.22In motion of rotation, a particle moves in a circular path. Its position at any instant is described by the angle θ covered by it from a fixed axis.

Angular displacement. θ : Change in angular position with respect to time.

Angular velocity ω : Rate of change of angular displacement. ω = d θ/ dt rad/sInstantaneous rectilinear velocity is given by v = ωr

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Linear motion Angular motion

v = u + a t ω= ω0 + αts = u t +(1/2) at2 θ= ω0 t + (1/2)αt2v2 = u2 +2as ω2 = ω02 + 2αθ

If N is the uniform speed of rotation in revolutions per minute (r.p.m.),Angular velocity ω= 2πN /60 rad/s.

Angular acceleration α : Rate of change of angular velocity.

α = d ω/ dt = d2 θ/ dt2 rad/s2

The acceleration can be resolved into two components. The tangential component and normal component.

Tangential component of acceleration at = r (d ω/ dt) = rα

Normal component of acceleration an = v2 /r = ω2 r.

Equations of linear motion and rotary motion.

2.2.1 PRACTICE PROBLEMS.

1 A wheel is rotating about its axis with a constant angular acceleration of 1 rad/s 2 . If the initial and final angular velocities are 5.25 rad/s and 10.5 rad/s respectively determine the total angle turned through during the time interval this change of angular velocity took place.

2 A Wheel accelerates uniformly from rest to 1500 r.p.m. in 15 seconds. What is the angular acceleration? How many revolutions does the wheel make in attaining the speed of 1500 r.p.m.?

3 A wheel of 1.2 m diameter is mounted on a shaft between two bearings. The wheel is set rotating from rest by applying a constant moment of 110Nm at the rim and the speed of the wheel becomes 100 r.p.m. in 10 minutes. Determine(i) The number of revolutions made by the wheel during this time.

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(ii) The angular acceleration of the wheel in rad/s2.

(iii) The work done in rotating the wheel during this time. (iv) Peripheral velocity of the wheel in metres/min. at 100 r.p.m.

4 The relation between the angle of rotation and the time in case of a rotating body is given by the equation θ = 2t3 + 3t2 +15Determine the angular velocity, displacement and acceleration when t = 10 seconds and state whether the acceleration is uniform.

5 A horizontal bar 1.5 metres long and of small cross section rotates about the vertical axis through one end. It accelerates uniformly from 1200 r.p.m. to 1500 r.p.m. in an interval of 5 seconds. What is the linear velocity at the beginning and at the end of the interval? What are the normal and tangential components of the acceleration of the mid point of the bar after 4 seconds of beginning of acceleration?ASSIGNMENTA flywheel is making 180 rpm and after 20 s it is running at 150 rpm. How many revolutions will it make and what time will it take before stopping, if retardation is uniform?

2.3 Oscillatory motion of a particle.

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A BC

Fig 2.31

Simple Harmonic Motion: A body having simple harmonic motion moves such that its acceleration is directed towards a fixed point on the path and is proportional to its distance from that point.Simple harmonic motion is a to and fro motion. Example: oscillations of a pendulum,Motion of piston in an engine cylinder, motion of a mass suspended by a spring.

Consider a particle moving from A to B and returning so that acceleration is always directed towards point C and proportional to distance from C, then it is in S,H,M.Acceleration is maximum at A and B , it is zero at C.Velocity is zero at A and B, it is maximum at C.

Oscillation: To and fro motion of a body from mean position to one extreme position, then to the other extreme position and back to the mean position.

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θ

r

ωA

B

B’C

x

Amplitude: Maximum displacement of the body from its mean position.

Time Period.: Time taken by the particle to complete one oscillation. T = 2π / ω Where ω is the angular velocity in rad/s.

Frequency: Number of oscillations per second.

f =1/T

Fig 2.32

If a particle is moving with uniform angular velocity in a circle of radius r , the projection of the particle B’ on the diameter moves with S.H.M. θ = ωt displacement x = r sin ωt velocity vx = dx/ dt = ωr cos ωt = ω BB’ vx = ω √(r2 – x2) Acceleration ax = dv/dt = d2x/dt2

= - ω2 r sin ωt ax = - ω2xIf f is the frequency, 2πf = ω or f = ω/ 2π time period T = 1/f

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W1 1

= 2π / ω = 2π√(1/ ω2) = 2π√( r sin ωt / ω2 r sin ωt) T = 2π√( x / ax)

Motion of a mass attached to a helical spring

Fig 2.33W = Weight attached Nk = Stiffness of the spring N/mx = displacement at any instant. ma = acceleration of the weight. m/s2

g = gravitational acceleration 9.81 m/s2

T = time period s.

