Kinetics Of A Particle: Jump To First Page

Chapter 3 Kinetics of a Particle

1 Jump to first page

■

■

constant and linear velocity

aA

 vB

A B

■

Kinetics is based on kinematics and   Newton's second law : F = m a A A reference system in which Newton's second law is valid is called an inertial system ∴ All systems moving with constant and linear velocities are inertial systems.

    F = m a A / B = m( a A − a B )   = m aA ( where aB = 0 )

O

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■

 F2

When a particle is subjected to forces :  

F1 , F2 ,⋅ ⋅ ⋅    FR = ∑ Fi = m a

 FR

 F1

Equation of motion

i

■

We can separate :

FRx = ma x FRy = ma y

■

* Free body diagram :  F1

FRz = ma z

 N  mg Fµ 3 Jump to first page

Example

■

o y

R=k v dv ∴ mg − kv = mv = m dt mdv ∴ ∫ dt = ∫ mg − kv m t = − n (mg − kv) + C k

when t = 0, v = 0 m ∴ C = n (mg ) k m  mg − kv   ∴ t = − n  k  mg 

R

■

mg 4

e

− kt / m

mg − kv = mg

Jump to first page

v

∴

mg/k

.....( 1 )

mg t→∞ v = vT = # k dy mg (b) v= = (1 − e − kt / m ) dt k h t mg − kt / m ∴ ∫ dy = ∫ (1 − e ) dt 0 0 k

t

∴h =

mg v= (1 − e − kt / m ) k

mg k

m2g t − 2 (1 − e− kt / m) # k

⋅ ⋅ ⋅ ⋅ ⋅ (2)

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Work ■

 F

Ft

α

Fn

2

 dr α

■ ■

■ ■

1

■ ■

The work done by a force on a particle in a displacement of dU-scalar  dU = F ; ⋅ if

 dr = Fds cos α

 dr

 ds = d r   

 then F = Fx i + Fy j + Fz k  :  work done ∴ The total dr F=dsdx i (F+cos dyαj =+Fdz k Now dU = )  F= ma ⋅ drds = Fx dx + Fy dy + Fz dz t

t

t

U =∫

  F ⋅ dr

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Kinetic energy K.E. Total work done by external force

 F

dv   t 2 2 2 2 W = ∫ 1 F ⋅ dr = ∫ 1 Fdr = ∫ 1 ma dr = ∫ 1 m dr 1− 2 t t t dt t dv dv  2 2 2 = ∫1 m dr = ∫ 1 m vdt = ∫ 1 mvdv dr dt t dt 1 2 1 2 = mv2 − mv1 = T2 − T1 = ∆T 2 2

 v

Total work done by external = change in K.E.

 F

Power

  dU F ⋅ dr   P= = =F ⋅ v dt dt

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Conservative force and potential energy ■

2

Path a

Path b ■

1

A conservative force is a force having the characteristic that the work done by the force on the particle depends on the net change in position and not on the actual path followed by the particle. e.g. ◆ ◆ ◆

gravitational force elastic force electrostatic force

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Path a

P2 Path b

1

• Work done to move P from 1 to 2 is independent of path     2 2 ∫ 1( path a) F ⋅ dr = ∫ 1( path b) F ⋅ dr = V − Vref where  V is the potential depending on position

Fis a conservative force

•The potential is written as  S   V (r ) = Vref − ∫ S F ⋅ dr o

at S0, V = Vref

9 Jump to first page

 r

 when d r is very small,

  S1   dV F ⋅ dr − (− ∫ S F ⋅ dr ) o   F ⋅ dr   = − F ⋅ dr   = −F x − F y − F z = − F • dr S2 =−∫S o S = − ∫ S2 1

2

 dr 1

x

y

z

Since V is a function of dV =

Fx = −

∂V ∂x

dx +

∂V ∂y

dy +

 r ∂V ∂z

dz

(exact differential)

∂V ∂V ∂V ; Fy = − ; Fz = − ∂x ∂y ∂z

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■

Another representation:

    − F = − i Fx − j Fy − k Fz ∂V ∂V  ∂V =i +j +k ∂x ∂y ∂z ∂ ∂ ∂   = i +j +k V  ∂x ∂y ∂z  ∂ ∂  ∂ +j +k Define gradient ; del : ∇ = i ∂x ∂y ∂z

 ∴ F = − ∇V

Note : For a closed path     2 1 ∫ 1( path a) F ⋅ dr + ∫ 2( path b) F ⋅ dr = 0 11 Jump to first page

