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Introduction

Mechanics

is that branch of science which deals with the state of rest or motion of bodies under the action of forces. The subject of Mechanics is logically divided into two parts.

Statics which concerns the equilibrium of bodies under the action of forces,and Dynamics concerns the motion of Bodies. Dynamics is again subdivided into a.

Kinematics b.

Kinetics www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

k2 Study of Kinematics concerns with the motion of a body, without referring to the forces causing the motion of that body.

Study of Kinetics concerns with the motion of the body considering the forces causing the motion.

Terms and definitions

space: is the geometric region occupied by bodies whose positions are described by linear and angular measurements relative to a coordinate system.

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k3 Time is the measure of succession of events and is basic quantity in dynamics.

Mass is a measure of the inertia of a body, which is its resistance to a change of velocity Particle.

A body of negligible dimensions is called a particle. In the mathematical sense a particle is a body whose dimensions approach to zero so that it may be analyzed as point mass.

Rigid body.

A body is considered rigid when the relative movements between its parts are negligible for the purpose at hand

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Newton’s Laws of Motion

First Law . A particle remains at rest or continues to move in a straight line with a uniform velocity if there is no unbalanced force acting on it.

Second Law.

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

Third Law.

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

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k5 Newton’s second law forms the basis for most of the analysis in dynamics. As applied to a particle of mass m, it may be stated as F =m a on the particle and Where a F is the resultant force acting is the resulting acceleration. This equation is a vector equation.

Rectilinear Motion: straight line .

It is the motion of a particle along a www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Rectilinear motion with uniform acceleration:

O X M N S t

∆s ∆t A particle moves along a straight line OX as shown in the figure. The moving particle is in position M at any time ‘t’ and it covers a distance ‘s’ from ‘O’. The particle moves to N, through a distance ∆s in a small interval of time ∆t.

The velocity v at the instant when the particle is at certain point M, at time ‘t’ is the rate of change of displacement ∆s as the increment of time ∆t approaches to zero as limit is known INSTANTANEOUS VELOCITY and is given by Lt v = ∆t  0 ∆s/∆t = dS/dt www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

k7 Average Velocity : It is the uniform velocity with which the particle may be considered to be moving in order to cover the total distance s in a total time t .

) When the particle moves with uniform velocity s = u.t

(ii) When particle moves with variable velocity v av = Total distance covered (s) Total time (t)

k8 (iii) When particle moves with initial velocity u and constant acceleration a, its velocity changes to v, then v av = (u+v)/2 s = (u+v)/2 x t s = distance covered in time ‘t’ (iv) If the distances moved by the particle from start are s 1 t 1 , s 2 in t 2 , then average velocity may also be found by in v avg = (s 2 - s 1 ) / (t 2 -t 1 ) www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Acceleration :

k9 If the velocity of a particle is v at M and it changes by ∆v, in a small interval of time ∆t then the acceleration of moving particle, a at the instant at which particle is at M i. e. the instantaneous acceleration is given by





 0 





dv dt



d dt ds dt



s dt

2 A particle may move in a straight line with constant acceleration or with variable acceleration.

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k10

Uniform Acceleration :

If the velocity of a body changes by equal amounts in equal intervals of time, the body is said to move with uniform acceleration .

Variable Acceleration: If the velocity of a body changes by unequal amounts in equal intervals of time, the body is said to move with variable acceleration.

Note: When the velocity is increasing the acceleration is reckoned as positive, when decreasing as negative (retardation or deceleration) .

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k11 Displacement – Time Variations: Time Fig. 1 In fig (1) The graph is parallel to the time axis indicating that the displacement is not changing with time. The slope of the graph is zero. The body has no velocity and is at rest

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k 12 ∆t ∆x Fig.2

Time

In fig (2) The displacement increases linearly with time. The displacement increases by equal amount in equal intervals of time. The slope of the graph is constant.

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k13 ∆t 2 ∆x 2 ∆t 1 ∆x 1

Fig. 3 Time

In fig(3) The displacement is not changing by equal amounts in equal intervals of time. The slope of the graph is different at different times.The velocity of the body is changing with time. The motion of the body is accelerated.

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Velocity – Time Variations:

k14 ∆t ∆v Constant velocity

O Time Fig. (a)

In fig. (a): The velocity of the body increases linearly with time. The slope of the graph is constant i.e. the velocity changes by equal amounts in equal intervals of time. (acceleration of the body is constant and at t = 0, the velocity is finite. Thus the body moving with a finite initial velocity, and has constant

Velocity – Time Variations:

k15 Uniform acceleration ∆v ∆t

O Time Fig. (b)

In fig. (b): The body has a finite initial velocity. As time passes, the velocity decreases linearly with time until its final velocity becomes zero i.e it comes to rest. Thus the body at a constant deceleration , since the slope of the graph is negative.