T = 1/f = 2π √ (W/kg)

2.3.1Practice problems.

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1 A body with simple harmonic motion has a velocity of 12 m/s when the displacement is 50 mm and 3 m/s when the displacement is 100 mm.. The displacement being measured from the mid point.Calculate the frequency and amplitude of motion. What is the acceleration when the displacement is 75 mm?

2 A piston of an engine moves with S.H.M. The crank rotates at 100 r.p.m. and stroke is 1.8 m. Find the velocity and the acceleration of the piston when it is at a distance of 0.6 m from the centre.

3 The amplitude of a body moving in simple harmonic motion is 0.6 m and its period of oscillation is 2.4 s. What would be its speed and acceleration 0.5 s after it has passed middle position. Also find its frequency.

4 The strength of a spring is such that a load of 25 N is required to elongate it by 10 mm. When a certain load is suspended from one end and caused to perform simple harmonic motion, the number of complete oscillations observed per minute is 80. Calculate the value of the load.

ASSIGNMENTA system, the amplitude of motion is 5 m and the time period is 4 s. Find the time required by the particle in passing between the points which are at distances of 4 m and 2 m from the centre of path and are on the same side of it

Chapter 3

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O BA

XB/AXA

XB

Relative motion of a particle

Consider two particles A and B moving along the same line.

XA – Displacement of A with reference to O

XB – Displacement of B with reference to O

XB/A – Displacement of B with reference to O , is known as relative position coordinate of B with reference to A.

XB = XA + XB/A

Differentiating, Relative Velocity VB = VA + VB/A

Differentiating again,Relative Acceleration AB = aA + aB/A

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Practice problems

1 A ball is thrown vertically upward from the 12 m level of an elevator shaft with an initial velocity of 18 m/s. At the same instant an elevator passes the 5 m level moving upward with a constant velocity of 2m/s.Determine (i) When and where the ball will hit the elevator? (ii) The relative velocity of the ball when it hits the elevator.

2 An automobile A starts from O and accelerates at the constant rate of 0.8 m/s2 .A short time later it is passed by bus B which is travelling in the opposite direction at a constant speed of 5 m/s. Knowing that bus B passes point O 22 s after automobile A started from there, Determine when and where the vehicles pass each other.

3 A balloon is rising up with constant acceleration of 1.6 m/s 2 from ground. After 5 seconds, a ball is thrown vertically up from ground towards balloon. Find maximum velocity of projection of ball so that it will just strike the balloon.

4 Planes A and B started from rest on straight parallel runways simultaneously with accelerations aA= 18t – 2 and aB = 15 t2 - 40.both in m/s2. Determine the distance between them when t = 6 s and speed of each plane at that instant.

5 In the arrangement shown in figure, the block A has a velocity of 3.6 m/s, to the right, determine the velocity of cylinder B

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6 In the figure shown, velocity of block A increases at the rate of 0.044 m/s every second, determine the acceleration of B.

7 At a certain instant, cylinder A has a downward velocity of 0.8 m/s and an upward acceleration of 2m/s2 . Determine the corresponding velocity and acceleration of cylinder B

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F = m a

Chapter 4Newton’s laws of motion

First law: A body continues in rest or in uniform motion until acted by external force.Second Law: Rate of change of momentum is proportional to applied force and acts in the same direction as the force.If m = mass of body kga = acceleration m/s2

F = applied force N Then according to Newton’s II law

Third law: Action and reaction are equal and oppositeExamples for III law:1 Jet propulsion2 recoiling of gun

Law of conservation of momentum:Total momentum of a system of objects remains same if no external force acts.

i.e., initial momentum = final momentum m1u1 + m2u2 + - - - = m1v1+ m2v2 + - - -

Impulse and impulsive force If a force F acts on a body for t seconds product Ft is called impulse of the force.

We have F = ma = m (v-u)/tImpulse = Ft = m(v-u) x t T = m (v –u) =mass x change in velocity

Rotary Motion

Torque: Turning moment of the force acting on the body. T = P x r

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T = Torque Nm , P = Force N , r = Perpendicular distance between the force and axis of rotation.