Example

m 2

g

h

  F = −mgi   dr = +dxi

  V2 = Vref − ∫ F ⋅ dr = Vref − (−mgh) ∴ V2 − Vref = mgh

1, ref

V = xy2z3

Example

 ∂V  ∂V  dV =   dx +   ∂x  y , z  ∂y

  ∂V   dy +   dz  ∂z  x , y  x, z

= ( y 2 z 3 ) dx + [ x(2 y ) z 3 ] dy + [ xy 2 (3z 2 )] dz dV = y 2 z 3 dx + [ x(2 y ) z 3 ] dy + [ xy 2 (3z 2 )] dz 12 Jump to first page

Example

■ ■

M

r R

m

Gravitational potential energy : Gravitational constant = 6.67 x 10-11 N m2/kg2     F = − G mM e ; d r = − dr e r r2 r    V (r ) = V∞ − ∫ ∞r [− G mM e ] ( − dr e ) 2 r [Set V∞ = 0 ] r = GmM / r ∞ = − G mM r On the surface of the earth, F = mg, r = radius of the earth R

mM mg = G 2 R 13 Jump to first page

Example : Spring ■

X X=0

X

Work done :

    F = − kx i , dx = dx i   e ∴Ve = − ∫ F ⋅ dx = ∫ kxdx 0

1 2 = kx 2

]

e 0

1 2 = ke 2

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Work-energy equation :   ∆K.E. = ∫ ∑ F ⋅ dr    (C ) = ∫ ∑ F + ∑ F (0) ⋅ dr

(

■ ■

 (C ) ∑ F - all conservative forces

 (0) ∑F - non-conservative forces (frictional force), depend on paths

∫ ∑F

■ ■

)

(C )

 ⋅ dr = − ∆V

If we specify

∫

( 0 )  ∑ F ⋅ dr = W , dissipative work

∴ ∆K.E. = − ∆V + W

15 Jump to first page

■

∴ ∆T + ∆(workdone) V =W

■

work-energy eq.

■ ■ ■

increase

decrease

If W = 0 then ∆T + ∆V = 0 The conservation of mechanical energy.

P.E. (Max)

∴

( T2 − T1 ) + (V2 − V1 ) = 0

∴

T2 + V2 = T1 + V1

K.E. (Max) 16 Jump to first page

Examples ■

Find v. Given r = 0.5m, ρ -mass per unit length

smooth

G y

G

r 0.5m At rest

rope ■

Lost of P.E.=

17

0.6m

0.6m

πr   ρ (πr ) g  y + 0.6 +  2   1 = [ ρ(πr + 0.6 ) ] v 2 2 =Gain in K .E.

G

πr 2

v

Jump to first page

y=?

■ ■ ■

■

ds = rdθ dm = ρ ds P.E of the small P.E. = dm g (r sin θ )element Total P.E. =

∫

π

0

g r sin θ r ρd θ

= r g ρ [ − cos θ ] = 2r ρ g = Mgy 2

ref

dθ

18

ds

r θ

M = πrρ

π 0

2

Mass of the semi-circular rope

2r ∴y = π ∴ v = 4.92 m / s

# Jump to first page

■ ■

■ ■

r

θ

r

Find : (a) the velocity of m when the rod passes the horizontal position (light rod). (b) the max velocity of the particles. (c) the max. value of θ .

θ

2m■ Change P.E.= ■

m The rod is released at rest when θ = 0

∴ ∴ 19

Gain in K.E. =

− 2mgr sin θ + mg (r − r cos θ ) 1 ( 2 m + m)v 2 2

1 (3m)v 2 = mgr (2 sin θ + cosθ − 1) 2 2 2 v / gr = (2 sin θ + cosθ − 1) 3 Jump to first page

■

(a) When the rod is horizontal, θ = 45o

∴ ■

■ ■ ■

(b)

v45o = 0.865 gr

#

d 2 2 ( v / gr ) = ( 2 cos θ − sin θ ) = 0 dθ 3 tan θ = 2 , ∴ θ = 63.4o

∴

2 vmax / gr = 0.824

∴

vmax = 0.908 gr

#

(c) max value of θ ⇔ v = 0 ∴ 2sinθ + cosθ - 1 = 0 ⇒ θ Since,

max

sin θ = 1 − cos 2 θ

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∴

2 1 − cos θ = 1 − cos θ 2

(

)

