Kinetics of Particles: Energy and Momentum Methods

(1)

13.1 Introduction

• Previously, problems dealing with the motion of particles were solved through the fundamental equation of motion,

Current chapter introduces two additional methods of analysis.

. a m F 



• Method of work and energy: directly relates force, mass, velocity and displacement.

• Method of impulse and momentum: directly relates force, mass, velocity, and time.

(2)

13.2 Work of a Force

• Differential vector dr is the particle displacement.

• Work of the force is

dz F dy F dx F

ds F

r d F dU

z y

x  







 cos

 

• Work is a scalar quantity, i.e., it has magnitude and sign but not direction.

force.

length 

• Dimensions of work are Units are

    

¹^N ¹^m

J

1 joule 

(3)

13.2 Work of a Force

• Work of a force during a finite displacement,

 









 

2

1

2

1 2

1

cos

2 1

A A

z y

x

s s

t s

s A

A

dz F dy F dx F

ds F ds

F r d F U



 

• Work is represented by the area under the curve of F_t plotted against s.

(4)

13.2 Work of a Force

• Work of a constant force in rectilinear motion,



F



x U₁_₂  cos 

• Work of the force of gravity,



^y ^y



^W ^y

W

dy W U

dy W

dz F dy F dx F dU

y y

z y

x



























1 2

2 1

2

1

• Work of the weight is equal to product of weight W and vertical displacement y.

• Work of the weight is positive when y < 0, i.e., when the weight moves down.

(5)

• Magnitude of the force exerted by a spring is proportional to deflection,



^N/m^or^lb/in.



constant spring



 k

kx F

• Work of the force exerted by spring,

22 2 2 1 2 1 2 1

1

2

1

kx kx

dx kx U

dx kx dx

F dU

x x





















• Work of the force exerted by spring is positive when x₂< x₁, i.e., when the spring is returning to its undeformed position.

• Work of the force exerted by the spring is equal to negative of area under curve of F plotted against x,



F F



x U _   ₁  ₂ 

2 2 1

1

(6)

13.2 Work of a Force

Work of a gravitational force (assume particle M

occupies fixed position O while particle m follows path shown),

1 2 2

2 1

2

1 r

G Mm r

G Mm dr

r G Mm U

dr r

G Mm Fdr

dU

r r





















(7)

Forces which do not do work (ds = 0 or cos   0:

• weight of a body when its center of gravity moves horizontally.

• reaction at a roller moving along its track, and

• reaction at frictionless surface when body in contact moves along surface,

• reaction at frictionless pin supporting rotating body,

(8)

13.3 Particle Kinetic Energy: Principle of Work & Energy

dv mv ds

F

ds mv dv dt

ds ds mdv

dt m dv ma

F

t

t t



• Consider a particle of mass m acted upon by force F

• Integrating from A₁ to A₂ ,

energy kinetic

mv T

T T

U

mv mv

dv v m ds

F

v v s

s t















2 2

1 1 2

2 1

12 2 2 1 2 2

2 1

1 2

1

• The work of the force is equal to the change in kinetic energy of the particle.

F

• Units of work and kinetic energy are the same:

J m N m

s kg m s

kg m

2 2 2

2

1    



 



 



 



 

 mv T

(9)

13.4 Applications of the Principle of Work and Energy

• Wish to determine velocity of pendulum bob at A₂. Consider work & kinetic energy.

• Force acts normal to path and does no work.

P

gl v

g v Wl W

T U

T

2 2 0 1

2

22 2

2 1 1







 _

• 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.

(10)

13.4 Applications of the Principle of Work and Energy

• Principle of work and energy cannot be

applied to directly 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 ,

l W gl g

W W P

l v g W W

P

a m F_n _n

2 3

22













gl v₂  2

(11)

13.5 Power and Efficiency

• rate at which work is done.

v F

dt r d F dt

dU Power

 





 



• Dimensions of power are work/time or force*velocity.

Units for power are

W s 746

lb 550ft

hp 1 s or

N m s 1

1J (watt) W

1      

•

input power

output power

input work k output wor efficiency



 

(12)

(13)

Sample Problem 13.1

An automobile weighing 4000 N is driven down a 5^o incline at a speed of 88 m /s when the brakes are applied causing a constant total breaking force of 1500 N.

Determine the distance traveled by the automobile as it comes to a stop.

SOLUTION:

• Evaluate the change in kinetic energy.

• Determine the distance required for the work to equal the kinetic energy change.

(14)

Sample Problem 13.1

SOLUTION:

• Evaluate the change in kinetic energy.



⁴⁰⁰⁰^/¹⁰

 

⁸⁸ ^1548800^N ^m

s m 88

2 2

2 1 2 1

1 1 1





mv T

v



1151N



0 m

N 1548800

2 2

1 1









 _

x T

U T

m 6 .

1345 x

• Determine the distance required for the work to equal the kinetic energy change.

    

 

x

x x

U

N 1151

5 sin N

4000 N

2 1500

1











 

0

0 ₂

2  T 

v

(15)

Sample Problem 13.2

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 m_k = 0.25 and that the pulley is weightless and frictionless.

