APPlied mechanics-II (Dynamics)

CHAPTER- 1: INTRODUCTION TO DYNAMICS Mechanics as the origin of Dynamics: Mechanics is defined as that science which describe and predicts the conditions of rest or motion of bodies under the action of forces. It is the foundation of most engineering sciences. It can be divided and subdivided as below:

(i) Newtonian Mechanics (Engineering Mechancis)

(ii) Relativistic Mechanics (It deals with the conditions involving seed of bodies close to the speed of light )

(iii) Quantum Mechanics (It deals with the conditions involving extremely small mass and size ie atomic distance)

(a) Mechancis of rigid bodies

(b) Mechanics of deformable bodies

(c) Mechanics of fluids

Mechanics of Compressible fluids

Dynamics

Statics

Kinematics

Mechanics of Incompressible fluids

Kinetics

Dynamics: It is which of Newtonian Mechanics which deals with the forces and their effects, while acting upon the bodies in motion. When we talk about the motion of the planets in our solar system, motion of a space craft, the acceleration of an automobile, the motion of a charged particle in an electric field, swinging of a pendulum, we are talking about Dynamics. Kinematics: It is that branch of Dynamics which deals with the displacement of a particles or rigid body over time with out reference to the forces that cause or change the motion. It is concerned with the position, velocity and acceleration of moving bodies as functions of time. Kinetics: It is that branch of Dynamics which deals with the motion of a particle or rigid body, with the reference to the forces and other factor that cause or influence the motion. For the study of motion Newton’s Second Law is widely used.

Downloaded from www.jayaram.com.np/ -1

Downloaded from www.jayaram.com.np

Chapter:- 2 Determination of motion of particles: • In general motion of particles (position, velocity and acceleration ) is expressed in terms of function as, X = f(x) , [ x = 6t2 +t3 ] * But in practice the relation of motion may be defined by any other equation with function of x, v,& t . a = f(t) a = f(x) a = f(v) etc. so these given relation are integrated to get the general relation of motion x = f(t) . Case-I: When acceleration is given as function of time [i.e a = f(t) ] [ a = 6t2+t3] We know, a = dv/dt → dv = adt or, dv = f(t) dt Now integrating both sides taking limit as time varies from 0 to t and velocity varies form vo to v.

∫

v

t

vo

dv = ∫ f (t )dt 0

t

v − v o = ∫ f (t )dt o

t

v = v o + ∫ f (t )dt .......(i ) o

Again, velocity is given by, V = dx/dt → dx = vdt Again integration both sides of equation similarly form time 0 to t and position xo to x. We get,

∫

x

xo

t

dx = ∫ vdt 0

⎡v + t f (t )dt ⎤ dt Putting value of V form equation (i) ∫o ⎢⎣ 0 ∫0 ⎥⎦ t t x = xo + ∫ ⎡vo + ∫ f (t )dt ⎤ dt .........(ii ) ⎥⎦ 0⎢ 0 ⎣ Thus position is obtained from equation of a = f(t)

x – xo =

t

# Find the velocity and position of a particles after its 5 sec from Rest, which moves with equation of a = 6t2-4t. Solution: Given equation a = f(t) → a = 6t2 – 4t xo = 0 , vo = 0 and t = 5. We know, V0 =

[

∫

t

o

f (t )dt = ∫

t

o

v = 2t 3 − 2t 2 Again,

]

5 0

⎡ 6t 3 4t 2 ⎤ − f (t )dt = ∫ (6t − 4t )dt = ⎢ ⎥ 0 2 ⎦ ⎣ 3 5

2

5 0

= 200 m / s

-By Er. Biraj Singh Thapa (Lecturer, Eastern College of Engineering, Biratnagar)/ -2

t

∫ ⎡⎢⎣v + ∫

X = xo +

o

0

0

=0+

t

f (t )dt ⎤ dt ⎥⎦

∫ [0 + 200]dt = ∫ t

5

0

0

200dt = [200t ]0

5

= 100 m Therefore, x = 100m and v = 200m/s after 5 second of motion. Case-II When the acceleration is a given function of position [ i.e a = f(x) eg. x2+4x] We know, a = dv/dt = dv/dx . dx/dt = v.dv/dx or, vdv = adx or, vdv = f(x)dx

[ ∵ a = f(x)]

Now , Integrating both sides of above equation , taking limit as velocity varies from Vo to v as position p varies form xo to x. t

vdv = ∫

x

⎡v2 ⎤ f ( x)dx ⇒ ⎢ ⎥ ⎣2⎦

i.e

∫

or,

x v 2 v 02 − = ∫ f ( x)dx x 0 2 2

vo

xo

v

x

v0

= ∫ f ( x)dx xo

1

x 2 ∵ v = ⎡v02 + 2∫ f ( x)dx ⎤ .......(1) ⎢⎣ ⎥⎦ x0 Again We know,

V = dx/dt ⇒ dx = vdt. Integrating both sides with limits as time varies from 0 to t and position from xo to x . i.e

∫

x

xo

t

dx = ∫ vdt ........(1) o

Putting value of v varies from equation (1) we get, x – xo =

∫

1 2

⎡v + 2 f ( x)dx ⎤ dt ∫xo ⎥⎦ ⎢⎣ 2 0

x

1

x 2 ∴ x = x0 + ∫ ⎡vo2 + 2∫ f ( x)dx ⎤ dt ⎥⎦ x0 0⎢ ⎣ t

Case III : When acceleration is a given function of velocity (i.e a =f(v) eg. a = v2+v) We know, a = v dv/dx ⇒ f(v) = v dv/dx Or , dx = v dv/f(v) Integrating both sides taking limit as velocity varies form vo to v and position varies from xo to x x v v dv dv dx v x x = ⇒ − = 0 ∫x0 ∫v0 f (v) ∫v0 v f (v) dv v0 f (v ) e.g The acceleration of a particle is defined as a = -0.0125v2, the particle is given as initial velocity v0, find the distance traveled before its velocity drops to half. Solution: Given, a = -0.0125v2 i.e a = f(v) , Initial velocity vo, final velocity vo/2 For motion a = f(v) xo = 0, x = ? v

∴ x = x0 + ∫ v


Downloaded from www.jayaram.com.np v

x = xo + ∫ v v0

dv f (v )

v0 2 v0

v

0 v −1 1 2 x= ∫ = dv dv 2 ∫ v 0.0125 0 v − 0.0125v v0 ⎡ ⎤ 1 [ln v]v20 = − 1 ln ⎢ v0 ⎥ =− 0.0125 0.0125 ⎣ 2v0 ⎦

Or, x = 24.08 m ans.