4 1 − cos θ = 1 − 2 cosθ + cos θ 2

2

∴ ∴

5 cos 2 θ − 2 cosθ − 3 = 0 cosθ = 0.2 ± 0.8 = 1 or − 0.6

∴

θ = 0 or θ = 126.9o

#

21 Jump to first page

Examp le O

(a) ν

h vB

vA

B v A = 50km / hr

Find

µN

N θ

µN − mg sin θ = mv θ B

22 mg

1 2 1 2  o  mv B − mv A  + ( − mg ( R − R cos 40 ) ) = 0 2 coefficient (b)ν B=?2If frictional  µ = 0.1. ∆ T∴ + v∆ V==28 W .9 m / s B #

v2 N − mg cos θ = m R

vB R

=? when friction = 0

∆T + ∆V = 0

R=140m 40o

A

B

 ( 2)

 (1)

 v2  ∴ v − µ   = µg cosθ − g sin θ R dv dθ ds dv  1   v = ⋅ ⋅ =  v dθ ds dt dθ  R  Jump to first page

 v2  v dv − µ  = µg cos θ − g sin θ R dθ R

∴ ■

■

Let

w=v

2

;

dw dv then = 2v dθ dθ

dw ∴ − 2 µw = ( 2µg cosθ − 2 g sin θ ) R dθ The solution : θ'

■

− 2µ θ 2 µ θ ' ∫ w(θ ' ) = e {C + e (2µgR cos θ − 2gR sin θ )dθ }

Formula :

θo

ax[a sin px − p cos px] e ∫ eax sin pxdx = a2 + p 2

ax[a cos px + p sin px] e ∫ eax cos pxdx = a2 + p2 2µ θ' ∴ w(θ ' ) = Ce + 23

2gR 1 + 4µ

2

2 2 [3µ sin θ '+(1 − 2µ ) cos θ ' ] = v Jump to first page

 π  Whenθ = 40 = 40  rad ,  180  50 ×103 v= m / s = 13.9 m / s 3600 o

2

π

[

0.2×40× 2 × 9.8 ×140  50  o o 180 = C e + 0 . 3 sin 40 + 0 . 98 cos 40   2 3 . 6 ( ) 1 + 4 0 . 1   ■ ■

∴ C = -2000 (m/s)2 when θ = 0o

w = vB2 = −2000 e 0.2×0 + 2641× (0.98) ∴

vB2 = 588.2 vB = 24.3 m / s

#

24 Jump to first page

]

Linear momentum and Impulse Newton's 2nd law :

■

Let

 Fi

  F2

FR

■

m

 v

 F1

■

 d   F = mv = (mv ) dt

 ;  -linear momentum G ≡ mv G  d   ∴F = G = G dt Impulse ≡   ∑ Fi = G

    t2 ∫ t ∑ Fi dt = ∫ dG = G2 − G 1

Impulse = change in linear momentum 25 Jump to first page

In component form :

∫

t2

∫

t2

∫

t2

t1

t1

t1

∑ Fx dt = ∆Gx ∑ Fy dt = ∆G y ∑ Fz dt = ∆Gz

Impulsemomentum equation

Problems on kinetics of a particle : (4), (5), (10), (13), (15), (18), (20), (21), (23), (24). 26 Jump to first page

Conservation of momentum   ∫ ∑ F ⋅ dt = ∆G t2

t1

 ∴ If ∑ F = 0 orthe time t1 → t2 is extremely

small, then ∆G = 0 B

A Fa FR

∫

t2

∫

t2

t1

t1

FA = Fa (action)

  ∑ FA ⋅ dt = ∆G A   ∑ FB ⋅ dt = ∆GB

FB = FR (reaction) = − FA

27 Jump to first page

∴∫

t2

t1

[

    ∑( FA ) + ∑( FB ) dt = ∆G A + ∆GB

]

◆

If no other forces acting on A and B, then

◆

Let

G S = GA + GB ,

◆

then

∆ GS = 0

◆ ◆

  0 = ∆G A + ∆GB

(Principle of conservation of linear momentum.)

