Kinetics Of A System Of Particles

Chapter 4 Kinetics of a system of particles

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Centre of mass ■

 Fi = external force

 : f i i th particle   Fi ri

O

 f i = internal force

o t

a



∴

t

F l 

i

F

 =

m

a

i

+

i

For the whole system :

f  =

■

i

(force exerted on ith particle by all the other particles in the system) i

■

 total   Fi = ∑ Fi + (∑ f i ) = ∑ mi ai

∑ i

∴∑ i

i

  Fi = ∑ mi ai i

i



i

0

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Centre of mass :

 a1

 rG O

C.M .

m m2 a2

 a3

∴∑ i

 m1 aG

m3

 rG ≡

∑

 mi ri

m = ∑ mi

i

m i  ∑i mi ri m1r1 + m2 r2 + ..... = ∑ mi m1 + m2 + ..... i

    mi ri ∑ mi ai ∑  i i   Then rG = = m m        mi ai = m rG = m aG ⇒ ∑ Fi = m aG i

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Momentum : ■

The total momentum of a system of particles :

  G = ∑ mi vi i

■

Now,

∑ i

■

Also

∑

  mi ri = m rG

∴∑

   mi ri = m rG

i

    ∴ G = m rG = m vG       Fi = G = m vG = m aG

i

Newton second law for a system of particles.

 t2 ∴ ∆G = ∫ t1

∑

 Fi dt

i

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■ ■

■

Usually: When then :

∑

 ∆G = 0

i

 Fi = or 0 t2-t1 is extremely small,

  ∴ G1 = G2 (conservation of linear momentum of a system of particles.)

■

where  

G1 =  ∑  i

   mi ri  1

    G2 =  ∑ mi ri   i 2

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Kinetic energy :       ri = rG + riG ; ri = rG + riG  G r iG  The total kinetic energy of the system rG

O

■

 ri

mi

of particles : 1  2  T =∑ mi ri ; 2 i

 r 2 =r  ⋅ r  i i i

1 2     = ∑ mi ( rG + riG ) 2 i 1     2 2    = ∑ mi ( rG + riG + 2rG ⋅ riG ) 2 i 1 1  2   2  =  ∑mi  vG + ∑ mi viG + rG ⋅∑mi riG 2 i 2 i i  Jump to first page

■

For the last term :

d      ∑i mi riG = dt  ∑i mi riG  = 0

1 1 1  2 2 2 ∴T = m vG + ∑ mi viG = ∑ mi ri 2 i 2 i 2 K.E. of the centre of mass

Total K.E. of individual particles w.r.t. the centre of mass

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Angular Momentum of a system of particles  rP O

P   

riP = ri − rP

 ri

mi

■

The angular momentum particle i about P :

of

         H iP = mi ri P × ri P = mi ( ri − rP ) × ( ri − rP ) ■

Take differentiation :

                H i p = mi ( ri − rp ) × ( ri − rP ) + mi ( ri − rp ) × ( ri − rp ) ⇑

0           ∴ H i p = ( ri − rp ) × ( mi ri ) − mi ( ri − rp ) × rp Jump to first page

■

Here

    mi ri = mi ai = Fi + f i External force

Internal force

           ∴ H i p = ( ri − rp ) × Fi + ( ri − rp ) × f i − mi ( ri − rp ) × rp ■

The change in total angular momentum per unit time : (about p)

    H p = ∑ Hi p i

=∑ i

         ( ri − rp ) × Fi + ∑ ( ri − rp ) × f i − ∑ mi ( ri − rp ) × rp i

i

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■

(i) The external torque about point P :

    M p = ∑ ( ri − rp ) × Fi i

■

■

(ii) (iii)

∑ i

   ( ri − rp ) × fi = 0

           ∑i mi ( ri − rp ) × rp =  ∑ mi ri  × rp −  ∑ mi  rp × rp  i   i       = m rG × rp − m rp × rp    = m( rG − rp ) × rp

      ∴ H p = M p − m( rG − rp ) × rp Jump to first page

■ ■ ■

■

  (a) rp = 0   (b) rGthat p is just the centre of mass G = rmeans p (c) The acceleration of P is in the direction of

  Either (rGone − rofp )the above

⇒

   Hp = Mp

 t2  ∴ ∆H p = ∫ M p dt t1

Exercises : 1,2,5,7,10 Due Date : 14/11/00 (Tue.) Jump to first page

m

Example shore

B

xB

shore

d

10m

xB '

x g'

before ■

B (d-2)

after

The centre of mass of both the boat and the girl will remain unchanged since there is no external force to this system. mg = 20 kg , mB = 80 kg

xB' = x g' − ( d − 2 )

xB = 10 − d ; ∴

mg ×10 + mB xB m g + mB

=

mg x g' + mB xB' m g + mB

∴

x g' + 4( xB' − xB ) = 10

∴

x g' + 4( x g' − (d − 2) − (10 − d )) = 10

∴

x g' = 8.4 m

#

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Another approach : Shore

mB x = m g ( 2 − x ) ∴ ∴

C.M .

B

before

10m B x

80 x = 20(2 − x) x = 0.4m

■

The girl will be away from the shore by m - (2 - x) = 10 - 2 + 0.4 = 8.4 m #

■

(6)

■

■

after 2-x

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= 10

Exampl e

v = 7500 m s

v

payload

booster (2000kg) … …

(3000 kg )

7500 + ∆vP

...

7500 + ∆vB ■

■

There is no external force to the whole system, so the linear momentum is conserved. ∆ G=0

∴ 3000 × 7500 + 2000 × 7500 = 3000 × (7500 + ∆vP ) + 2000 × (7500 + ∆vB ) Jump to first page

⇒ ■ ■

∆vB = −1.5∆vP

6 x 400J x 0.4 = Mechanical energy added to the system. By conservation of energy : (Mechanical)

1 1 2 × 3000 × (7500 + ∆vP ) + × 2000 × (7500 × ∆vB ) 2 2 2 1 1 2 = × 3000 × 7500 + × 2000 × 7500 2 + 6 × 400 × 0.4 2 2

∴

∆vP = 0.506 m s #

∴

∆vB = −0.759 m s # Jump to first page

Exam ple

4.5 × 10 −3 kg

u

Find u.

rest

2kg

rest

v µ = 0 .2

1.8m

There is no external force to the system of  the bullet and the block. ∴∆G =0

i.e. 4.5 x 10 u = (4.5 x 10 + 2) v -3

-3

Lost of K.E. :

1 × (4.5 ×10 −3 + 2)v 2 = (4.5 ×10 −3 + 2) × 9.8 × 0.2 ×1.8 2

∴ v = 7.06(m s ) 2

2

⇒ u = 1184 m s# Jump to first page