2.2 Uniform Rectilinear motion: * Uniform motion means covering equal distance over equal intervals of time. ie velocity = constant. We have, V = dx/dt = v [ v = constant velocity of body]

∫

∴ dx = udt ⇒

x

x0

t

dx = ∫ vdt 0

[ Integrating both sides under limits as position varies from xo to x and time 0 to t]

∴ x – xo = vt x = xo +vt

∴ Change in position (or displacement) is equal to uniform velocity x change in time [ i.e s = vt] 2.3 Uniform Accelerated Rectilinear motion: If constant acceleration be ‘a’ then, dv/dt = a = constant ⇒ dv = adt

Integrating both sides with limit v0 to v and 0 to t . We get,

∫

v

v0

t

dv = a ∫ dt ⇒ v − v0 = at 0

v = v 0 + at ........(1) Again for position , we have v = dx/dt …..(2) from 1 and 2. dx = (vo +at) dt , Integrating both sides over the limits

∫

x

x0

dx = ∫ (v 0 + dt )at ⇒ x − x0 = v0 t + t

0

x = x0 + v0 t + Also, a = v

1 2 at 2

1 2 at 2

dv dx

Or, vdv = adx Integrating both sides under limits

∫

v

v0

x

vdv = a ∫ dx x0

1 2 (v − v02 ) = a(x − x0 ) 2 v 2 = v 02 + 2a( x − x 0 )

xo = Initial position x = Final position v0 = Initial velocity. v = Final velocity 0 = Initial time t = Final time

2.4 Motion of several particles: Two or more particles moving in straight line. Equations of motion may be written for each particles as:


(a) Relative motion of two particles (b) Dependent motion. (a) Relative motion of two particles: Consider two particles A and B moving along the same straight line as follows: A

B

xA

[ xB/A =xB - xA ]

xB/A xB

Position co-ordinates of A = xA Position co-ordinates of B = xB Relative position co-ordinate of B w.r.t A = xB-xA = xB/A

∴ xB = xA+xA/B ………(1) [ + xB/A → B is right to A in position ] [ -xB/A → B is left to B in position ]

Differentiating equation (1) w.r.t time we get, VB = VA+VB/A ------(2)

[ VA , VB → absolute velocity of pt. A and B] [VB/A → velocity of B observed from pt. A]

Differentiating equation (2) w.r.t time we get, aB = aA +aB/A ………… (3) [ aB , aA → absolute velocity of pt A and B and aB/A acceleration of pt B w.r.t pt A. ] (b) Dependent Motion: When the position of a particles will dependent upon the position of another or several other particles, the motions are said to be dependent. eg :- pulley systems, Gear system etc. Pulley system as Dependent motion: Consider the pulleys system in which the position of Block ‘B’ depends upon the position of Block A as follows: From Figure: IH = Constant G

H

JB = Constant Arc CD = Constant Arc EF = Constant

xA

D

C

D’ xB

AC+DE+FG = Constant Now, XA = AC +IH = AC + Constant ………. (i) XB = FG +JB = FG + Constant ………….(ii)

A

Multiplying equation (ii) by 2 and Adding to (ii) we get, XA +2xB = AC+2FG+ Constant = AC + FG+FG +Constant = AC +FG+DE+D’G+ constant [ Since FG = DE+D’G] =AC+FG+DE+ Constant [ Since D’G = Constant ]

E

J

F

B


Downloaded from www.jayaram.com.np ∴ xA+2xB = Constant ……..(iv) [ AC+DE+FG = Constant ]

If A block ‘A’ is given ∆xA motion it will produce xB = (-∆xA/2) as the motion of Block B.

Differentiating equation (iv) w.r.t time , we get, VA+2VB = 0 , ………(v) Or, VB = -(VA/2) Similarly differentiating equation (v) w.r.t time we get, AA +2aB = 0 ………(vi) Or, aB = - (aA/2) [ Negative sign denotes opposite in direction ] * In this case displacement, velocity and acceleration of one body gives the displacement, velocity and acceleration of other body. This arrangement is called 1-degree freedom . # Derive the equation of motion of the given pulley system. Solution: 2xA +2xB +xc = Constant ……….(1) 2vA +2vB +vc = 0 …………..(2) 2aA +2aB +ac = 0 ………….(3)

xA

I

G

D

E

N

M

o

xC xB

F

C J

K

L

A

B

2.6 Graphical solution of Rectilinear motion problems: * Graphical solution are very helpful to simply and solve the problems of Dynamics. * Using the motion graphs (i.e x –t , v-t and a-t ) the missing value at any point can be obtained . * If any on equation of motion is known all the three graphs can be obtained as follows. If equation of displacement x = f(t) ……..(i) is known, Then , V = dx/dt …….(ii) a = dv/dt …….(ii) i.e velocity is slope of x-t curve and acceleration is slope of v-t curve. a

V

x x= f(t)

Slo e = dx/dt = a1

x1

V1

Slo e = dx/dt = v 1

t1

t

a= f(t)

V= f(t)

t1

a1

t

t1

t


If x-t curve is given, then computing slope at each point of x-t curve corresponding v-t curve can be generated and computing slope at each point of v-t curve a-t curve can be generated. Again, From equation (ii)

dx = vdt

∫

x2

x1

t2

t2

t1

t1

dx = ∫ vdt ⇒ x 2 − x1 = ∫ vdt ...........(iv)

And from equation (iii) , dv = adt

∫

v2

v1

t2

t2

t1

t1

dv = ∫ adt ⇒ v 2 − v1 = ∫ adt •

...........(v)

This means change in position in given by the area under curve v-t and change in velocity is given by area under the curve a-dt. a

x

V

a1

t2

V2

x2 vdt = x2 - x1 x 1 t1

V1 t1

t

t2

t1

t

t1

t2

t

Tutorial Examples:

) r t3 ˆ 2 ˆ 1) The motion of a particles is defined by the position vector r = 6ti + 4t j + k where r in meter and t in 4 second. At the instant when t = 3 sec, find the unit position vector, velocity and acceleration. Solution:

r

)

We have , r = 6ti + 4t 2 ˆj +

t3 ˆ k 4

r

At time t = 3 sec. r = 18iˆ + 36 ˆj +

r r= r =

(18)2 + (36)2 + ⎛⎜ 27 ⎞⎟ ⎝ 4 ⎠

27 ˆ k 4

2

= 40.81m

Now unit position vector at t = 3 sec.

27 ˆ r 18iˆ + 36 ˆj + k r 4 rˆ = r = r 40.81 ∴ rˆ = 0.44iˆ + 0.88 ˆj + 0.165kˆ

(

)

Ans

Again,


r r dr d ⎛ ˆ t3 v= = ⎜⎜ 6ti + 4t 2 ˆj + 4 dt dt ⎝ r 3 v = 6iˆ + 8tˆj + t 2 kˆ 4

Downloaded from www.jayaram.com.np ⎞ kˆ ⎟⎟ ⎠

At time t = 3 sec.

r ⎛ 24 ˆ ⎞ v = ⎜ 6iˆ + 24 ˆj + k⎟ 4 ⎠ ⎝ 1

2 r ⎡ 2 ⎛ 24 ⎞ ⎤ 2 2 Velocity (v) = v = ⎢6 + 24 + ⎜ ⎟ ⎥ = 25.64 m / s ⎝ 7 ⎠ ⎦⎥ ⎣⎢

V = 25.64 m/s

r r dv d ⎛ ˆ 3 ⎞ Again, acceleration (a ) = = ⎜ 6i + 8tˆj + t 2 k ⎟ 4 dt dt ⎝ ⎠ r 3 a = 8 ˆj + tkˆ 2 r At t = 3 sec , a = 8 ˆj + 4.5kˆ

[

r 2 Acceleration, (a) = a = 8 2 + (4.5)

]

1 2

= 9.18 m / s 2

Ans

2) A ball is thrown vertically upward with a velocity of 9.15m/s. After 1s another ball is thrown with the same velocity. Find the height at which the two ball pass each other? Solution: Let the initial velocity of both balls V01 = v02 = vo = 9.15 m/s h be the height at which two balls pass each other t1 be the time elapsed by the first ball before passing second and t2 be the time elapsed by second. From the given condition: t1 – t2 = 1 ……..(i) for 1st ball , n = v0t1 – ½ 9t12 ……..(ii) For 2nd ball n = v0t2 – ½ 9t22 ………(iii) Substituting equation (iii) form equation (ii) , we get,

1 2 2 9(t1 − t 2 ) = v0 (t1 − t 2 ) 2 1 9(t1 + t 2 )(t 2 − t1 ) = v 0 (t1 − t 2 ) 2 2v 2 × 9.15 ∴ (t1 + t 2 ) = 0 = = 1.865 9 9.81 (t1 + t 2 ) = 1.865 ...........(iv)

Adding equation (i) and (ii) , we get t1 = 1.43 sec and t2 = 0.43 sec

1 ∴ h = v 0 t1 − 9t12 = 3.05m 2 h = 3.05 m Hence , two balls pass each other at 3.05m above the ground. 3) In the following pulley system, Block 2 has velocity 2m/s upward and its acceleration is 3m/s2 downward while block 3 has velocity and acceleration 2m/s up ward and 4m/s2 downward respectively. Find the velocity and acceleration of block 1.


