Kinetics Of Particles Study Of The Relations Existing Between The

Kinetics of particles study of the relations existing between the forces acting on a body, the mass of the body, and the motion of the body • Newton’s first law and third law are sufficient for studying bodies at rest (- statics) or bodies in motion with no acceleration Ist law: A particle remains at rest or continues to move with uniform velocity (in a straight line with const. speed) if there is no unbalanced force acting on it

• When a body accelerates (change in velocity magnitude or direction), Newton’s second law is required to relate the motion of the body to the forces acting on it • Newton’s Second Law: If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to the magnitude of resultant and in the direction of the resultant. R.Ganesh Narayanan, IITG

Consider a particle subjected to constant forces, F1 F2 F3 constant = = = = mass, m a1 a 2 a 3 Characteristic of particle considered When a particle of mass m is acted upon by a force ‘F’, the acceleration of the particle must satisfy F = ma Magnitudes of F & a are proportional; F & a vectors are in same direction Body subjected to several forces, ΣF = ma If force acting on particle is zero, particle will not accelerate, i.e., it will remain stationary or continue on a straight line at constant velocity – R.Ganesh Narayanan, IITG Newton’s Ist law

Linear momentum of a particle Replacing the acceleration by the derivative of the velocity yield Σ F = m dv dt d dL = (mv ) = Mass is constant dt dt L = linear momentumof theparticle ‘L’ - Same direction as that of velocity of particle – mv (vector)

L = mv => ΣF = dL/dt

Unit: kg. (m/s)

The rate of change of linear momentum (dL/dt) is zero when ΣF = 0 => If the resultant force acting on the particle is zero, linear momentum of the particle ‘L’ is constant, in magnitude and direction This is called as Principle of conservation of momentum => another version R.Ganesh Narayanan, IITG of Newton’s 1st law

Equations of motion In rectangular components, Newton’s second law provides

∑F = ma

∑(Fxi + Fy j + Fz k )= m(axi + ay j + az k ) ∑ Fx = max ∑ Fy = may ∑ Fz = maz ∑ Fx = mx ∑ Fy = my ∑ Fz = mz 0

-W

0

For tangential and normal components,

∑ F t = mat

∑ F n = man

dv F = m ∑ t dt

∑Fn = m

v2

ρ Narayanan, IITG R.Ganesh

For a projectile without air resistance

Dynamic equilibrium • Alternate expression of Newton’s second law, ∑ F − ma = 0 − ma = inertial vector

Particle is in equilibrium under given forces and inertia vector – called Dynamic equilibrium

In coplanar force system, all forces can be made into closed vector polygon and can be solved (OR) Sum of force components including inertia vector can be equated to zero

ΣFx = 0; ΣFy = 0 R.Ganesh Narayanan, IITG

Rectangular components

Tangential components can also be used, -mat & -man Tangential component: Measure of resistance of particle to change in speed Normal component: Tendency of particle to leave its path

Inertia vectors or inertia forces are measure of resistance that particles offer for motion These forces are not considered like g, contact forces etc. and hence dynamic equilibrium method is not used

R.Ganesh Narayanan, IITG

Beer/Johnston, 12.3

The two blocks shown start from rest. The horizontal plane and the pulley are frictionless, and the pulley is assumed to be of negligible mass. Determine the acceleration of each block and the tension in the cord. Write the kinematic relationships for the dependent motions and accelerations of the blocks. yB =

1x 2 A

aB =

x

1a 2 A

Apply Newton’s II law to blocks A, B, pulley C Write equations of motion for blocks and pulley:

∑ Fy = m B a B :

y

∑ Fx = m A a A :

m B g − T2 = m B a B

T1 = (100 kg )a A

(300 kg )(9.81 m s 2 )− T2 = (300 kg )a B T2 = 2940 N - (300 kg )a B R.Ganesh Narayanan, IITG

T2 = 2940-150 aA

T2 − 2T1 = 0 Put T1, T2 in above equn., 2940-150aA-2 (100aA) = 0; aA = 8.4 m/s2 aB = 8.4/2 = 4.2 m/s2; T1 = 100 (8.4) = 840 N T2 = 2 (840) 1680 N

R.Ganesh Narayanan, IITG

Beer/Johnston, 12.14

88 km/h

A light train made up of two cars is traveling at 88 km/h when the brakes are applied to both cars. Knowing that car A has a weight of 24947.56 kg and carB has a weight of 19958 kg and that the braking force is 31137.5 N on each car, determine (a) the distance traveled by the train before it comes to a stop, (b) the force in the coupling between the cars while the train is slowing down. Consider Fb as braking force

Fb

A

B Fb

Fb

Find ax (a) Relate ‘ax’ to ‘Xf’ by ax = v (dv/dx); Apply BCs and by integration of ‘x’ and ‘v’; Find ‘Xf’ Fb

(b) Find Fc from this

R.Ganesh Narayanan, IITG

A

Fc

12.36 A 2-kg ball revolves in a horizontal circle as shown at a constant speed of 1.5 m/s. Knowing that L = 600 mm, determine (a) the angle θ that the cord forms with the vertical, (b) the tension in the cord.