Work done by a torque: W = T x θ W = Work done, Nm θ = Angle of rotation, radians.Newton’s II Law of motion applied to rotary motion;The rate of change of angular momentum is proportional to the external torque and takes place about the axis of the torque in the same direction of rotation.T = I x α I = Moment of inertia, α = Angular acceleration, T = Torque

Law of conservation of angular momentumIf the external torque acting on a system is zero, the angular momentum of the system remains constant.I1ωO1 + I2ωO2 = I1ω1 + I2ω2

I1 , I2 Moment of inertia of the bodiesωO1, ωO2 Initial angular velocities, ω1, ω2 Final angular velocities

Practice Problems 1 A horse pulls a cart of mass 500 kg and produces an acceleration of 4 m/s2. Find the force exerted by the horse. 2 The weight of a body on earth is 490 N. The acceleration due to gravity on earth is 9.8 m/s2 . What will be the weight of the body on (i) moon where acceleration due to gravity is 1.4 m/s2 and (ii) sun where the acceleration due to gravity is 270 m/s2. 3 A body of mass 200 kg is moving with a velocity of 20 m/s when a force of 100 N acts on it for 90s. Determine the velocity of the body (i) when the force acts in the direction of motion and (ii) when the force acts opposite to the direction of motion. 4 A train of mass 20000 kg is moving at 10 km/h and after 20 seconds, it is moving at 50km/h. What is the average force acting upon it during this time in the direction of motion? 5 A train weighing 3500kN has a frictional resistance of 5N per kN of weight. What average pull will be required if it is to attain a speed of 72 km/h from rest in 5minutes on a level track? 6 A man of mass 80 kg dives into a swimming pool from a tower of height 18m. He was found to go down in water by 2.2 m and then started rising. Find the average resistance of water. Neglect the resistance of air.Take value of g = 9.8 m/s2. 7 A gun of mass 3x 104 kg fires projectile of mass 456 kg with a velocity 305 m/s. With what initial velocity the gun will recoil/ If the recoil is overcome by an average force of 60kN, how far the gun will travel? How long will it take? 8 Two railway wagons of masses 10000kg and 8000 kg moving in the same direction at speeds of 10m/s and 15m/s respectively collide and subsequently move together. Calculate the common speed due to impact.9 To determine the moment of inertia of a wheel about its axis, a string of length 6 m is wrapped round its shaft. The string is pulled with a constant force of 100N. It is observed that , when the string leaves the axle, the wheel is rotating 3 times every second. Calculate the moment of inertia of the wheel.

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10 An uniform homogeneous cylinder rolls without slipping along a horizontal level surface with a transitional velocity of 0.2 m/s. If its mass is 0.1 kg and the radius is 100 mm what is its total kinetic energy?

ASSIGNMENT

A train weighing 3500kN has a frictional resistance of 5N per kN of weight. What average pull will be required if it is to attain a speed of 72 km/h from rest in 5minutes on a level track?

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Chapter 5

Work Power and Energy

Work: If force acting on a body moves it through a distance, work is done. Work = Force x Displacement W = F x s Nm or Joule

Power: Rate of doing work is Power. Power = work done Time taken P = Fs/t = Force x velocity = F v Nm/s or Joule/s or Watt.Energy: It is the capacity to do work.Units of energy and work are same. Nm or Joule

Forms of energy: 1 Mechanical Energy 2 Electrical energy 3 Heat energy 4 light energy 5 Chemical energy 6 sound energy 7 magnetic energy 8 nuclear energyThe mechanical energy can exist in two forms.1 Potential energy – energy due to the position of the body2 Kinetic energy – energy due to motion of the body Potential energy P.E. of a body of mass raised through a height of h from reference level is given by P.E. = m g h.Kinetic energy of a body of mass m moving with a velocity of v is given by K.E. = mv 2 2Law of conservation of energy:Statement: Total energy in the universe is constantOrEnergy cannot be created or destroyed but it can be converted from one form to the other.K.E. + P.E. = Constant. Practice problems 1 An engine pulls a train of mass 400 tones, including its own mass on a level ground with uniform acceleration until it acquires a velocity of 54km/h in 5 minutes. If the frictional resistance is 40 Newtons per tone of mass and the air resistance varies with the

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square of the velocity, find the power of the engine. Take air resistance as 500N at 18km/h. 2 A vehicle of weight 1500KN is drawn up an incline of 1 in 150 with uniform speed of 24km/h. If the resistance due to friction, air etc., is 4N per kN calculate the power of the engine.