Solution: Given, V2 = 2m/s (↑) a2 = 3m/s2 (↓) v3 = 2m/s (↑) a3 = 4m/s2 (↓) E

Here , AB +CF = constant GI +HJ = Constant DE = Constant Portion of rope around the pulley is also constant. Now, X1 = AB + constant ……..(i) X2 = GI +CF+ constant …….(ii) X3 = HJ+CF+ Constant ……(iii) Multiplying equation (i) by (2) and adding (i), (ii) and (iii) , we get 2x1 +x2+x3 = 2AB+CF+CF+GI+HJ+Const. 2x1+x2+x3 = const. ………..(iv)

x1

C

B

x2 A 1 x3 G I 2

Differentiating equation (iv) w.r.t time. 2v1+v2+v3 = 0 ………(v) 2a1+a2+a3 = 0 ……..(vi) From equation ‘v’

v1 = −

v 2 − v3 − 2 − 2 = = −2 m / s 2 2

Therefore, velocity of block 1 (v1) = 2 m/s (↓) From equation (vi)

a1 =

H

F

− a 2 − a3 + 3 + 4 = = 3 .5 m / s 2 2 2

Therefore, acceleration of block 1 (a1) = 3.5 m/s2 (↑)


J 3


Chapter – 3 Curvilinear Motion of Particles 3.1

Position vector, Velocity and Acceleration: When a particle moves along a curve path other than a straight line, it is said to be in curvilinear motion. Vector analysis is used to analyze the change in position and direction of motion of particles.  Y

Y p ∆r

r

p

∆s p

r

r O

(a)

Y

∆r ∆t

∆r

∆r≈∆ s p

p s

r X

O

r o

X

(b)

X

(c)

Let, at time ‘t’ the position vector of particle be r and at another time (t + ∆t ) the particle takes a new position p' and its position be r !. Then ∆ r represents the change in directoin as well as magnitude of the position vector r . (fig. a) The average velocity of the particles at time interval ∆t =

∆r ∆t

( in magnitude and direction of ∆ r ) ∆r d r = ∆t →0 ∆t dt

∴Instantaneous Velocity, v = lim

As ∆ r and ∆t becomes shorter, P & P ′ gets closer and v is tangent to the path of the particle. (fig c) And, As ∆t decreases, length of PP ′ ( ∆ r ) equals to length of arc ∆s (fig b) ∆s ds PP ′ = lim ∴ν = lim = ∆t → 0 ∆t ∆t →0 ∆t dt Y Change in position ( ∆ r ) can be resolved into two components, i One parallel to x-axis and ( PP ′′ ) ii Other parallel to y-axis ( P ′′P ′ ) ∴ ∆ r = PP ′′ + P ′′P ′ ∆r PP ′′ P ′′P ′ or , lim = lim + lim ∆t → 0 ∆t ∆t →0 ∆t ∆t →0 ∆t

V α

p ∆r

∆Y

Y P

vy

v

O

α

x

p

∆x

vx P v -By Er. Biraj Singh Thapa (Lecturer, Eastern College of Engineering, Biratnagar)/ -10 O

X

y X

∆x ˆ ∆y ˆ i + lim j ∆t →0 ∆t ∆t →0 ∆t

or ,ν = lim

dx ˆ dy ˆ i+ j dt dt

or ,ν =

or ,ν = v x iˆ + v y ˆj or ,ν = xiˆ + yˆj

Then, V = v + v [Magnitude of Velocity] 2 x

tan α =

2 y

Positive Value of v x → Right Direction Positive Value of v y → Upward Direction

vy

∴α = tan

vx v −1 y vx

vx =

dx dy = x;vy = =y dt dt

[Direction of Velocity]

Similarly, acceleration by curvilinear motion can be computed as: Y

Y v

vy

∆s v

v

(a)

∆v

ay

Q

q

v Q x

X

O

(b)

X

If v and v ′ be the velocities at time ‘t’ 4 (t + ∆t ) i.e. tangents at acceleration of the particle over the timer interval is given by ∆t ( a ) =

β ax

q

v

r O

q

Q

∆r

r

Y

O

(c)

X

P and P ′ , then the average

∆v ∆t

∆v d v = ∆t →0 ∆t dt Again, ∆v can be resolved into QQ ′′ & Q ′′Q ′ parallel to x & y-axes respectively. Then, or , a = lim

v = QQ ′′ + Q ′′Q ′ ∆v QQ ′′ Q ′′Q ′ = lim + lim ∆ t → 0 ∆ t → 0 ∆t ∆t ∆t ∆ v ∆v y ˆj or , a = lim x iˆ + lim ∆t → 0 ∆t ∆t →0 ∆t dv y dv ˆj or x iˆ + dt dt or , a x iˆ + a y ˆj or , lim

∆t →0

or , a = xiˆ + yˆj

or , a = (a x ) 2 + (a y ) 2 ⇒ acceleration in magnitude

ax = ay =


dv x dt dv y dt

Downloaded from www.jayaram.com.np tan β = 3.2

ay

ax

⇒ β = tan −1 (a y a x ) ⇒ in direction

Derivatives of a vector function: Y

Y dp du ∆p

O

O

X

(a)

Ζ

X

(b)

p= f(u)= 2u2 +4 u +3

Ζ

Let, P (u ) be a vector function of scalar variable u. If value of ‘u’ is varied, ‘ P ’ will trace a curve in space. Considering change of vector P corresponding to the values u 4 ( u + ∆u ) as shown in figure(a). Then ∆ p = p (u + ∆u ) − p (u ) i.e.

⎧⎪ p(u + ∆u ) − p (u ) ⎫⎪ ∆p dp = lim = lim ⎨ ⎬ ∆u du ∆u →0 ∆u ∆u →0 ⎪⎩ ⎪⎭

- (1)

As ∆u → 0, ∆ p becomes tangent to the curve. Thus

dp is tangent to the curve as shown in du

figure(b). Again, Considering the sum of two vector functions p (u ) & Q(u ) of the same scalar variable u. Then the derivative of the vector ( P + Q) is given by: ⎡ ∆{P) ∆{Q) ⎤ ∆( P + Q) d ∆P ∆Q = lim ⎢ + ( P + Q) = lim = lim + lim ⎥ ∆u → 0 ∆u → 0 ∆u ∆u ⎥⎦ ∆u →0 ∆u ∆u →0 ∆u du ⎣⎢ ∆u ∴

d d P dQ ( P + Q) = + du du du

-

(2)

Again, product of scalar function f(u) and pf a vector function P (u ) of the same scalar variable u. Then, derivative of f P is given by: df . p du

= lim

∆u → 0

(f

(

)