θ T

ΣFy = m (ay) = 0 =>T cos θ-w = 0 ΣFx = m (ax) => T sinθ = ma

ma W

a is towards circle center = v2/ρ = v2/Lsinθ Solve these equns. and get θ, t

R.Ganesh Narayanan, IITG

12.53 A curve in a speed track has a radius of 200 m and a rated speed of 180 km/h. Knowing that a racing car starts skidding on the curve when traveling at a speed of 320 km/h, determine (a) the banking angle θ, (b) the coefficient of static friction between the tires and the track under the prevailing conditions

Rated speed is the speed at which a car should travel if no lateral friction force is to be exerted on its wheels ma

F

Y

θ N

w X

Put F=0 and find θ

FindNarayanan, µ from µIITG = F/N R.Ganesh

Angular momentum of particle • H O = r × mV = moment of momentum or the angular momentum of the particle about O. • H O is perpendicular to plane containing r and mV

HO

i = x mv x

j y mv y

k z mv z

We have Hx, Hy, Hz

• Derivative of angular momentum with respect to time, H O = r × mV + r × mV = V × mV + r × ma = r×∑ F = ∑ MO

0

• It follows from Newton’s second law that the sum of the moments about O of the forces acting on the particle is equal to the rate of change of the angular momentum of the particle about O R.Ganesh Narayanan, IITG

Eqs of Motion in Radial & Transverse Components • Consider particle at r and θ, in polar coordinates,

2 ∑ Fr = mar = m(r r θ ) ∑ Fθ = maθ = m(rθ +2rθ )

Motion under central force • When only force acting on particle is directed toward or away from a fixed point O, the particle is said to be moving under a central force. • Since the line of action of the central force passes through O, ∑ M O = H O = 0 and r × mV = H O = constant R.Ganesh rmv sinφ = r0mv 0 sinφ0Narayanan, IITG

The 2.7 kg collar B slides on the frictionless arm AA′. The arm is attached to drum D and rotates about O in a horizontal plane at the rate θ = 0.8t, where θ and t are expressed in rad/s and seconds, respectively. As the arm-drum assembly rotates, a mechanism within the drum releases cord so that the collar moves outward from O with a constant speed of 0.457 m/s. Knowing that at t = 0, r = 0, determine the time at which the tension in the cord is equal to the magnitude of the horizontal force exerted on B by arm AA′. r

t

dr/dt = r = 0.457 m/s; ∫dr = ∫0.457 dt => r = 0.457t; r = 0 0

0

(

∑ Fr = mar = m r r θ

)

2

∑ Fθ = maθ = m(rθ +2rθ ) ar = -0.292 t3 m/s2; aθ = 1.09 m/s2

R.Ganesh Narayanan, IITG

M aθ M ar =

T Q

T = Q => -m ar = m aθ 0.292 t3 = 1.09 => t = 1.66 s

R.Ganesh Narayanan, IITG

Energy and momentum methods • Previously, problems dealing with the motion of particles were solved through the fundamental equation of motion, F = ma • We introduce two additional methods of analysis • Method of work and energy: directly relates force, mass, velocity and displacement • Method of impulse and momentum: directly relates force, mass, velocity, and time

R.Ganesh Narayanan, IITG

Method of Work & energy Work of a force • Work of the force is dU = F • dr = F ds cos α = Fx dx + Fy dy + Fz dz • Work is a scalar quantity, i.e., it has magnitude and sign but not direction. • Dimensions of work are length × force. Units are 1 J ( joule ) = (1 N )(1 m )

•Work of finite displacement •Work of force of gravity •Work of the force exerted by a spring R.Ganesh Narayanan, IITG

Kinetic energy of particle • Consider a particle of mass m acted upon by force, F dv Ft = mat = m dt dv ds dv =m = mv ds dt ds F t ds = mv dv • Integrating from A1 to A2 , s2

v2

s1

v1

1 mv 2 − 1 mv 2 F ds = m v dv = ∫ t ∫ 2 2 1 2

U1→2 = T2 − T1

T = 12 mv 2 = kinetic energy

KE: Unit: Nm or Joule Scalar quantity

• The work of the force F is equal to the change in kinetic energy of the particle => Principle of work & energy T2 = T1 + U (1-2)

R.Ganesh Narayanan, IITG

T1

• To determine velocity of pendulum bob at A2. Consider work & kinetic energy. • Force P acts normal to path and does no work.

T2

T1 + U 1→ 2 = T 2 1W 2 0 + Wl = v2 2 g v2 =

2 gl

Advantages: • Velocity found without determining expression for acceleration and integrating • All quantities are scalars and can be added directly • Forces which do no work are eliminated from the problem. R.Ganesh Narayanan, IITG

• Principle of work and energy cannot be applied to directly to determine the acceleration of the pendulum bob • Calculating the tension in the cord requires supplementing the method of work and energy with an application of Newton’s second law

• As the bob passes through A2 ,

∑ Fn = m a n

ΣFt = m at = 0

W v 22 P −W = g l W 2 gl P =W + = 3W g l

Problem with more particles: KE can be obtained for each particle separately and KE can be added for all particles T1 + U (1-2) = T2 T – arithmetic sum of the KEs of particles involved; U (1-2) = work of all the forces acting on the particles

R.Ganesh Narayanan, IITG

Power: Rate at which work is done Power = dU/dt = F.dr/dt = F.v Unit: Nm/s

Efficiency = Power output / Power input < 1

R.Ganesh Narayanan, IITG

Two blocks are joined by an inextensible cable as shown. If the system is released from rest, determine the velocity of block A after it has moved 2 m. Assume that the coefficient of friction between block A and the plane is µk = 0.25 and that the pulley is weightless and frictionless. Apply the principle of work and energy separately to blocks A and B. + ve

(

)