28 Jump to first page

 v1

Initially : ■

Collison

Finally

■

m2

 v2

m1

   Gi = m1v1 + m2 v2

no external force

m2

' v2

m1

' v1

    ∴ G f = m1v1, + m2 v2, = Gi

? 1 1 1 1 2 2 ,2 ,2 Whether : m1v1 + m2 v2 = m1v1 + m2 v2 2 2 2 2

■

elastic collision

K.E.(i) = K.E.(f)

■

inelastic collision 29

K.E.(i) > K.E.(f) Jump to first page

Impact : ■

(a) Direct central impact :

 v1

 v2 m2

m1

' v1

' v2

■

By conservation of linear momentum

■

The coefficient of restitution e :

m1

m2

Before impact After impact

m1v1 + m2 (−v2 ) = m1 (−v1' ) + m2 (v2' )

e≡

relative velocity of separation relative velocity of approach 30

=

v2' − (−v1' ) (−v2 ) − v1 Jump to first page

Impact : ■

■ ■ ■ ■ ■

0≤ e≤ 1 e = 1 : perfectly elastic impact (no energy loss) e = 0 : perfectly plastic impact (Max. energy loss) e depends on the geometry and materials of the impact bodies as well as the impact velocities.

31 Jump to first page

(b) Oblique central impact m1

 v1

n

θ1 θ2

θ1' θ 2'

 v2

m2

m2

m1

t ' v2

during

before

' v1

After (impact)

✦

Assume no friction; only impact in n-direction is considered. (v1)n = -v1sin θ 1 , (v1)t = v1cos θ 1

✦

(v2)n = v2sin θ

✦

2

,

(v2)t = v2cos θ

2

32 Jump to first page

Oblique central impact ✦ By conservation of linear momentum m1 (v1 ) n + m2 (v2 ) n = m1 (v1' ) n + m2 (v2' ) n m1 (v1 )t = m1 (v )

' 1 t

m2 (v2 ) t = m2 (v2' )t ✦

Also

en =

(v ) − (v ) ' 2 n

' 1 n

(v2 ) n − (v1 ) n

✦ Four

equations, four unknow numbers Hence v1’, v2’, θ 1’ and θ 2’ can be found.

33 Jump to first page

Angular Momentum :  vθ

 r o

■

 v

■ ■

β

 vr

■

take moment about O The angular momentum about O : HO = r m vθ = r m v sinβ

m

θ

34 Jump to first page

■

Also

H O = mr 2θ

; vθ = rθ

dH O d = (mr 2θ ) = mr (rθ + 2rθ ) dt dt = m r aθ = rFθ = M O (Torque, moment of force about O)   dG F= dt

dH O ∴MO = dt

∫

t2

t1

 Ho

M O dt = H O 2 − H O1 = ∆H O

 v β 35

∫

t2

t1

  ∑ Fdt = ∆G

    H O = r × G = mr × v = mrv sin βHˆ O

 r Jump to first page

 Ho

 v β

 r

    r = xi + yj + zk     v = vx i + v y j + vz k    i j k  HO = m x y z

vx

vy

vz

   = m ( yvz − zv y ) i + ( zv x − xvz ) j + ( xv y − yvx ) k

[

∴ ∴ 36

]

 dH O    = r × F = M0 dt   t2  ∫ M 0 dt = H O 2 − H O1 t1

Jump to first page

Central force motion :  v m

mo ■

■

37

 F

r θ

directrix

d

Due to central gravitational attraction :

 mm0  F = −G 2 er r : By polar coordinates

(

)

  2      m a = m r − rθ er + m(rθ + 2rθ ) eθ Jump to first page

(

■

■

■ ■

■

and

Gmm0 .....(A) 2    ∴− = m r − rθ 2 r

Thus and Take

(

….(B)

) )

0 = m rθ + 2rθ 1 d 2  = mr θmomentum) (angular r dt is a constant.

(

HO

)

H O = mr θ

2 (constant)

r θ = h 2

38 Jump to first page

■

The rate of area swept by radius vector :

■

(Kepler's second law of 2 planetary motion.)

■

A = 1 r θ = h = Constant 2 2 The solution to equation (A) :

■

C, δ are the integration constants. Choose r be a minimum on x-axis (θ = 0) then δ = 0, we get

…...(C)

..…….(E)

■

Gm0 1 = C cos(θ + δ ) + 2 r h

39

Gm0 1 = C cos θ + 2 r h

…….(D)

Jump to first page

■

A conic section is formed by the locus of a point moving with the ratio e of its distance from a point (focus) to a line (directrix) being constant.

r e= d − r cosθ

■

i.e.