+ ∆f ) P + ∆P − f P ∆u

⎡ ∆f ∆P ⎤ = lim ⎢ P+ f ⎥ ∆u → 0 ∆u ∆u ⎥⎦ ⎣⎢ ∴

df P df dP = P+ f du du du

-

(3)


Similarly, scalar product and vector product of two vector functions P (u ) and Q(u ) may be obtained as: d dP dQ ( P.Q) = .Q + P. du du du

- (4) [Scalar Product]

d dP dQ ( P × Q) = ×Q + P× du du du

- (5) [Vector Product]

Again, P = Px iˆ + Py ˆj + Pz kˆ

-

(6)

where, Px , Py & Pz are the rectangular scalar components of vector P & iˆ, ˆj , kˆ are the unit vector. ∴

d P dPx ˆ dPy ˆ dPy ˆ = ×i + × j+ ×k du du du du

-

∴

d P dPx ˆ dPy ˆ dPy ˆ = ×i + × j+ ×k dt dt dt dt

-

(7) [where, P = f (u )]

And,

3.3

(8) [where, P = f (t )]

Rectangular Components of Velocity and Acceleration: When, the position of a particle is defined by at any instant by its rectangular co-ordinates x, y,

z as: r = x&iˆ + y&ˆj + z&kˆ -

(i)

Then, differentiating both sides w.r.t. time, we get,

ν= where,

dr = &x&ˆi + &y&ˆj + &z&kˆ dt

x& = v x , y& = v y & z& = v z &x& = a x , &y& = a y & &z& = a z

So,

∴ V = (v x2 + v y2 + v z2 )

V = v x iˆ + v y ˆj + v z kˆ a = a x iˆ + a y ˆj + a z kˆ

∴ a = ( a x2 + a y2 + a z2 )

Y vy

vx

vz v o Z

X



When the motion in each axis can be represented independent with each other then the use of rectangular components to describe the position, velocity and acceleration of a particle is effective i.e. motion in each axis can be considered separately. For e.g. for projectile motion, neglecting air resistance, the components of acceleration are: a x = 0 and a y = − g or , v x = 0 and v y = − g

ay

or , dv x = 0 and dy = − gdt On Integrating both sides under the limits,

∫

vx

vy

dv x = 0

and

∫

vx − vxo = 0

and

v y − v yo = −gt

v xo

v x = v xo

v yo

dy = − g ∫ dt 0

v y = v yo = − gt − (v)

and

or , x& = v xo

ax

t

y& = v yo − gt

and

dx dy = v xo and = v yo − gt dt dt or , dx = v xo dt and dy = v yo dt − gtdt Integrating both sides under limits considering motion starts from origin by co-ordinates i.e. x at t=0 ; x=0 ; y=0 and at t=t0, x=x0 and y=y0 or ,

∫

x0

0

t

dx = v x0 ∫ dt

∫

and

0

yo

0

t

t

0

0

dy = v y0 ∫ dt − g ∫ tdt

1 2 gt − (vi) 2 Thus motion under projectile can be represented by 2-independent rectilinear motion.

x 0 = v x0 t

and

y 0 = v y0 t −

Y uyo Xo

Vx O

Vxo Vxo

V

uy

yo

X

Problems: 1. A bullet is fired upward at an angle of 30° to the horizontal from point P on a hill and it strkies a target which is 80m lower than the level of projection as shown in figure. The initial velocity of the bullet is 100m/s. Calculate: a. The maximum height to which the bullet will rise above the horizontal. b. The actual velocity with which it will strike the target Vo = 100m/s c. The total time required for the flight of the bullet. A nmax Solution: P V0 = 100m / s B V x0 = V0 cos 30 = 86.60m / s V y0 = V0 sin 30 = 50m / s

Hill 80m


V02 sin 2 30 = 127.42 hmax = a) 2g ∴hmax = 127.42m ( Ans ) b) Let, V1 y = vertical component of velocity at highest point A = 0 V2 y = vertical component of velocity striking target H = vertical distance between point A and target = 127.42+80 = 207.42m Then, V22y − V12y = 2 gH

[V1 y = 0]

or , ⇒ V2 y = 63.79m / s V2 x = V1x = V x0 = 86.60m / s

[V x0 = cons tan t ]

∴V2 = V22x + V22y = 107.55m / s (Ans) &

c)

θ = tan −1

V zy V xy

h = (V y ) 0 t 2 −

Vo= A A

= 36.37°

nmax P

B 80m

1 2 gt 2 2

1 or , − 80 = 50 × t 2 − × 9.8 × t 22 2 2 or , 4.905t 2 − 50t 2 − 80 = 0

Total time of flight is the sum of time to reach B from A & to C from B T=t1+t2 2V sin α V 2 sin α t1 = 0 & PB=Range= 0 g g

Solving , xv = t 2 = 11.60 sec ( Ans)

2) The motion of a vibrating particle is defined by the equation x=100sin πt and y=25cos2 πt , where x & y are expanded in mm & t in sec. a) Determine the velocity and acceleration when t=15 b) Show that the path of the particle is parabolic. Solution: a) We have, x=100sin πt ⇒ V x = x& = 100π cos πt a x = &x& = −100π 2 sin πt Again, y=25cos2 πt ⇒ V y = y& = −50π sin 2πt a y = &y& = −100π 2 cos 2πt

Then, for t=2sec, V x = [100π cos π × 1]:V y = [−50π 2 sin π × 1] V = V = V x2 + V y2

= (100π cos πt ) 2 + (−50 sin 2π ) 2 V = 100π mm / s

α = tan −1 (

Vy Vx

)=0


Downloaded from www.jayaram.com.np And, For t=1sec a x = [(−100π 2 sin π ]: a y = −100π 2 cos 2π × 1 a = a = a x2 + a y2 = 100π 2 mm / s 2

β = tan −1 (

ay ax

) = 270°

b) Since, x=100sin πt

∴

x = sin πt 100

x 2 ∴( ) = sin 2 πt 100

− (i )

Again, y = 25 cos 2πt y = 2 cos 2 πt − 1 25 ⎞ ⎛ y or , ⎜ + 1⎟ = 2 cos 2 πt ⎝ 25 ⎠ y + 25 = cos 2 πt − (ii ) ∴ 50 or ,

Adding equations (i) and (ii), we get ; x2 y + 25 + = sin 2 πt + cos 2 πt = 1 10000 50 2 or , x + 200 y + 5000 = 10000 or , x 2 + 200 y + 5000, which is the equation of parabola = [ax 2 + by = c] 3) The motion of a particle is given by the relation Vx=2cost & Vy=sint. It is known that initially both x & y co-ordinates are zero. Determine a) Total acceleration at the instant of 25 b) Equation of the parabola a) Here, Vx=2 cost & Vy =sint At t=2sec dv Then, ax= x = −2 sin t ax=-2sinz=-1.82 dt ay=cos2=-0.42 dv y and, a y = = cos t α = a x2 + a y2 = 1.865m / s 2 dt ay ∴a = a x iˆ + a y ˆj β = tan −1 = 193° ax a = −2 sin t iˆ + cos t ˆj or , y = − cos t + 1 ⇒ ( y − 1) = cos 2 t 2

− (ii )

For

t

=

Adding (i ) and (ii ), x 2 4 + ( y − 1) 2 = 1 -By Er. Biraj Singh Thapa (Lecturer, Eastern College of Engineering, Biratnagar)/ -16 or , x 2 + 4 y 2 − 8 y = 0, which is the required equation of parabola

2 sec

a = −2 sin 2iˆ + cos 2iˆ

[where, 2 is in radian]