W A = (200 kg ) 9.81 m s 2 = 1962 N

FA = µ k N A = µ k W A = 0.25(1962 N ) = 490 N T1 + U1→2 = T2 : 0 + FC (2 m ) − FA (2 m ) = 12 m A v 2 FC (2 m ) − (490 N )(2 m ) = 12 (200 kg )v 2

R.Ganesh Narayanan, IITG

(

)

WB = (300 kg ) 9.81 m s 2 = 2940 N T1 + U1→2 = T2 : 0 − Fc (2 m ) + WB (2 m ) = 12 m B v 2 − Fc (2 m ) + (2940 N )(2 m ) = 12 (300 kg )v 2

• When the two relations are combined, the work of the cable forces cancel. Solve for the velocity

(2940 N )(2 m ) − (490 N )(2 m ) = 12 (200 kg + 300 kg )v 2 4900 J = 12 (500 kg )v 2

v = 4 .43 m s R.Ganesh Narayanan, IITG

A 40.8 kg package is at rest on an incline when a constant force P is applied to it. The coefficient of kinetic friction between the package and the incline is 0.35. Knowing that the speed of the package is 0.6m/s after it has moved 0.9 m up the incline, determine the magnitude of the force P.

R.Ganesh Narayanan, IITG

400 20°

P 50°

F = 0.35 N

T1 = 0 N = 400 Cos 20 + P sin 50

Put N in U (1-2)

U (1-2) = T2 – T1 => P = 737.9 N

R.Ganesh Narayanan, IITG

N

A 2000-kg automobile starts from rest at point A on a 6° incline and coasts through a distance of 150 m to point B. The brakes are then applied, causing the automobile to come to a stop at point C, 20 m from B. Knowing that slipping is impending during the braking period and neglecting air resistance and rolling resistance, determine (a) the speed of the automobile at point B, (b) the coefficient of static friction between the tires and the road

6°

w

VA = 0; Vc = 0; w = 19620 N F

U (A-B) = TB – TA 19620 (150) Sin 6 = ½ (2000) (VB2) VB =17.5 m/s U (A-C) = TC – TA 19620 (170) Sin 6 – F (20) = 0 19620 (170) Sin 6 – 19620 (µ) cos 6 (20) = 0 µ = 0.89

R.Ganesh Narayanan, IITG

N

Car B is towing car A with 4.6 m cable at a constant speed of 9 m/s on an uphill grade when the brakes of car B are fully applied causing it to skid to a stop. Car A, whose driver had not observed that car B was slowing down, then strikes the rear of car B. Neglecting air resistance and rolling resistance and assuming a coefficient of kinetic friction of 0.9, determine the speed of car A just before the collision.

0

Car B

mg

F = 0.9 N

F = 0.9 mg cos 5

N = mg cos 5

=> d = 4.2 m

mg

For Car A: initial to contact

Car A

(-mg sin 5) (4.6+4.2) = ½ m VA2 – ½ (m) 92 => VA = 8.22 m/s R.Ganesh Narayanan, IITG

N

Potential energy PE due to displacement; U (1-2) = Wy1 – Wy2 = V1-V2 PE due to spring deformation; U (1-2) = ½ k(x1)2 – ½ k (x2)2 Conservative forces Forces whose work done is independent of path followed; depends only on initial and final positions => conservative forces • Concept of potential energy can be applied if the work of the force is independent of the path followed by its point of application. U1→2 = V ( x1 , y1 , z1 ) − V ( x2 , y 2 , z 2 )

For any conservative force applied on closed path

∫ F. dr = 0 R.Ganesh Narayanan, IITG

• Elementary work corresponding to displacement between two neighboring points, dU = V ( x , y , z ) − V ( x + dx , y + dy , z + dz ) = − dV ( x , y , z )

du = -dv (x, y, z)

The elementary work of a conservative force is exact differential

 ∂V ∂V ∂V  Fx dx + Fy dy + Fz dz = − dx + dy + dz  ∂y ∂z   ∂x r  ∂V ∂V ∂V  F = − + +  = −grad V Differential of a function  ∂x ∂y ∂z  of several variables

Fx = -∂v/∂x; Fy = -∂v/∂y; Fz = -∂v/∂z F = -grad V

Depends only on position of point of application

For conservative R.Ganesh Narayanan, IITGforce

Conservation of energy U (1-2) = V1 – V2; V – POTENTIAL ENERGY U (1-2) = T2 – T1; T – KINETIC ENERGY V1 – V2 = T2 – T1 T1+V1 = T2+V2 When a particle moves under the action of conservative forces, the sum of kinetic energy and potential energy of particle remains constant

T + V = CONSTANT = MECHANICAL ENERGY, E

R.Ganesh Narayanan, IITG

At A1; T1 = 0; V1 = Wl; T1 + V1 = Wl At A2; T2 = ½ (W/g) v22 = ½ (W/g) (2gl) = Wl; V2 = 0 T2 + V2 = Wl E = T + V = CONSTANT

Energy: A1 = only potential; A2 = Only kinetic Only PE depends on elevation and not KE. So, speed is same in A, A’ R.Ganesh Narayanan, IITG

A

B

C

Particle will have same speed at A, B, C as long as weight of particle and normal reaction of path are the two forces acting; without friction Weight of particle, forces exerted by spring – conservative forces Friction force => Non-conservative force; work done by friction force can not be expressed as change in potential energy as it depends on path followed by point of application A mechanical system involving friction will have decrease in Narayanan,as IITGheat etc. total mechanical energy;R.Ganesh dissipated