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k16 Variable acceleration

∆v 2 ∆t 2 ∆t 1 ∆v 1

Time Fig. (c)

In fig. (c): The velocity –Time graph is a curve. The slope is, therefore, different at different times. In other words, the velocity is not changing at constant rate. The body does not have a uniform acceleration since acceleration is changing with time.

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Equations of Motion Under Uniform Acceleration

k17

Equation of motion (Relation between v,u, a & t) If we assume a body starts with an initial velocity u and uniform acceleration a . After time t , it attains a velocity v . Therefore the change in velocity in t seconds is (v – u) Change in velocity / sec. = v – u / t = a v = u + at -----(1) www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

k18 2. Equation of Motion : (Relation between s, u, a and t) Let a body moving with an initial uniform velocity u be accelerated with a uniform acceleration a for time t . If v is the final velocity, the distance s which the body travels in time t is determined as follows.

Now since acceleration is uniform it is obvious that the average velocity = (u + v) /2  Distance traveled = v av x t = (u + v)/2 x t = (u + (u + at))/2 x t (Substituted from 1) s = ut + ½ at 2 -----(2) www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Equation of motion: (Relation between u, v, a and s) s = average velocity x time = (u + v)/2 x t = (u + v)/2 x (v - u)/a for t = (v – u)/a therefore s = (v 2 - u 2 )/2a v 2 = u 2 +2as-----------(3) www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS k19

k20

Motion under Gravity

It has been observed that bodies falling to the earth (through distances which are small as compared to the radius of the earth) experience entirely unrestricted increase in their velocity by 9.81 m/s for every second during their fall. This acceleration is called the acceleration due to gravity and as conventionally denoted by g .

For downward motion:

a = +g v = u + gt h = ut + ½ gt 2 v 2 = u 2 + 2gh

For upward motion:

a = – g v = u – gt h = ut – ½ gt 2 v 2 = u 2 – 2gh h = Distance moved in vertical direction www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Comparison between equations of motion under uniform acceleration and variable acceleration:

Acceleration 1

Equations

2 3 k21 Uniform v = u + at S = ut + ½ at 2 v 2 = u 2 + 2as t t s Variable v = u+∫ a(t)dt S = ut + 1/2∫ a (t 2 )dt v 2 – u 2 = 2 ∫ a (s)ds 0 0 0

Graphical Representation:

v = ds/dt

s-t curve S

ds dt t 1 The slope ds/dt at any point gives the velocity

t 2 at that point.

k22

v dt dv v – t curve v

t 1

dt t 2

t The slope dv/dt at any point gives the acceleration a at that point. The shaded area under v – t curve shown above gives the incremental displacement ds during the small interval of time dt .

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k23

dv=adt a t 1 t 2 t dt

The shaded area under a – t curve shown above gives the incremental velocity dv during the small interval of time dt . www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

PROJECTILES

INTRODUCTION Assumptions:

1. Mass of the projectile is not considered. 2. Air resistance is neglected.

3. The trajectory of the particle is in the vertical plane.

k24 www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

k25

A Projectile

is a particle moving in space under the action of gravity.

The velocity with which the particle is projected into space has horizontal and vertical components. The combined effect of both components is to move the particle along a

parabolic

path. The parabolic path traced by the projectile is known as T

rajectory

of the Projectile.

The horizontal component remains constant ( as air resistance is ignored) while the vertical component of motion is always subjected to acceleration due to gravity.

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k26

DEFINITIONS:

Velocity of Projection

is the velocity with which a body is projected into space.

Angle of projection

is the angle which the initial velocity vector makes with the horizontal, or the angle at which a projectile is projected with respect to horizontal.

Range

is the distance along the reference plane between the point of projection and the point at which it strikes the plane.

Time of Flight

is the total time during which the particle remains in motion.

Maximum Height

is the maximum vertical distance covered by the projectile from the point of projection. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

MOTION OF A PROJECTILE

k27 u trajectory P(x,y) Q  h x O R M Consider a particle thrown upwards from a point initial velocity u , at an angle  O , with an with the horizontal as shown in the figure above. After attaining maximum height h , it descends and finally hits the reference plane.

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Equation of the Trajectory

k28 From the figure above x = u x . t = u (Cos  ) t t = x / (u Cos  ) --------------(1) y = u y . t – ½ gt 2 =u (Sin  )t – ½ gt 2 ------------------(2) Sub (1) in (2) we get

y = x tan



- gx 2 / (2u 2 Cos 2



)

This is an equation for a parabola. Hence the path of the projectile is a parabola.

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k29 The horizontal distance covered by the projectile is known as

Range of Projectile

denoted by R .