Q a = −1.82iˆ − 0.42 ˆj ∴α = Total acceleration

α = α = (−1.82) 2 + (−0.42) 2 = 1.865m / s 2 ay

⎛ − 0.42 ⎞ = tan −1 ⎜ ⎟ = 193° ( Ans) ax ⎝ − 1.82 ⎠ x t dx = 2 cos t ⇒ ∫ dx = 2 ∫ cos dt b) V x = 2 cos t ⇒ 0 0 dt or , x = 2 sin t

β = tan −1

x2 or , = sin 2 t − (i ) 4 y t dy Again, V y = = sin t ⇒ ∫ dy = ∫ sin tdt 0 0 dx

3.4

Motion Relative to a Frame in Translation:

Let A and B be the particles moving is a same plane with rA & rB be their position with respect to XY axis. Considering New axes (X'-Y') centered at ‘A’ and parallel to original axes X-Y, the motion of particle ‘B’ can be defined with respect to motion of particle ‘A’ such that: From vector triangle OAB where, rB = rA + r B / A − (i ) • XA,YA & XB & YB are co-ordinates of A & B and similarly , w.r.t. XY axes XB = X A + X B/ A • XB/A, YB/A are co-ordinates of ‘B’ .r.t. X'-Y' axis - (ii) YB = YA + Y B / A Differentiating equ(i) w.r.t. time, we get: V B = V A + VB − (iii ) r → Equation of motion in both x and y axes In scalar form: X& B = X& A + X& B / A X → Equation of motion in x - axis - (iv) Y&B = Y&B + Y&B / A Y → Equation of motion in y - axis V( B )x

OR = V( A) x + V( B / A) x

V( B )y = V( A) y + V( B / A) y

where, X& A , Y&A are X & Y components of VA X& B , Y&B are X & Y componens of VB X& B / A Y&B / A are X & Y components of V B / A

Again, differentiating (iii) with respect to time, we get: aB = a A + a B / A − (v ) In scalar, Downloaded from www.jayaram.com.np/ -17

aB =

(aBx )2 + (a By )2

β = tan −1

a By a Bx

X&& B = X&& A + X&& B / A Y&& = Y&& + Y&& B

3.4

A

Downloaded from www.jayaram.com.np a ( B ) x = a ( A) x + a ( B / A) x - (vi) a ( B ) y = a ( A) y + a ( B / A) y

OR

B/ A

Tangential and Normal Components: The velocity of particle is vector tangent to the path of particle. But acceleration may not be tangent in curvilinear motion. The acceleration vector may be resolved into two components perpendicular with each other in directions i First component along the tangent of path of particle (at) ii Second component along the normal of path of particle (an) Let eˆt and eˆn be the unit vectors directed along the tangent and normal of the path

respectively. Then, in curvilinear motion, eˆt and eˆn would change the direction as particle moves from one point to another. y

y

et=dr ds

r/ s p

p’ r

r

o

p x o

(a)

From fig (a)

[

∆r = r ' − r & lim ∆r = ∆s

Then, eˆt = lim

∆s →0

∴eˆt =

∆s →0

x

]

∆r dr = ∆s ds

dr ds

− (i )

Again, eˆt = lim

∆s →0

Therefore, eˆt =

(b)

∆r ∆s

=

∆s =1 ∆s

[lim ∆r = ∆s] ∆s →0

dr is the unit vector along the tangetn to path. ds

Let, ρ be the radius of curvature of the path at the point P and eˆt & eˆt ' be the tangent unit vectors at P and P'. ∆eˆt be the change in unit vector while the particle moves from P to P'.


y

ex y p’

e’ r o

et e’x

en

a =at et + an en

a t =dVet dt

ex

p

p x

o

(a)

(b) o

Now, from fig, ∆s = PP' = ρ∆θ ∆eˆt = eˆt ' − eˆt ≡ ∆θ eˆt = ∆θ eˆn

x

(c)

[As ∆s → 0

eˆ t = eˆn → 1 in magnitude]

∆eˆt deˆ ⎤ ⎡ ⇒ ⎢eˆn = t ⎥ − (ii ) ∆s →0 ∆θ dθ ⎦ ⎣ Similarly, deˆt ⎡ dθ 1 ⎤ = and = eˆn ⎥ − (iii) ⎢ ds ρ dθ ⎣ ⎦ Also, ∆s ρ∆θ ds dθ = lim = lim =ρ V = ∆ t → 0 ∆ t → 0 ∆t ∆t dt dt dθ ⎡ ⎤ ⎢⎣∴ V = ρ dt = ρθ ⎥⎦ − (iv)

∴ eˆn = lim

d r d r ds = . = eˆt v = veˆt dt ds dt ∴ V = Veˆt − (v ) And, V =

[

]

deˆ deˆ dθ ds dv d dv dv eˆt + v t = eˆt + v t . . = (Veˆt ) = dt dt dt dt dt dθ ds dt 1 or , a = V&eˆt + V (eˆn )( ) a=

ρ

⎡ V2 ⎤ or , ⎢a = V&eˆt + eˆn ρ ⎥⎦ ⎣

− (vi)

which can be represented as in fig(c). a = at eˆt + a n eˆn , where dv = v& dt V2 = ρθ& 2

at = Tangential component of acceleration = a n = Normal component of acceleration =

ρ

Notes: For increasing velocity at will be in the direction of velocity and fx decreasing velocity at will be in opposite to the direction of velocity. If the speed is constant at=0 but an≠0. [an=0 fx Recti ρ = α ] 3.6

an is always directed towards the centre of curvature For higher velocity and smaller radius higher is an. Radial and Transverse Components:


Downloaded from www.jayaram.com.np 9 For the motion described by polar co-ordinates. 9 Position of particles P is defined by the co-ordinates r & θ , where r is the length and θ is the angle in radians. 9 The unit vectors in radial and transverse direction are denoted by eˆr and eˆθ respectively along radius and 90° clockwise to the radius in direction. fig (a) y

y

eo

e

er e eo

r’ P(r, )

r = rer

er r x

x

(b)

(a)

As the particle moves from P to P '. The unit vectors er , & eθ

change to eˆ' r & eˆ'θ by

∆er and ∆eθ respectively. Here, deˆ deˆ dθ deˆ ⎡ ⎤ = eˆθ θ& - - - - - (i) ⎢direction of r is in direction of eˆθ ⎥ eˆ&r = r = r . dt dθ dt dθ ⎣ ⎦ deˆ deˆ dθ deˆ ⎤ ⎡ e&ˆθ = θ = θ . = −eˆrθ& - - - - - (ii) ⎢direction of θ is in direction of - eˆ r ⎥ dθ dθ dt dθ ⎦ ⎣ Now, r = reˆr deˆ dr d dr = (reˆr ) = eˆr + r r = r&eˆr + re&ˆr dt dt dt dt & ∴ V = r&eˆr + rθ&eˆθ - (iii) eˆ r = eˆθ θ& which can expressed as V = Vr eˆr + Vθ eˆθ , where Vr = Radial component of velocity = r& And, Vθ = Transverse component of velocity = r θ& Similarly, Then, V =

[

]

[

(

]

)

dv d = r&eˆr + rθ&eθ dt dt = &r&eˆr + r&eˆ&r + r&θ&eˆθ + rθ&&eˆθ + rθ&eˆ&θ = &r&eˆr + r&θ&eˆθ + r&ϑ&eˆθ + rθ&&eˆθ − rθ& 2 eˆr ∴e&r = eˆθ θ& & eˆ&θ = −eˆrθ& ∴ a = &r& − rθ& 2 eˆr + rθ&& + 2r&θ& eˆθ − (v ) which can be represented as, a = a r eˆr + aθ eˆθ where, a r = Radial component of acceleration = &r& − rθ& 2 and, aθ = Transverse component of acceleration = rθ&& + 2r&θ& a=