A 20 N collar slides without friction along a vertical rod as shown. The spring attached to the collar has an undeflected length of 4 cm and a constant of 3 N/cm. If the collar is released from rest at position 1, determine its velocity after it has moved 6 cm. to position 2. Position 1: Ve = 12 kx12 = 12 (3N / cm )(8 cm − 4 cm ) = 24 N ⋅ cm 2

PE: V1 = Ve + V g = 24 N ⋅ cm + 0 KE: T1 = 0 Position 2: Ve = 1 kx22 = 1 (3N / cm )(10 cm − 4 cm )2 = 54 N ⋅ cm 2 2

Vg = Wy = (20 N )(− 6 cm ) = −120 N ⋅ cm

PE:

V2 = Ve + Vg = 54 − 120 = −66 N ⋅ cm

KE:

T2 = 12 mv22 =

1 20 2 v2 2 10 R.Ganesh Narayanan, IITG

Conservation of Energy:

T1 + V1 = T2 + V2 0 + 24 N ⋅ cm = v22 − 66 N ⋅ cm v2 = 9.5 m s ↓

R.Ganesh Narayanan, IITG

13.55 A force P is slowly applied to a plate that is attached to two springs and causes a deflection x0. In each of the two cases shown, derive an expression for the constant , ‘ke’ in terms of k1 and k2 , of the single spring equivalent to the given system, that is, of the single spring which will undergo the same deflection x0 when subjected to the same force P.

Series connection:

Force in both springs is the same = P X0 = X1 + X2 P/Ke = P/K1 + P/K2 => Ke = (K1.K2)/(K1+K2)

Parallel connection:

Deflection in both springs is the same = X0 P = P1 + P2 = K1 X0 + K2 X0 P = Ke X0 => Ke = K1 + K2 R.Ganesh Narayanan, IITG

13.59 A 750-g collar can slide along the horizontal rod shown. It is attached to an elastic cord with an undeformed length of 300 mm and a spring constant of 150 N/m. Knowing that the collar is released from rest at A and neglecting friction, determine the speed of the collar (a) at B, (b) at E.

R.Ganesh Narayanan, IITG

R.Ganesh Narayanan, IITG

A 2.7 kg collar can slide without friction on a vertical rod and is held so it just touches an undeformed spring. Determine the maximum deflection of the spring (a) if the collar is slowly released until it reaches an equilibrium position, (b) if the collar is suddenly released. 2.7 kg

a) Collar in equilibrium ΣF = 0 => -26.5+2627X = 0 Xmax = 0.01 m b) Collar in suddenly released T1+V1 = T2+V2 0+0 = 0 + (-Wh+1/2x2627xh2) h = 2 (26.5)/2627 = 0.02 m R.Ganesh Narayanan, IITG

K = 2627 N/m

A spring is used to stop a 90.7 kg package which is moving down a 20° incline. The spring has a constant k = 22 kN/m and is held by cables so that it is initially compressed 15 cm. Knowing that the velocity of the package is 2.4 m/s when it is 7.6 m from the spring and neglecting friction, determine the maximum additional deformation of the spring in bringing the package to rest.

Position 1 is at the top of the incline; position 2 is when the spring has maximum deformation T1 + V1 = T2 + V2 ½ (90.7) (2.4) + [1/2 (22000) (0.15)2 + 890 (7.6+X) sin 20] = 0 + [0 + ½ (22000) (X+0.15)2] Where x = Deformation of the spring

Solving this, X = 0.367 m R.Ganesh Narayanan, IITG

Method of impulse and momentum • Method of impulse and momentum: directly relates force, mass, velocity, and time. F = d (mv) / dt

FX

Fdt = d (mv)

t2

∫ F dt = mv2 - mv1 t1 t t1

t2

t2

mv1 + ∫F dt = mv2 t1

Integral => Impulse of force, F (Imp1-2)

Vector quantity; Unit : N.s

Imp1-2 = i ∫FX dt + j ∫Fy dt + k ∫Fz dt, from t1 to t2 R.Ganesh Narayanan, IITG

t2

mv1 + ∫F dt = mv2 t1

mv1 + Imp1-2 = mv2 The final momentum of the particle can be obtained by adding vectorially its initial momentum and the impulse of the force during the time interval.

Note: KE, work => scalar quantities; Momentum, impulse => vector quantities R.Ganesh Narayanan, IITG

Several forces acting on one particle mv1 + Σ Imp1-2 = mv2 Two or more particles

Σ mv1 + Σ Imp1-2 = Σ mv2

• Impulse of action and reaction forces exerted by particles cancel out; Only the impulses of external forces need be considered • If sum of external forces is zero, then Σ mv1 = Σ mv2. This says that total momentum of the particles is conserved. (discussed later)

R.Ganesh Narayanan, IITG

Impulsive motion Force acting on a particle during a very short time interval that is large enough to cause a significant change in momentum is called an impulsive force and resulting motion is called impulsive motion When impulsive forces act on a particle,

m v1 + ∑ F ∆ t = m v 2 When a baseball is struck by a bat, contact occurs over a short time interval but force is large enough to change sense of ball motion

Non-impulsive forces are forces for which (F ∆t) is small and therefore, may be neglected; Eg., weight R.Ganesh Narayanan, IITGof body, spring force etc.