Resolving the initial velocity u components u x = u Cos  into horizontal and vertical u y = uSin  (constant)

Time of Flight:

We know that height.

v y = u y + at At Q v Where y  t m = 0 t 0 = u y m - gt = u y / g = u Sin  Time of flight T = 2t m / g = 2 (u Sin  /g) is the time taken in seconds to reach maximum www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Maximum height attained (h):

v y 2 - u y 2 =- 2gh (upward motion) v y = 0 h = u y 2 /2g ; h = u 2 (Sin 2  ) / 2g

Range:

R= u x x T (time of flight) = 2u 2( Sin  Cos  ) / g R = u 2 Sin2  /g (Sin2  = 2Sin  Cos  ) www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS k30

k31 From the above equation it is clear that range will be maximum if Sin2  = 1 i.e. 2  = 90 or  = 45 R max = u 2 /g Projectile will cover a maximum range when it is directed at an angle of 45°.

Two angles of Projections for a given range: We know that

Range R = u 2 Sin2  /g Sin  = Sin (   )   u 2 Sin (  - 2  )/g = u 2 Sin (2  1 )/g (say) where 2  1 = (  - 2  ) www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

k32 Thus for same range R = u 2 Sin2  /g = u 2 Sin (  - 2  )/g = u 2 Sin2  1 /g Which shows that the horizontal range remains the same when  is replaced by  1 .

Or  1 =  - 2  =(  / 2 )  2 www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

  k33

Projection on an Inclined Plane

u Q B  P 1 P

R(Sin  )

R= Range along incline; R(cos  ) α=angle of projection; β =angle of inclined plane S x =u x + ½ a x t 2 www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

k34

Projection on an Inclined Plane

R(cos β)=u(cosα)(t)+(0) ; a x =0 t= Rcos β / ucosα…..(1) S y =u y + ½ a y t 2 R sin β=u sinα (t)- ½ gt 2 Substituting eqn.(1) and simplifying we get The Range along the inclined plane

R={2u

cos

α [sin(α- β)] } / gcos

β ..(2)

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k35

Projection on an Inclined Plane:

To get maximum range on the incline, Differentiating R w.r.t

α and equating it to zero we get α= π/4 + β/4 Substituting this value of α in eqn.(2) We get maximum Range

R

max

= u

/g(1+sin β)

To find the time of flight: using the relation v=u + at www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Projection on an Inclined Plane:

k36 0=u x sin( α- β)-gx(cos β) x t  t= {usin( α- β)}/ g(cos β) a=(g cos β) is the acceleration due to gravity along inclined plane t=time taken by the projectile to reach Q where QP is perpendicular distance to the incline plane

Time of Flight T= 2xt T={2 u sin( α- β)} / g(cos β)

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k37

Projection on an Inclined Plane:

Maximum Height attained h, (PQ) using the relation v 2 -u 2 =2as 0-u 2 sin 2 ( α- β)=-2 gcos β x h

h={ u

sin

( α- β)} / 2gcos β

Vertical height P 1 Q=h/ cos β ={u 2 sin 2 ( α- β)} / 2gcos 2 β www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Practice Problems:

k38 1. On turning a corner, a motorist rushing at 15m/s, finds a child on the road 40m ahead. He instantly stops the engine and applies brakes,so as to stop the car within 5m of the child, calculate: (a) retardation the car (b) time required to stop Ans: (a) -3.21m/s 2 , (b) 4.67s

2. A stone is dropped from the top of a tower 100m high.

Another stone is projected upward at the same time from the foot of the tower, and meets the first stone at a height of 40m. Find the velocity,with which the second stone is projected upwards.

Ans: www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS u=28.6 m/s

Practice Problems:

k39 3. A train starting from rest is accelerated and acceleration at any instant is 3/(v+1) m/s. where v is the velocity of the body in meters per second at any instant. Find the distance in which the train attains velocity of 48kmph.

Ans: S=293m 4. A projectile is fired from the edge of a 150m high cliff with an initial velocity of 180m/s at an angle of elevation of 30 o with the horizontal. Neglecting air resistance, find (a) the horizontal distance from the gun to the point where the projectile strikes the ground,and (b) the greatest elevation above the ground reached by the projectile.

Practice Problems:

k40 3. A train starting from rest is accelerated and acceleration at any instant is 3/(v+1) m/s. where v is the velocity of the body in meters per second at any instant. Find the distance in which the train attains velocity of 48kmph.

(a)

Practice Problems:

k41 5. An aero plane is flying at a height of 300m,with a velocity of 360kmph. 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 ?

Ans:( 126m/s, 37.47

o ) 6. A projectile is fired from a point ‘o’ at a velocity of 125m/s has to strike a point located on the top of a tower of 200m high. The horizontal distance betweem the point ‘o’ and the tower is 1000m. Neglect the air resistance and take g= 9.8m/s 2 www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS

Practice Problems:

k42 Contd: Calculate: a) The angle, to the horizontal,at which the projectile must be fired in order to strike the point on tower in minimum time.

b) the time taken for flight c) The maximum height above ‘o’ reached by the projectile.

Ans: (a) 68.55

0 or 32.99

0 ,( b) 9.54 s, (c) 233.8m

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