[ (

)

(

)

]

[

(

(

)

]

)


In case of a particle moving along a circular path with its centre at the origin O, we have r=constant &r& = 0 or, r& = 0 & Then, v = rθ& eˆθ − (vi) a = − rθ& 2 eˆ + rθ&&eˆ − (vii) θ

r

Problems:

1) The motion of a particle is defined by the position vector, r = 3t 2 iˆ + 4t 3 ˆj + 5t 4 kˆ, where r is in m and t is in sec. At instant when t=4 sec, find the normal and tangential component of acceleration and the radius of curvature. Solution, we have r = 3t 2 iˆ + 4t 3 ˆj + 5t 4 kˆ ∴V =

dr = 6tiˆ + 12t 2 ˆj + 20t 3 kˆ dt

&a=

dv = 6iˆ + 24tˆj + 60t 2 kˆ dt

Again,

(

V = V = 36t 2 + 144t 4 + 400t 6

[

(

a = a = 36 + (24t ) + 60t 2

)

1 2

− (i )

)]

1 2 2 2

− (ii )

Now, At t = 4sec V=1294.54m/s [putting t=4 in equ-(i)] a=964.81m/s2 [putting t=4 in equ-(ii)] Again, Tangential component of acceleration,

(

)

1 dv d 36t 2 + 144t 4 + 400t 6 2 = dt dt 1 1 = . × 72t + 576t 3 + 2400t 5 1 2 36t 2 + 144t 4 + 400t 6 2 At time t=4 sec, at=963.56m/s2 (Ans) Now,

at =

(

a n = a 2 − at2 =

)

(

)

(964.81)2 − (963.56)2

∴a n = 49.1m / s 2 ( Ans) Again,

V 2 (1294.54) ρ= = = 34131.03m ( Ans ) an 49.1 2

2. A car is traveling on a curved section of the road of radius 915m at the speed of 50km/hr. Brakes are suddenly applied causing the car to slow down to the 32 km/hr after 6 sec. Calculate the acceleration of the car immediately after the brake have been applied. Downloaded from www.jayaram.com.np/ -21

Downloaded from www.jayaram.com.np Solution: Given, ρ = 915m V0 = 50km/hr = 13.88 m/sec V1 = 32km/hr = 8.88 m/sec

At the instant when the brake is applied, 2 V 2 (13.88) an = = = 0.210m / s 2 ρ 915 V − V0 = −0.833m / s 2 at = 1 ∆t a = a n eˆn + at eˆt a = 0.210eˆn − 0.833eˆt a= a =

(0.21)2 + (− 0.83)2

β = tan −1

an ⎛ 0.21 ⎞ = tan −1 ⎜ − ⎟ = 14.2° ( Ans ) at ⎝ 0.83 ⎠

= 0.856m / s 2 ( Ans )

3. The plane curvilinear motion of the particle is defined in polar co-ordinates by r=t3/4+3t and θ =0.5t2 where r is in m, θ is in radian and t is in second. At the instant when t=4 sec, determine the magnitude of velocity, acceleration and radius of curvature of the path. Solution: We have, t3 3t 2 + 3t ⇒ r& = + 3 ⇒ &r& = 3t / 2 4 4 Again, θ = 0.5t 2 ⇒ θ& = t ⇒ θ&& = 1 Now, we have r=

⎛ 3t 2 ⎞ ⎛ t3 ⎞ + 3 ⎟⎟eˆr + ⎜⎜ + 3t ⎟⎟teˆθ v = r&eˆr + rθ&eˆθ = ⎜⎜ ⎝ 4 ⎠ ⎝4 ⎠

− (i )

Again, at t = 4 sec v = 15eˆr + 112eˆθ ∴v = v = Again,

(

(15)2 + (112)2

= 133m / s 2

( Ans)

)

a = &r& − rθ& 2 eˆr + (rθ&& + 2r&θ&)eˆθ ⎧ 3t ⎛ t 3 ⎧⎛ t 3 ⎞ ⎫ ⎞ ⎛ 3t 2 ⎞⎫ = ⎨ − ⎜⎜ + 3t ⎟⎟t 2 ⎬eˆr + ⎨⎜⎜ + 3t ⎟⎟ × 1 + 2⎜⎜ + 3 ⎟⎟t ⎬eˆθ ⎠ ⎭ ⎠ ⎝ 4 ⎠⎭ ⎩2 ⎝ 4 ⎩⎝ 4

At t = 4 sec, a = −442eˆ + 148eˆ r θ 1 a = a = ⎡(− 442 )2 + (148)2 ⎤ 2 = 466.12m / s 2 ( Ans ) ⎥⎦ ⎢⎣

Again, from equ(i) [for ρ ]


1 2 2⎤ ⎡ 2 ⎞ ⎥2 ⎛ t4 ⎞ ⎢⎛⎜ 3t 2 ⎟ ⎜ ⎟ + 3t v= v =⎢ +3 + ⎟ ⎥ ⎜ 4 ⎟ ⎜ 4 ⎠ ⎥⎦ ⎝ ⎠ ⎢⎣⎝ 1 ⎤2 dv d ⎡ t 6 33t 4 27t 2 Qa = = ⎢ + + + 9⎥ t dt dt ⎢ 16 16 2 ⎥⎦ ⎣ Qa =

1 × 2

⎞ ⎛ 3t 5 33t 3 ×⎜ + + 27t ⎟ ⎟ 1 ⎜ 8 4 ⎠ ⎝ ⎞2 ⎛ t 6 33t 4 27t 2 ⎜ + + + 9⎟ ⎟ ⎜ 16 16 2 ⎠ ⎝ 1

At t = 4 sec,

at = 16.055m / s 2

[

Q a n = a − (at ) 2

] = [(466.12)

1 2 2

2

− (16.005)

]

1 2 2

∴ a n = 465.84m / s 2 ∴ρ =

(113) = 27.41m v2 = a n 465.84 2

Hence, Radius of curvature = 27.41m (Ans)



Chapter – 4 KINETICS OF PARTICLES NEWTONS SECOND LAW 4.1 Newton’s Second Law of Motion: Newton has given his understanding of motion of particles and their causes and effects in 3 laws. The first and third law of motion deals with the bodies at rest or moving with uniform velocity i.e. without any acceleration. For the bodies under the motion with acceleration the analysis of motion and forces producing it is done by the application of Newton’s Second Law. Statement of Newton’s 2nd Law: “If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to the magnitude of the resultant and in the direction of this resultant force. a2 F2

a1 F1

a2 F2

If F1 , F2 , F3 , etc be the resultant forces of different magnitude and direction acting on the particle. Each time the particle moves in the direction of the force acting on it and if a1 , a 2 , a3 , etc be the magnitude of the accelerations produced by the resultant forces. Then,

F1 ∝ a1 , F1 a1

=

F2 a2

F2 ∝ a 2 , =

F3 a3

F3 ∝ a3

........

etc

= .......... = constant = mass of particle (m)

So, when a particle of mass ‘m’ is acted upon by a force F and acceleation a, they must satisfy the relation, F = ma

i.e.