SOLUTION:

• Apply the principle of work and energy separately to blocks A and B.

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

Solve for the velocity.

(16)

Sample Problem 13.2

SOLUTION:

• Apply the principle of work and energy separately to blocks A and B.

   

 

   

    

₂¹

 

²

2 2

1 2

2 1 1

2

kg 200 m

2 N 490 m

2

m 2 m

2 0

:

N 490 N

1962 25

. 0

N 1962 s

m 81 . 9 kg 200

v F

v m F

F

T U

T

W N

F W

C

A A

C

A k A

k A

A



















m m

   

    

₂¹

 

²

2 2

1 2

2 1 1

2

kg 300 m

2 N 2940 m

2

m 2 m

2 0

:

N 2940 s

m 81 . 9 kg 300

v F

v m W

F

T U

T W

c

B B

c B





















(17)

Sample Problem 13.3

A spring is used to stop a 60 kg package which is sliding on a horizontal surface.

The spring has a constant k = 20 kN/m and is held by cables so that it is initially compressed 120 mm. The package has a velocity of 2.5 m/s in the position shown and the maximum deflection of the spring is 40 mm.

Determine (a) the coefficient of kinetic friction between the package and surface and (b) the velocity of the package as it passes again through the position shown.

SOLUTION:

• Apply the principle of work and energy between the initial position and the

point at which the spring is fully

compressed and the velocity is zero.

The only unknown in the relation is the friction coefficient.

• Apply the principle of work and energy for the rebound of the package. The only unknown in the relation is the velocity at the final position.

(18)

Sample Problem 13.3

SOLUTION:

• Apply principle of work and energy between initial

position and the point at which spring is fully compressed.



⁶⁰^kg



²^.⁵^m ^s



² ¹⁸⁷^.⁵^J ₂ ⁰

2 2 1 2 1

1  1mv   T 

T

 

       

k

f kW x

U

m m

m

J 377 m

640 . 0 s m 81 . 9 kg

60 ²

2 1











 

  

    

   



²⁴⁰⁰^N ³²⁰⁰^N



⁰^.⁰⁴⁰^m



¹¹²^.⁰^J

N 3200 m

160 . 0 m kN 20

N 2400 m

120 . 0 m kN 20

2 1

max 2 min

2 1 1

0 max

0 min

























 P P x

U

x x

k P

kx P

e



₁ ₂

 

₁ ₂

 

³⁷⁷^J



¹¹²^J

2

1_  U _ _f  U _ _e   _k 

U m



³⁷⁷^J



¹¹²^J ⁰

- J 5 . 187

2 :

2 1 1







 _

k

T U

T

m m_k  0.20

(19)

Sample Problem 13.4

A 2000 N car starts from rest at point 1 and moves without friction down the track shown.

Determine:

a) the force exerted by the track on the car at point 2, and

b) the minimum safe value of the radius of curvature at point 3.

SOLUTION:

• Apply principle of work and energy to determine velocity at point 2.

• Apply Newton’s second law to find normal force by the track at point 2.

• Apply Newton’s second law to find minimum radius of curvature at point 3 such that a positive normal force is

exerted by the track.

(20)

Sample Problem 13.4

SOLUTION:

 



⁴⁰^m



²



⁴⁰^m

 

¹⁰^m ^s



²⁸^.²⁸^m ^s

2

2 m 1

40 0

: m 40

2 0 1

2 2

2 2 2

2 1 1

2 1

2 2 2

2 2 2 1 1













v g

v

g v W W

T U

T

W U

g v mv W

T T

• Apply Newton’s second law to find normal force by the track at point 2.

n :

n ma F 



 

 

W N

g g

W v

g a W

m N

W _n

5

m 20

m 40 2

2 2 2





 

N 10000

 N

(21)

Sample Problem 13.5

The dumbwaiter D and its load have a combined weight of 600 N, while the counterweight C weighs 800 N.

Determine the power delivered by the electric motor M when the dumbwaiter (a) is moving up at a constant speed of

8 m/s and (b) has an instantaneous

velocity of 8 m/s and an acceleration of 2.5 m/s², both directed upwards.

SOLUTION:

Force exerted by the motor cable has same direction as the dumbwaiter velocity.

Power delivered by motor is equal to Fv_D, v_D = 8 m/s.

• In the first case, bodies are in uniform motion. Determine force exerted by motor cable from conditions for static equilibrium.

• In the second case, both bodies are accelerating. Apply Newton’s

second law to each body to

determine the required motor cable force.

(22)

Sample Problem 13.5

• In the first case, bodies are in uniform motion.

Determine force exerted by motor cable from conditions for static equilibrium.

  

s m N 1600

s m 8 N 200





 Fv_D Power



¹⁶⁰⁰^N^m ^s



 Power Free-body C:

:

 0







Fy 2T 800N  0 T  400N Free-body D:

:

 0







Fy

N 200 N

400 N

600 N

600

0 N 600















T F