PDF Kinetics of Particles: Energy and Momentum Methods

Preview of 13.5- 13.9

13.6 Potential Energy gy

• Work of the force exerted by a spring depends only on the initial and final deflections of the spring,

22 2 2 1 2 1 2 1

1 kx kx

U _→ = −

• The potential energy of the body with respect to the elastic force

to the elastic force,

( ) ( )

2 2 e 1

V V

U

kx V =

( ) ( )

2

1 Ve Ve

U _→ = −

• Note that the preceding expression for V_e is valid only if the deflection of the spring is measured from its undeformed position.

13 - 2

13.7 Conservative Forces

• Concept of potential energy can be applied if the work of the force is independent of the path

f ll d b it i t f li ti followed by its point of application.

(

) (

)

2

1 V x , y , z V x , y , z

U _→ = −

Such forces are described as conservative forces.

Such forces are described as conservative forces.

• For any conservative force applied on a closed path,

= 0

∫

∫

• Elementary work corresponding to displacement between two neighboring points,

( ) ( )

( ) ( )

(

)

dV

dz z

dy y

dx x

V z

y x V dU

, ,

, ,

, ,

−

=

+ +

+

−

=

V V

V ⎞

⎛∂ ∂ ∂

V V

V

z dz dy V

y dx V

x dz V

F dy F dx

F_{x y z}

⎟⎞

⎜⎛∂ ∂ ∂

⎟⎠

⎜ ⎞

⎝

⎛

∂ + ∂

∂ + ∂

∂

− ∂

= +

+

r V

z V y

V x

F V ⎟ = −grad

⎠

⎜ ⎞

⎝

⎛

∂ + ∂

∂ + ∂

∂

− ∂ r =

13.8 Conservation of Energy gy

• Work of a conservative force,

2 1 2

1 V V

U _→ = −

• Concept of work and energy,

1 2 2

1 T T

U _→ = −

• Follows that

constant

2 2

1 1

= +

=

+

= +

V T E

V T

V T

constant

= +

=T V E

• When a particle moves under the action of conservative forces, the total mechanical l

l W

V T

W V

T

= +

=

=

1 1

1

1 0

energy is constant.

1 1

(

)

g mv W

T = ₂² = = ₂ =

2

2 1 2 0

2

1 • Friction forces are not conservative. Total l

W V

T

g

= + ₂

2

2 mechanical energy of a system involving

friction decreases.

M h i l i di i t d b f i ti

13 - 4

• Mechanical energy is dissipated by friction into thermal energy. Total energy is constant.

13.9 Motion Under a Conservative Central Force

• When a particle moves under a conservative central force, both the principle of conservation of angular

t momentum

φ

φ sin

sin ₀

0

0mv rmv

r =

and the principle of conservation of energy GMm

GMm

V T V

T₀ + ₀ = +

may be applied.

r mv GMm

r

mv − GMm = ² −

2 1 0

02 2

1

may be applied.

• Given r, the equations may be solved for v and ϕ.

• At minimum and maximum r, ϕ = 90^o. Given the launch conditions, the equations may be solved for r_min, r_max, v_min, and v_max.

13-2-6

Sample Problem 13.6 p

SOLUTION:

l h i i l f i f

• Apply the principle of conservation of energy between positions 1 and 2.

• The elastic and gravitational potential

• The elastic and gravitational potential energies at 1 and 2 are evaluated from the given information. The initial kinetic

i A 20 N collar slides without friction

along a vertical rod as shown The

energy is zero.

• Solve for the kinetic energy and velocity along a vertical rod as shown. The at 2

spring attached to the collar has an undeflected length of 4 cm and a

t t f 3 N/

at 2.

constant of 3 N/cm.

If the collar is released from rest at position 1 determine its velocity after position 1, determine its velocity after it has moved 6 cm. to position 2.

Sample Problem 13.6 p

SOLUTION:

• Apply the principle of conservation of energy between positions 1 and 2

positions 1 and 2.