− (i ) [ where direction of F & a are same]

(

Fxiˆ + Fy ˆj + Fz kˆ = m axiˆ + a y ˆj + az kˆ

)

which is Newton’s Second Law. When a particle is subjected simultaneously to several forces equation(i) is modified as: i.e. F iˆ + F ˆj + F kˆ = m a iˆ + a ˆj + a kˆ F = ma

∑(

∑

x

y

z

) (

x

y

z

)

∑ F = sum of resultant of all forces acting. ∑ F ,∑ F , ∑ F , a , a , a are x, y and z component of the forces and acceleration acting on

where,

x

y

z

x

y

z

the particle respectively. Notes: (i) When the resultant force is zero, the acceleration of the particle is zero. (ii) When V0=0 and

∑ F = 0 , Then particle would remain at rest.


∑ F = 0 , Then particle would move with constant velocity, V along the

(iii)When V0=V and

straight line. (iv) All the above cases defines the first law, hence the Newton’s 1st Law of Motion is a particular case of Newton’s 2nd Law of Motion. 4.2

Linear Momentum and Rate of Change [Impulse Momentum Theorem}:

From Newton’s 2nd Law, F = ma or, F = m

dv dt

d (mv) − (i ) dt Multiplying both sides by dt and integrating under the limits, we get:

∴F =

∫

t2

t1

v2

Fdt = ∫ dmv ⇒ I 1− 2 = mv 2 − mv1

− (ii )

v1

Improper Path Function t2

The term ∫ F dt is called the impulse ( I ) of the force during time interval (t2-t1) whereas mv is t1

the linear momentum vector of the particle. So, equation (ii) states that “ The impulse ( I ) over the time interval (t2-t1) equal the change in linear momentum of a particle during that interval.” [Impulse Momentum Theorem] The impulse of force is known even when the force itself may not be known. Again, from equatin(ii) mv 2 = mv1 + I 1− 2

− (iii )

( ) of the particle may be obtained by adding vectorically its initial

i.e. Final momentum mv 2

momentum mv1 and the impulse of the force F during the time interval considered. Or, showing in vector form. mV2

I1-2

I1-2

mV1 mV1

When several forces act on a particle, the impulse produced by each of the forces should be considered. i.e. mv1 + ∑ I 1 − 2 = mv 2 − (iv) where,

∑I

1− 2

=∫

t2

t1

(∑ F )dt = ∫ (F + F t2

t1

1

2

)

t2

t2

t1

t1

+ F3 ..... dt = ∫ F1 dt + ∫ F2 dt + .....


Downloaded from www.jayaram.com.np F2

a1

F

m

F2 =0

F1

m

m

F1

(ma)rev

4.3

System of Unit: Units of measurement should be consistent and one of the standards should be followed. Generally 2 standard units are taken: a) System de' International Unit (SI unit) b) U.S. Customary Units (used by American Engineers)

SI Units: SI stands for System de’ International. SI units are the world-wide standards for the measuring system. SI units are fundamental or derived.

Fundamental and Derived Units: Fundamental and Derived units are the SI units. Fundamental units are independent of any other measuring units and are the basic units for all other system whereas Derived units are the units which are expressed in terms of powers of one or more fundamental units. Fundamental Units Derived Units Length = metre (m) Velocity = L/T = m/s Mass = kilogram (kg) Acceleration = V/T = L/T2 = m/s2 Time = second (s) Force = ma = ML/T2 = kgm/s2 (N) SI units are the absolute system of units and results are independent upon the location of measurement. US Customary Units: This system is not absolute system of unit. They are gravitational system of units. Base Units length = foot(ft) force = pound (lb) time = second (s) Conversion from US Customary Units to SI Units: length : 1 ft = 0.3048 m force : 1 lb = 4.448 N mass : 1 slug = 14.59 kg : 1 pound = 0.4536 kg

4.4

Equations of Motion and Dynamic Equilibrium


Considering a particle mass ‘m’ acted upon by several forces. Then from second law, F = ma − (i ) ⇒ F = m a iˆ + a ˆj + a kˆ

∑

(

∑

x

y

)

z

Using rectangular components, the equation of motions are

∑F

x

= ma x , ∑ F y = ma y ,

∑F

z

= ma z

− (ii )

(i) and (ii) gives the equation of motion of particle under the force F or,

∑F

x

= m&x&, ∑ Fy = m&y&,

∑F

z

= m&z&

Integrating these equation as done in 3.3, the equation of motion can be obtained. Again, the equation(i) may be expressed as

∑ F − ma = 0 i.e., if we add vector − ma to the resultant force in opposite direction, the system comes under the equilibrium state. This force ( − ma ) opposite to the resultant force is called Inertial Force or Inertia Vector. This equilibrium state of a particle under the given forces and the inertia vector is said to be dynamic equilibrium. At the dynamic equilibrium, ∑ Fx = 0, ∑ Fy = 0 &

∑F

z

=0

Inertia vector measure the resistance that particles offer when we try to set them in motion or when we try to change the condition of their motion.

• Angular Momentum and Rate of Change (Angular Momentum Theorem) Statement “The rate of change of angular momentum of the particle about any point at any instant is equal

()

to the moment of the force F acting on that particle about the same point.” Let a particle of mass ‘m’ moving in the XY-plane and the linear momentum of the particle is equal to the vector mv.

Y mVy

mV

The moment about O of the vector mv. (linear momentum) is called angular momentum of the particle

mVy

at that instant and is denoted by H 0

y x X

o

Now, mvx and mvy are components of mv. in x & y direction. Then, from definition, (+ H0=x(mvy)-y(mvx)

[∴ H

o

= m( xv y − yv x )

]

− (i )

Differentiating equ(i) with respect to time, we d H& 0 = m (xv y − yv x ) dt

about O

mVy

Total momentum about a point is sum of moment about the point Varignon' s Theorem

Q x& = v x

&

y& = v y

v& x = a x

&

v& y = a y


get:

Downloaded from www.jayaram.com.np = m(x&v y + xv& y − y& v x − yv& x )

= m(xa y − ya x )

QH& = xma y − ymax H& = xF − yF y

x

∴H& = moment of Force about O ∴ H& = m0 − (ii )

[

]

Thus, the rate of change of angular momentum of the particle about any point to any instant is equal

()

to the moment of force F acting on that particle about the same point. i.5 (a)

Equation of Motion Rectilinear motion of particles: If a particle of mass ‘m’ is moving in a straight line under the action of coplanar forces

F1 , F2 , F3 , etc Then the motion of particle can be written as

∑ F = ma

− (i ), where ∑ F = F1 + F2 + F3

For Rectilinear motion, motion is only along the single coi.e. ax=a & ay=0 ∴ Equn-(i) may be written as

∑F ∑F

x

= ma x

y

=0

F1

ordinate,

F3

F2

m

ma p

- (ii)

m

These are the equation of motion for the particle moving in the straight line. (b)

Curvilinear motion of particles: i Rectangular components ii Tangential and Normal components iii Radial and Transverse Components

i. Rectangular components From Newton’s second law,

∑F

x

= ma x

;

∑F

y

= ma y

Fy

For Projectile motion, neglecting air resistance ∑ Fx = 0 ⇒ ∑ Fx = ma x = 0 ⇒ a x = 0

∑F

y

= ma y = − w = −mg

or , a y = − ∴a x = 0

Fx = 0

mg = −g m

(i) a y = −g These are the equations of motion.