Impulse motion of several particles

∑m v1 + ∑ F ∆ t =∑ m v 2 Second term involves only impulse and external forces If all the external force are non-impulsive, then Σ m v1 = Σ m v2 This indicates that total momentum of particle is conserved, and not energy; Eg., Two particles moving freely collide each other

R.Ganesh Narayanan, IITG

A 0.5 kg baseball is pitched with a velocity of 80 m/s. After the ball is hit by the bat, it has a velocity of 120 m/s in the direction shown. If the bat and ball are in contact for 0.15 s, determine the average impulsive force exerted on the ball during the impact. x component equation:

− mv1 + Fx ∆t = mv2 cos 40° 0.5 (80)+ Fx (0.15) = 0.5 (120 cos 40°) 9.81 9.81 Fx = 58.42 N

−

y component equation:

0 + Fy ∆t = mv2 sin 40° 0.5 ( ) (120 sin 40°) = Fy 0.15 9.81

Fy = 26 .21 N F = 64.03 N; Θ = 24.16° R.Ganesh Narayanan, IITG

y

x

The initial velocity of the block in position A is 9 m/s. Knowing that the coefficient of kinetic friction between the block and the plane is 0.30, determine the time it takes for the block to reach B with zero velocity, if (a) θ = 0, (b) θ =20°. A) θ = 0

t = 3.05 s R.Ganesh Narayanan, IITG

b) θ = 20

t = 0.96 s R.Ganesh Narayanan, IITG

A light train made of two cars travels at 72 km/h. The mass of car A is 18 Mg and the mass of car B is 13 Mg. When the brakes are suddenly applied, a constant braking force of 19 kN is applied to each car. Determine (a) the time required for the train to stop after the brakes are applied, (b) the force in the coupling between the cars while the train is slowing down. (a) Entire train: 72 km/h = 20 m/s

R.Ganesh Narayanan, IITG

72 km/hr

18Mg

13Mg

b) Car A: mA = 18 Mg = 18x103 kg; t1-2 = 16.32 s

R.Ganesh Narayanan, IITG

Impact • Impact: Collision between two bodies which occurs during a small time interval and during which the bodies exert large forces on each other. • Line of Impact: Common normal to the surfaces in contact during impact. Direct Central Impact

• Central Impact: Impact for which the mass centers of the two bodies lie on the line of impact; otherwise, it is an eccentric impact.. • Direct Impact: Impact for which the velocities of the two bodies are directed along the line of impact.

Oblique Central Impact

• Oblique Impact: Impact for which one or both of the bodies move along a line other than the line of impact. R.Ganesh Narayanan, IITG

Direct central impact

Before impact

At maximum deformation VA’

VB’

• Bodies moving in the same straight line, vA > vB . • Upon impact the bodies undergo a period of deformation, at the end of which, they are in contact and moving at a common velocity. • A period of restitution follows during which the bodies either regain their original shape or remain permanently deformed. • Wish to determine the final velocities of the two bodies. The total momentum of the two body system is preserved,

mAvA + mBvB = mAvA’ + mBvB’ After impact

(In scalar, same direction)

• A second relation between the final velocities is R.Ganesh Narayanan, IITG required.

Particle A • Period of deformation: m A v A − ∫ Pdt = m Au

During period of deformation, P is the impulse force exerted on A by B • Period of restitution:

m Au − ∫ Rdt = m Av′A

During period of restitution, R is the impulse force exerted on A by B In general, R and P are different; Impulse of R < Impulse of P e = coefficient of restitution Rdt u − v′A =∫ = ∫ Pdt v A − u 0 ≤ eR.Ganesh ≤1 Narayanan, IITG

Particle B v′ − u e= B u − vB

• A similar analysis of particle B yields

• Combining the relations leads to the desired second relation between the final velocities.

v′B − v′A = e(v A − v B )

The relative velocity of two particles after impact can be obtained by multiplying relative velocity of two particles before impact with ‘e’. => This property is used to find ‘e’ of particles experimentally

FINAL:

mAvA + mBvB = mAvA’ + mBvB’ v′B − v′A = e(v A − v B ) R.Ganesh Narayanan, IITG

+ ve sign = right motion - ve sign = left motion

mAvA + mBvB = mAvA’ + mBvB’

v′B − v′A = e(v A − v B )

1. Perfectly plastic impact, e =0 => vB’ = vA’ => No period of restitution; Particles stay together after impact Let vB’ = vA’ = v’ => mAvA + mBvB = (mA+ mB) v’ ; => v’ can be solved 2. Perfectly elastic impact, e =1 => vB’ - vA’ = vA - vB => Relative velocities before and after impact are equal; Impulses received by each particle during period of deformation and restitution are same; Particles move with same velocity but opposite in nature => Total energy and total momentum conserved (not in general case) => vA’ and vB’ can be solved

R.Ganesh Narayanan, IITG

Oblique central impact Velocities of particles are not directed along the line of impact => Oblique impact

VB’ VA’

Final velocities VA’ and VB’ and directions are un-known n axis => along line of impact; t axis => common tangent -F ∆t

F ∆t

Frictionless, smooth surface; Impulses due to internal forces directed along the line of impact – n axis R.Ganesh Narayanan, IITG

1) Tangential component of momentum (v A )t = (v ′A )t for each particle considered separately is conserved.

(v B )t = (v ′B )t

2) Normal component of total m A (v A )n + m B (v B )n = m A (v′A )n + mB (v′B )n momentum of the two particles is conserved.