3 A hammer weighing 70 N is arranged to swing downwards in a circular path. It is released at a point 1.2m higher than the lowest point of the circle. Find the kinetic energy and the speed of the hammer at the lowest point. If at that point it breaks a piece of metal and so parts with 42Nm of energy, to what height will it rise on the outer side?4 A bullet of mass 0.03 kg is fired horizontally into a body of mass 10 kg, which is suspended by a string of 1m length. Due to the impact, the body swings through an angle of 300 with the vertical. Find the velocity of the bullet at impact.

5 A pile of mass 500kg is driven into ground by dropping freely a hammer of mass 318 kg through a height of 2.7m. If the pile is driven into the ground by 0.15m, calculate the average resistance of the soil.

6 A block weighing 260N lying on a horizontal plane is pushed by applying a force P =260 N on it. The force is removed after the block covers a distance of 3.9m. Taking the coefficient of kinetic friction between the block and the ground as 0.35, determine the following.i Velocity of the block after it covers a distance of 3.9m.ii Further distance covered by the block, from the moment the force is removed.

ASSIGNMENTA block weighing 260N lying on a horizontal plane is pushed by applying a force P =260 N on it. The force is removed after the block covers a distance of 3.9m. Taking the coefficient of kinetic friction between the block and the ground as 0.35, determine the following.i Velocity of the block after it covers a distance of 3.9m.ii Further distance covered by the block, from the moment the force is removed.

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Chapter 6Collision of Elastic bodies

Impact: A collision between two bodies over a small interval of time exerting large forces.Line of impact: The common normal to the surfaces in contact during impactDirect impact: The velocities act along the line of impact.Oblique impact: The velocities not acting along the line of impact.Restitution: regaining the shape after impactTime of restitution: time taken by the body to regain original shape after compression during impact.Time of compression: Time taken by the bodies in compression after the collisionTime of collision: sum of time of compression and time of restitution.Law of conservation of momentum as applied to collision of bodies:Total momentum of the two bodies before collision is equal to the total momentum of the bodies after collision. m1u1 + m2u2 = m1v2 + m2v2

Newton’s Law of collision of elastic bodies:When two elastic bodies collide, their velocity of separation is proportional to their velocity of approachv2 – v1 = e(u1 - u2 )Where u1 , u2 initial velocities of two bodies v1, v2 , final velocities of two bodiese constant of proportionality called as coefficient of restitutione = 0 for inelastic bodiese = 1 for perfectly elastic bodies

Practice Problems1 A ball of mass 8 kg moving with a velocity of 10 m/s collides directly on another of mass 24 kg moving at a speed of 2 m/s moving in the opposite direction. If e = ½, find the velocities of the balls after impact.

2 A ball of mass 2 kg strikes a ball of mass 4 kg which is moving in the same direction as the first. If the coefficient of restitution is ¾ and the first ball is reduced to rest after impact, find the ratio between the velocities of the balls before impact.3 A ball of mass 4 kg moving with velocity of 2 m/s impinges on another ball of mass 5 kg moving with velocity of 1 m/s in the opposite direction. If e = ½ , find the velocities of the balls after impact.

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Chapter 7

KINEMATICS OF RIGID BODIES

D’Alembert’s principle;Statement.- External forces acting on a rigid body are equivalent to the effective forces of the various particles forming the body.∑(FX)eff = māx

∑(Fy)eff = māy

General plane motion of a rigid bodyCombined motion of a rigid body which is which is neither pure translation nor pure rotation is known as general plane motion.

Example 1

A simple pendulum of mass m and length r is mounted on a flat car, which has a constant horizontal acceleration of aᶿ as shown in figure. If the pendulum is released from rest relative to the car at the position ᶿ = 0, determine the expression for the tension T in the supporting light rod for any value of ᶿ . Also find T for ᶿ = π/2 and ᶿ = π.

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A

B

40o

120o

o

B

A

1 For the rod AB 600 mm long shown in figure, which slides without loosing contact, find the velocity of A, if velocity of point B is 1m/s. Also find the angular velocity of the rod.

2 A motor gives a disc of 0.3m diameter an angular acceleration of α =0.6t2 + 0.75 rad/s2, where t is in seconds. If initial velocity of the disc is ωo = 6 rad/s find magnitude and velocity op block B when t =2s.

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References

1 R.K.Rajput, Engineering Mechanics, Dhanpat Rai, 2003

2 Ferdinand J Beer, E. Russel Jonsten, Mechanics for Engineers- Dynamics, Mc Graw-Hill International, 1987

3 Bhavikatti S.S, Engineering Mechanics, New age International,2008

4 Mariam , Craig , Engineering Mechanics Dynamics. New age International,2008

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