W


ii. Tangential and Normal components: From Newton’s 2nd law, dv ∑ Ft = mat = m dt v2 = = F m a m ∑ n n

FT at = dV

Fn

dt

an = V2

ρ

m

∑F = ∑F +∑F t

x

These are the equations of motion. iii. Radial and Transverse components: From Newton’s 2nd law, F = ma = m(&r& − rθ& 2 )

∑ ∑ Fθ = maθ = m(rθ&& + 2r&θ&) ` ∑ F = ∑ F + ∑ Fθ r

F = ma

r

Fr = mar

a

r

m

These are the equations of motion. Note: In case of Dynamic Equilibrium all the components of forces are balanced by Inertial Vector or Inertia force. So, for dynamic equilibrium condition, the equation of motion becomes

∑ F = ma = 0 ∑ F = ma = 0 ∑ F = ma = 0 ∑ F = ma = 0 ∑ F = ma = 0 ∑ Fθ = maθ = 0 i.6

r

x

y

y

t

t

n

n

r

r

- (i) - (ii)

- (iii)

Motion due to Central Force-Conservation of Angular Momentum

When the force F acting on a particle P is directed towards or away from the fixed point O, the particle is said to be moving under a central force. The fixed point ‘O’ is called the center of force. As shown in the figure, particle P moves along the curve path. O = origin of co-ordinates Now, Fr = Radial component of force F Fθ = Transverse component of force F For central motion Fθ = 0 Q Fθ = rθ&& + 2r&θ& = 0

(

)

1 2 && r θ + 2rr&θ& = 0 r d 2& or , r θ = 0 ⇒ d r 2θ& = 0 dt

or ,

( )

( )


Downloaded from www.jayaram.com.np Integrating both sides we get r2 θ& =constant=h - (i) Now, if elementary section are swept in time ‘dt’ be dA 1 [θ = S / R ; S = 1 for dθ → 0] Q dA = r.rdθ 2 1 or , dA = r 2 dθ 2 dA 1 2 dθ 1 2 & [Dividing both sides by dt ] or , = r = r θ dt 2 dt 2 dA Here, = Rate of change of sweeping Area or Area Velocity ( A.V .) dt 1 1 Q A.V . = r 2θ& = h 2 2 ∴ h = 2 A.V . − (ii ) Thus, when a particle moves under the central force, the areal velocity is constant. This is also called Kepler’s Law Again, Angular momenum = momentum of linear momentum about the fixed point. H 0 = mvθ × r Now, vθ = rθ& ∴H 0 = mrθ& r ∴H 0 = mr 2θ& − (iii ) or , H 0 = mh

[Qr θ& = h] 2

or , H 0 = 2m( A.V .) ∴ H 0 = cons tan t

[Qh = 2 A.V .] [sin ce, A.V . = cons tan t & m = cons tan t ]

Hence, when a particle is moving under a central force, Angular momentum is always conserved. i.7 Newton’s Universal Law of Gravitation: Statement: Every particle in the universe attracts every other particle with a force, which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Y Mathematically,

Mm d2 GMm ∴F = d2 F∝

Po

dA

− (i )

⎡ where, m and M are masses in kg ⎤ ⎢ ⎥ d is the distance in m ⎣ ⎦

Also, G is the Universal Gravitation constant with its value 6.673×10-11Nm2/kg2 and Z F is the force of attraction between them.

rd F r

d

rP

F


X

For a body of mass ‘m’ located on or near the surface of earth, force exerted by the earth on a body equals to the weight of the body i.e. F = mg and d = R (radius of the earth). GMm F = mg = R2 GM − (ii ) Qg = 2 R where, g is the acceleration due to gravity with its standard value 9.81 m/s2 at the sea level. Since, earth is not perfectly spherical so the value of R is different and hence g varies according to the variation of altitude and latitude. i.8 Application in space mechanics: Earth satellite and space vehicles are subjected only to the gravitational pull of the earth after crossing the atmosphere. The gravitation force acts as a central force on them and hence their motions can be predicted as follows: From central force motion, r 2θ& = h − (i ) Trajectory of a particle under a central force:

Considering a particle P under central force F (i.e. directed towards center ‘O’) Then we have Radial component of force, F = ma = m &r& − rθ& 2 = − F − (i )

∑

r

r

(

)

And, Transverse component of force F = ma = m(rθ&& + 2r&θ&) = 0

∑

θ

− (ii )

θ

From equ(ii) since m≠0 ∴ rθ&& + 2r&θ& = 0

( )

1 d 2& r θ =0 r dt On integrating, r 2θ& = cons tan t = h or ,

∴θ& =

− (iii )

dθ h = 2 dt r

Again, r& =

dr dr dθ h dr d = . = 2. = −h dt dθ dt r dθ dθ

∴&r& =

⎛1⎞ ⎜ ⎟ ⎝r⎠

dr& dr& dθ h dr& h d = = 2 = 2. . dt dθ dt r d θ r dθ

− (iv)

⎡ d ⎛ 1 ⎞⎤ ⎢− h dθ ⎜ r ⎟⎥ ⎝ ⎠⎦ ⎣

h2 d 2 ⎛ 1 ⎞ ⎜ ⎟ − (v ) r 2 dθ 2 ⎝ r ⎠ Putting values of θ& & &r&in equ (i ) we get ∴&r& = −


Downloaded from www.jayaram.com.np ⎡ h d ⎛1⎞ h ⎤ m ⎢= 2 r = −F − ⎜ ⎟ 2 4 ⎥ r r d r θ ⎝ ⎠ ⎣ ⎦ 2

2

2

1 Putting v = , we get r ⎡ ⎤ d 2u − m ⎢h 2 u 2 + h 2u 2 ⎥ = −F 2 dθ ⎣ ⎦

d 2u F ∴ 2 +u = − (vi ) dθ mh 2 u 2 This is second order differential equation, which is the trajectory followd by the particle which is moving under a central force F . Note: i.

F is directed towards ‘O’

ii.

Magnitude of F is +ve if F is actually towards O (i.e. attractive force)

iii.

F should be ‘-ve’ if F is directed away from O.

The trajectory of a particle under a central force is d 2u F +u = − (ii ) 2 mh 2 u 2 dθ Again, GMm F= = GMmu 2 − (iii ) r2 where, M = mass of earth m = mass of the space vehicle r = distance from the centre of earth to the space vehicle, u =

1 r

From equ (ii) & (iii) d 2u GMmu 2 GM +u = = 2 = constant mh 2 u 2 h dθ 2

−(iv)

⎛ GM ⎞ This equn(iv) is second order differential equation with constant co-efficient ⎜ 2 ⎟. The general ⎝ h ⎠ solution of the differential equation is equal to the sum of the complementary i.e. U = Uc+Up Uc = complementy solution i.e. for tangient condition wherem, Up = particular solution i.e. for steady state condition Uc = A sinθ + B cosθ GM Up = 2 h Again, Uc = A sinθ + B cosθ = C (cosθ cosθ0 + sinθ sinθ0) = C cos(θ-θ0) GM ∴ U = C cos(θ − θ 0 ) + 2 h -By Er. Biraj Singh Thapa (Lecturer, Eastern College of Engineering, Biratnagar)/ -32

Now, choosing θ0 = 0 and U =

1 [i.e. inital line is axis of symmetry] r

we, get: 1 GM = 2 + c cos θ − (v ) r h Again, we have the equation of conic section, l r= l + e cos θ 1 1 e ∴ = + cos θ − (vi ) r l l Comparing equn (v) & (vi), we get: e c = ⇒ e = cl l Again, 1 GM h2 = 2 ⇒l = l GM h ch 2 ∴e = which is eccentricity of the conic section. GM So, three cases may arise: a) If e>1 (i.e. conic is a hyperbola) ch 2 GM >1 or , c > 2 GM h b) If e=1 (i.e. conic is a parabola) i.e.

e