(v′B )n − (v′A )n = e[(v A )n − (v B )n ] 3) Normal components of relative velocities before and after impact are related by the coefficient of restitution. We have 4 independent equations to solve for components of velocities of A and B after impact Particles canR.Ganesh move Narayanan, in space;IITGno constraints

One or two particles have constraints • Block constrained to move along horizontal surface. • Impulses from internal forces F and − F along the n axis and from external force Fext exerted by horizontal surface and directed along the vertical to the surface. • Three unknowns: i) Final velocity of ball in direction and magnitude and ii) unknown final block velocity magnitude. Three equations required.

R.Ganesh Narayanan, IITG

1) Tangential momentum of ball is conserved.

(v B )t = (v′B )t

2) Total horizontal momentum of block and ball is conserved.

m A (v A ) + m B (v B ) x = m A (v′A ) + m B (v′B ) x

3) Normal component of relative velocities of block and ball are related by coefficient of restitution.

(v′B )n − (v′A )n = e[(v A )n − (v B )n ]

Not extension of central impact case mava - ∫ Pdt cos θ = mau; mAu - ∫ Rdt cos θ = mAvA’ Period of deformation

Period of restitution

R.Ganesh Narayanan, IITG

e = coefficien t of restitutio n Rdt u − v ′A =∫ = ∫ Pdt v A − u 0 ≤ e ≤1

Multiply all velocities by cos θ to obtain their projections on line of impact e = un – (vA’)n / (va)n - un Similar to the central impact case, ‘e’ can be derived between initial and final velocities

R.Ganesh Narayanan, IITG

A 20 Mg railroad car moving at a speed of 0.5 m/s to right collides with a 35 Mg car which is at rest. If after collision 35 Mg car is observed to move to the right at a speed of 0.3 m/s. find the coefft. of restitution between the two cars mAvA + mBvB = mAvA’ + mBvB’ (20 x 0.5) + 0 = 20 x vA’ + (35 x 0.3) vA’ = -0.025 m/s v′B − v′A = e(v A − v B )

e = 0.3 – (-0.025)/0.5-0 = 0.65

R.Ganesh Narayanan, IITG

The magnitude and direction of the velocities of two identical frictionless balls before they strike each other are as shown. Assuming e = 0.9, determine the magnitude and direction of the velocity of each ball after the impact.

(v A )n = v A cos 30° = 26.0 m s (v A )t = v A sin 30° = 15.0 m s (vB )n = −vB cos 60° = −20.0 m s (v B )t = v B sin 60° = 34.6 m s We know that,

(v A )t = (v′A )t

(v B )t = (v′B )t

Find vA’ t, vB’ t

m A (v A )n + m B (v B )n = m A (v′A )n + mB (v′B )n

(v′B )n − (v′A )n = e[(v A )n − (v B )n ] R.Ganesh Narayanan, IITG

Find vA’n & vB’n

Two steel blocks slide without friction on a horizontal surface; immediately before impact their velocities are as shown. Knowing that e = 0.75, determine (a) their velocities after impact, (b) the energy loss during impact.

a) Put given input values and solve these two equations; find final velocities

b) ∆E = T1 – T2

R.Ganesh Narayanan, IITG

Application of three methods • Three methods for the analysis of kinetics problems: - Direct application of Newton’s second law - Method of work and energy - Method of impulse and momentum • Select the method best suited for the problem or part of a problem under consideration.

R.Ganesh Narayanan, IITG

(b)

(c)

(d)

(b) Pendulum A from A1 to A2: Apply conservation of energy principle, find vA2 at A2 (c) A hits B: Total momentum of two pendulums is conserved; use relation between relative velocities; find vA3, vB3 after impact (d) B from B3 to B4: Apply conservation of energy principle to pendulum B R.Ganesh Narayanan, IITG

A 7.9 kg sphere A of radius 11.4 cm moving with a velocity v0 of magnitude 1.8 m/s strikes a 0.73 kg sphere B of radius 5 cm which was at rest. Both spheres are hanging from identical light flexible cords. Knowing that the coefficient of restitution is 0.8, determine the velocity of each sphere immediately after impact. F∆t Ball A

θ

+

=

mAvA

0.064

mAvA’

0.164

Θ =22.9°

mAvA - F∆t cos θ = mAvA’ θ

Ball B F∆t

θ

mBvB’ F∆t = mBvB’

=

vA’ cosθ – vB’ = e (-v0 cos θ)

θ

Solve three equns,

R.Ganesh Narayanan, v IITG A’ =

+1.56 m/s; vB’ = 2.76 m/s

System of particles Motion of large number of particles considered together Application of Newton’s second law for system of particles • Newton’s second law for each particle Pi in a system of n particles, Fi +

n

∑ fij = mi ai j =1

ri × Fi +

∑ (ri × fij )= ri × mi ai n

j =1

Fi = external force

fij = internal forces

mi ai = effectiveforce Repeat this for ‘n’ number of particles & will get ‘n’ equations. The vectors miai are called effective forces of the particles; The external forces Fi and internal

forces fij acting on the various particles form a system equivalent to the system of the effective forces miai. R.Ganesh Narayanan, IITG

The external forces Fi and internal forces fij acting on the various particles form a system equivalent to the system of the effective forces miai.