■

Rearrange the above equation :

1 1 1 = cosθ + r d ed

40

…..(F)

1 h 2C ∴d = , e= C Gm0 Jump to first page

2a a (1 + e)

2b

Apogee

m

v r

circle e = 0

a (1 − e) θ

Perigee

mO

ellipse e < 1 parabola e = 1 hyperbola e > 1

Three cases : 41

e<1,

e=1,

e>1. Jump to first page

Case 1 : ellipse (e < 1) ed ■ From (E) : rmin = ; (θ = 0 )

rmax

1+ e ed = ; (θ = π ) 1− e

(

∴

rmin + rmax = 2a and a = ed / 1 − e

i.e.

1 1 + e cosθ = r a 1 − e2

(

2

)

∴ rmin = a (1 − e) , rmax = a (1 + e) ■

This is the Kepler's first law : the planets move in elliptical orbits around the sun as a focus. When e42= 0, r = a ⇒ circle. Jump to first page

)

■

Then the period τ for the elliptical orbit :

A 2( πab ) 2πa τ=  = = A h R g

3/ 2

■

■

where b = a 1 − e 2 is the semiminor axis, h2=GmOa(1-e2) and GmO=gR2. This is the Kepler's third law : the square of the period of motion is proportional to the cube of the semimajor axis.

43 Jump to first page

Case 2 : parabola (e = 1) ■

1 1 = (1 + cosθ ) r d θ →π , r → ∞

(F) becomes :

as

r → ∞ as θ =(e θ >and Case 3 : hyperbola 1) θ ■ ■

1

= -θ

1

where

1 cosθ1 = − e

θ1

mo

hyperbola

44 Jump to first page

The total energy of the mass m : 2 1 mgR 2 2 2 E = m ( r + r θ ) − 2 r To evaluate E, we use the property of the perigee that Then we get θ = 0 , r = 0

■ ■

1 gR =C+ 2 r h Hence

■

2

, r 2θ = h and h 2C = egR 2

2 (elliptical) 2 4 E < 0, ⇒ e < 1 ∴ e = 1 + 2 Eh / mg R ◆ E = 0, ⇒ e = 1 (parabolic) ◆ E > 0, ⇒ e > 1 (hyperbolic) ◆

45 Jump to first page

■

Since :

■

and

(

)

− 1 − e 2 mg 2 R 4 E= 2 2h 2 1 a (1 − e ) =d = C e

∴ E = − gR m / 2a 2

■

■

1 2 2 Also E = mv − mgR / r 2 1  2 2 1 ∴ v = 2 gR  −   r 2a 

velocity of m at different position can be found by the above equation. 46 Jump to first page

Assumptions : ■

(1)The bodies have spherical mass symmetry.

■

(2)Only gravitational force exists.

■

(3)Mass mo is fixed in space.

47 Jump to first page

Example 3R

2R

R

The satellite moves in a circular orbit of radius 2R initially.

Find the min. velocity boost ∆ v for the satellite to reach B. At what B position of the circular orbit the booster should be introduced? The min. velocity boost ∆ v is obtained when the point B is at the apogee of an elliptical orbit while the booster point is at the opposite direction of B and becomes the perigee of the elliptical orbit.

48 Jump to first page

The velocity of the satellite in the circular orbit is expressed by : (e=0 , r=a)



1 1  v = 2 gR  −  and a = 2 R  r 2a  2

2

g ∴ vc = R = a

gR 9.8 × 6371×103 3 = = 5.59 ×10 m s 2 2

For an elliptical orbit with 2a=5R, the velocity at the perigee : 1  3 2 2 1 vP = 2 gR  −  = gR  2 R 5R  5

3 × 9.8 × 6371×103 3 ∴ vP = = 6.12 ×10 m s 5 ∴49 ∆v = vP − vc = 5.3 ×10 2 m s #

Jump to first page

Example

A R 2

 vB B

β

R

Find the magnitude of the necessary launch velocity at B.

θ

β = 90 θ = 135 o

1 1 + e cosθ Since = r a (1 − e 2 ) 1 1 + e cos135o ∴ At point B : = R a (1 − e 2 )

o

3R Also rmax = a (1 + e) = 2 The above two equations give :

e = 0.6306

a = 0.9199 R

50 Jump to first page

1 1  v = 2 gR  −   r 2a  2

Use :

2

1 1  ∴ v = 2 gR  −   R 2 × 0.9199 R  2 B

2

∴ vB = 0.913 gR = 0.913 × 9.8 × 6371×10

3

= 7.55 ×103 m s # 51 Jump to first page