Position 1:

( )( )

0 cm N

24

cm N

24 cm

4 cm 8 cm / N 3

1

2 2

2 1 2 1 1

+

⋅

= +

=

⋅

=

−

=

=

V V

V

kx V_e

0

0 cm N

24

1 1

=

+

= +

= T

V V

V _{e g}

Position 2: ^V_e = ₂^1kx₂² = ¹₂

( (

)( )(

) )

( )( )

2

2 2 2

cm N

66 120

54

cm N

120 cm

6 N

20

cm N

54 cm

4 cm 0 1 cm / N 3

V V

V

Wy V

kx V

g e

g e

⋅

−

=

−

= +

=

⋅

−

=

−

=

=

2 2 2

2 2 2 1

2

10 20 2 1

cm N

66 120

54 v mv

T

V V

V _{e g}

=

=

+

Conservation of Energy:

N 66 N

24

0 ²

2 2

1

1 +V = T +V T

13 - 8

cm N

66 cm

N 24

0+ ⋅ = v₂² − ⋅

↓

= 9.5m s v2

Sample Problem 13.7 p

SOLUTION:

• Since the pellet must remain in contact p with the loop, the force exerted on the pellet must be greater than or equal to zero Setting the force exerted by the zero. Setting the force exerted by the loop to zero, solve for the minimum velocity at D.

The 0.5 N pellet is pushed against the

• Apply the principle of conservation of energy between points A and D. Solve for the spring deflection required to

p p g

spring and released from rest at A.

Neglecting friction, determine the smallest deflection of the spring for

for the spring deflection required to produce the required velocity and kinetic energy at D.

smallest deflection of the spring for which the pellet will travel around the loop and remain in contact with the loop at all times

loop at all times.

Sample Problem 13.7 p

SOLUTION:

• Setting the force exerted by the loop to zero, solve for the minim m elocit at D

minimum velocity at D.

n :

n ma F =

↓ +

∑

( ) (

)

2

2

s m 20 s

m 0 1 m

2 =

=

=

=

= rg v

r v m mg

ma W

D

D n

( )

(

)

=

= rg v_D

• Apply the principle of conservation of energy between points A and D

points A and D.

( )

0

5 . 1 cm

/ N 3 0

1

2 2

2 2 1

2 1 1

=

=

= +

= +

= T

x x

kx V

V

V _{e g}

1

( )( )

( )

N 5 0 1

m N 2 m

4 N 5 . 0 0

2 2 1 2

2 = V_e +V_g = +Wy = = ⋅

V

(

)

s m 10

N 5 . 0 2

1 _{2 2}

2 2

2

2 = 1 mv_D = = ⋅

T

2 2

1

1 +V = T +V T

13 - 10

2 5 . 0 5

. 1

0 ²

2 2

1 1

+

=

+ x x =1.3m

Sample Problem 13.9 p

SOLUTION:

• For motion under a conservative central force, the principles of conservation of energy and conservation of angular

momentum may be applied simultaneously.

A satellite is launched in a direction

momentum may be applied simultaneously.

• Apply the principles to the points of minimum and maximum altitude to A satellite is launched in a direction

parallel to the surface of the earth with a velocity of 36900 km/h from

ltit d f 500 k

determine the maximum altitude.

• Apply the principles to the orbit insertion i t d th i t f i i ltit d t an altitude of 500 km.

Determine (a) the maximum altitude reached by the satellite and (b) the

point and the point of minimum altitude to determine maximum allowable orbit

insertion angle error.

reached by the satellite, and (b) the maximum allowable error in the direction of launching if the satellite is to come no closer than 200 km to is to come no closer than 200 km to the surface of the earth

Sample Problem 13.9 p

• Apply the principles of conservation of energy and conser- vation of angular momentum to the points of minimum and maximum altitude to determine the maximum altitude

maximum altitude to determine the maximum altitude.

Conservation of energy:

2 2 1 2 1

2 0 1

r mv GMm r

mv GMm V

T V

T_A + _A = _A′ + _A′ − = −

2 1

2 r0 r

Conservation of angular momentum:

0 0 1 1

1 0 0

v r v mv

r mv

r = =

1 0 1 1

1 0

0 r

Combining,

0 0

2 2 0

1 ⎜⎛ r ⎟⎞ GM ⎜⎛ r ⎟⎞ r 2GM

2 0 1 0

0 1

0 2 0

1 2 0

2 0

1 2

1 1

1

v r

GM r

r r

r r

GM r

v r ⎟⎟⎠ + =

⎜⎜ ⎞

⎝

⎛ −

⎟ =

⎟

⎜ ⎠

⎜

⎝ −

0 = 6370km +500km = 6870km r

(

)(

)

2 0 6

s m 10 398 m

10 37 . 6 s m 81 . 9

s m 10 25 . 10 h

km 36900

×

=

×

=

=

×

=

=

gR GM

v

13 - 12

km 60400 m

10 4 .

60 ⁶

1 = × =

r