• Summing over all the elements, n

n n

n

i =1

i =1 j =1

i =1

∑ Fi + ∑ ∑ f ij = ∑ mi ai ∑ (ri × Fi )+ ∑ ∑ (ri × f ij )= ∑ (ri × mi ai ) n

n n

n

i =1

i =1 j =1

i =1

R.Ganesh Narayanan, IITG

• Since the internal forces occur in equal and opposite collinear pairs, the resultant force and couple due to the internal forces are zero, ∑ Fi = ∑ m i a i ∑ (r i × F i ) = ∑ ( r i × m i a i )

The system of the external forces acting on the particles and the system of the effective forces of the particles are equipollent

NOTE: Equipollent system of vectors: ΣF = ΣF’; ΣMo = ΣMo’

R.Ganesh Narayanan, IITG

Linear & angular momentum Linear momentum of the system of particles,

n

L = ∑ mi vi i =1 n

n

i =1

i =1

L = ∑ mi vi = ∑ mi ai Angular momentum about fixed point O of system of particles, n

H O = ∑ (ri × mi vi ) i =1 n

collinear n

H O = ∑ (ri × mi vi )+ ∑ (ri × mi vi ) i =1 n

ΣF = L; ΣMo = Ho

i =1

= ∑ (ri × mi ai ) i =1

Resultant of the external forces is equal to rate of change of linear momentum of the system of particles; Moment resultant about fixed point O of the external forces is equal to the rate of change of angular momentum of the system of particles R.Ganesh Narayanan, IITG

Motion of the mass center of a system of particles Mass center G of system of particles is defined by position vector rG which satisfies n

mrG = ∑ mi ri i =1

Differentiating,

Where ‘m’ is the total mass Σmi of the particles

n

mrG = ∑ mi ri i =1

n

mvG = ∑ mi vi = L i =1

Differentiating, ma = L = ∑ F G This defines motion of mass center G of the system of particles; The mass center of a system of particles moves as if the entire mass of the system and all the external forces were concentrated at that point; MOTION OF EXPLODING SHELL

Similarly angular momentum of a system of particles about mass center can be obtained R.Ganesh Narayanan, IITG

Conservation of momentum for a system of particles • If no external forces act on the particles of a system, then the linear momentum and angular momentum about the fixed point O are conserved.

L = ∑F = 0 L = constant

HO = ∑ M O = 0 H O = constant

• Concept of conservation of momentum also applies to the analysis of the mass center motion

L = mv = constant G

vG = Constant

Similarly, HG = Constant

R.Ganesh Narayanan, IITG

Kinetic energy of system of particles KE of system of particles is defined as sum of KEs of various particles of the system T

=1 2

n

∑ mi vi2

i =1

It is convenient to consider separately the motion of mass center G of the system and the motion of the system relative to a moving frame attached to G T=

1 mv 2 G 2

+

n 1 ′2 m v ∑ i i 2 i =1

Kinetic energy is equal to kinetic energy of mass center ‘G’ plus kinetic energy relative to the centroidal frame Gx’y’z’. R.Ganesh Narayanan, IITG

Work-energy principle for system of particles Principle of work and energy can be applied to each particle Pi ,

T1 + U1→2 = T2 where U1→2 represents the work done by the internal forces, fij and the resultant external force Fi acting on Pi . Principle of work and energy can be applied to the entire system by adding the kinetic energies of all particles and considering the work done by all external and internal forces Although fij and fji are equal and opposite, the work of these forces will not, in general, cancel out. If the forces acting on the particles are conservative, the work is equal to the change in potential energy and T1 + V1 = T2 + V2

which expresses the principle ofR.Ganesh conservation of energy for the system of particles. Narayanan, IITG

Principle of impulse & momentum for system of particles ΣF = L; ΣMo = Ho

t2

∑ ∫ Fdt = L2 − L1 t1

t2

∑ ∫ M O dt = H 2 − H1 t1

t2

L1 + ∑ ∫ Fdt = L2 t1

t2

H1 + ∑ ∫ M O dt = H 2 t1 R.Ganesh Narayanan, IITG

t2

∑ ∫ Fdt = L2 − L1 t1

t2

∑ ∫ M O dt = H 2 − H1 t1

t2

L1 + ∑ ∫ Fdt = L2 t1

t2

H1 + ∑ ∫ M O dt = H 2 t1

• The momenta of the particles at time t1 and the impulse of the forces from t1 to t2 form a system of vectors equipollent to the system of momenta of the particles at time t2 . With no external forces acting, L1 = L2; H1 = H2 The linear momentum & angular momentum of system of particles are conserved R.Ganesh Narayanan, IITG

Variable systems of particles • Kinetics principles established so far were derived for constant systems of particles, i.e., systems which neither gain nor lose particles. • A large number of engineering applications require the consideration of variable systems of particles, e.g., hydraulic turbine, rocket engine, etc. • For analyses, consider auxiliary systems which consist of the particles instantaneously within the system plus the particles that enter or leave the system during a short time interval. The auxiliary systems, thus defined, are constant systems of particles.

R.Ganesh Narayanan, IITG

Steady stream of particles

Air flow through duct or blower • System consists of a steady stream of particles against a vane or through a duct. • Define auxiliary system which includes particles which flow in and out over ∆t. • The auxiliary system is a constant system of particles over ∆t. t2

L1 + ∑ ∫ Fdt = L 2 t1

[∑ mi vi + (∆m )v A ]+ ∑ F ∆t = [∑ mi vi + (∆m)vB ] dm (v B − v A ) ∑ F = R.Ganesh Narayanan, IITG dt

Applications

• Fluid Stream Diverted by Vane or Duct

• Fan

• Fluid Flowing Through a Pipe

• Jet Engine • Helicopter R.Ganesh Narayanan, IITG

Systems gaining or losing mass

• Define auxiliary system to include particles of mass m within system at time t plus the particles of mass ∆m which enter the system over time interval ∆t. • The auxiliary system is a constant system of particles. L1 +

t2

∑ ∫ F dt = L 2 t1

[mv + (∆m )va ]+ ∑ F ∆t = (m + ∆m )(v + ∆v ) ∑ F ∆t = m∆v + ∆m(v − va )+ (∆m )∆v

u = va - v

R.Ganesh Narayanan, IITG

negligible

∑ F ∆ t = m ∆ v - (∆ m) u

ΣF = m (dv/dt)-u (dm/dt) ΣF + u (dm/dt) = ma In the case of losing mass case, rate of change of mass is negative

R.Ganesh Narayanan, IITG

A 20 N projectile is moving with a velocity of 100 m/s when it explodes into 5 and 15 N fragments. Immediately after the explosion, the fragments travel in the directions θA = 45o and θB = 30o. Determine the velocity of each fragment. y x

R.Ganesh Narayanan, IITG

• Write separate component equations for the conservation of linear momentum. m Av A + mB vB = mv0 (5 g )v A + (15 g )vB = (20 g )v0 x components:

5v A cos 45° + 15vB cos 30° = 20(100) y components:

5v A sin 45° − 15vB sin 30° = 0 • Solve the equations simultaneously for the fragment velocities.

v A = 207 m s

v B = 97.6 m s

R.Ganesh Narayanan, IITG

Grain falls onto a chute at the rate of 240 N/s. It hits the chute with a velocity of 20 m/s and leaves with a velocity of 15 m/s. The combined weight of the chute and the grain it carries is 600 N with the center of gravity at G. Determine the reactions at C and B. ∆m 240 N/s = = 24 kg/s 2 ∆t 10 m /s

The sum Σmivi of the momenta of the particles supported by the chute is same at ‘t’ and ‘t+∆t’ R.Ganesh Narayanan, IITG

Apply impulse-momentum principle, L1 + In x-axis: In y-axis:

∑ ∫ F dt = L 2

C x ∆t = (∆m )vB cos10°

(

)

− (∆m )v A + C y − W + B ∆t = −(∆m )vB sin 10°

H C ,1 + ∑ ∫ M C dt = H C ,2 − 3(∆m )v A + (− 7W + 12 B )∆t

= 6(∆m )vB cos10° − 12(∆m )vB sin 10°

Put input values in three equations and solve for B, Cx, Cy r r r B = 584 .75 N C = ( 354 .5 i + 432 .75 j ) N

R.Ganesh Narayanan, IITG

A bullet is fired with a horizontal velocity of 500 m/s through a 3-kg block A and becomes embedded in a 2.5-kg block B. Knowing that blocks A and B start moving with velocities of 3 m/s and 5 m/s, respectively, determine (a) the mass of the bullet, (b) its velocity as it travels from block A to block B (a) the mass of the bullet

mv0+mA(0)+mB(0) = mvB+mAvA+mBvB m = mAvA+mBvB / (v0-vb) = (3) (3) + (2.5) (5) / (500-5) = 43.434 x10-3 kg (b) velocity as it travels from block A to block B

mv0+mA(0) = mv1+mAvA => v1 = mv0 - mAvA / m = = 43.434 x10-3 (500) – (3) (3) / 43.434 x10-3 = 292.8 m/s

R.Ganesh Narayanan, IITG

A system consists of three particles A, B, and C. We know that WA = WB = 17.79 N and Wc = 124.55 N and that the velocities of the particles, expressed in m/s are, respectively, vA = 42i + 63j, vB = −42i + 63j, and vC = −9j − 6k. Determine the angular momentum HO of the system about O.

H0 = ZERO

R.Ganesh Narayanan, IITG

In a game of pool, ball A is moving with a velocity v0 of magnitude v0 = 4.57 m/s when it strikes balls B and C, which are at rest and aligned as shown. Knowing that after the collision the three balls move in the directions indicated and assuming frictionless surfaces and perfectly elastic impact (that is, conservation of energy), determine the magnitudes of the velocities vA, vB and vC.

vC

vA

vB v0 j i

X-axis:

Find VB and Vc in terms of V0 and VA

y-axis:

Put VB and Vc

R.Ganesh Narayanan, IITG

The nozzle shown discharges a stream of water at a flow rate Q= 1.8 m3/min with a velocity v of magnitude 18.29 m/s. The stream is split into two streams with equal flow rates by a wedge which is kept in a fixed position. Determine the components of the force exerted by the stream on the wedge.

v1

j i

v2

Q = 1.8 m3/min = 0.03 m3/s

Impulse-momentum principle

∆m (v) + F ∆t = ∆m/2 (v1) + ∆m/2 (v2) F = ∆m / ∆t (1/2 v1+1/2 v2 – v)

Velocity vectors: R.Ganesh Narayanan, IITG

F

F = - 117.17 i - 56.8 j N

R.Ganesh Narayanan, IITG

Sand falls from three hoppers onto a conveyor belt at a rate of 40 kg/s for each hopper. The sand hits the belt with a vertical velocity v1 = 3 m/s and is discharged at A with a horizontal velocity v2 = 4 m/s. Knowing that the combined mass of the beam, belt system, and the sand it supports is 600 kg with a mass center at G, determine the reaction at E.

a

a

a

a

h

E

2

4

F

W = mg = 600 x 9.81 = 5886 N

R.Ganesh Narayanan, IITG

R = 4040 N (+)