To predict the launch from a catapult, you must apply the principle of work-energy.

Chap 17

Dynamics

(EMEA1120)

Introduction

To predict the launch from a catapult, you must apply the principle of work-energy.

To determine the forces acting on the stopper pin when the

catapult reaches its final position, angular impulse momentum

equations are used.

Introduction

• Method of work and energy and the method of impulse and momentum will be used to analyze the plane motion of rigid bodies and systems of rigid bodies.

• Principle of work and energy is well suited to the solution of problems involving displacements and velocities.

2 2

1

1 U T

T  _ 

• Principle of impulse and momentum is appropriate for problems involving velocities and time.

 

 

2 1

2

1 2

1

O t

t

O O

t t

H dt

M H

L dt F

L     









   

• Problems involving eccentric impact are solved by supplementing the principle of impulse and momentum with the application of the coefficient of restitution.

Introduction

Forces and Accelerations

Velocities and Displacements

Velocities and Time

Approaches to Rigid Body Kinetics Problems

Newton’s Second Law (last chapter)

Work-Energy Impulse-

Momentum

2 2

1

1 U T

T  _ 

G

G G

F ma

M H









2

1

1 2

t

mv 



2 1

1 2

t

G t G G

I  



Principle of Work and Energy for a Rigid Body

• Work and kinetic energy are scalar quantities.

• Assume that the rigid body is made of a large number of particles.

2 2

1

1 U T

T  _ 

2 

1,T T

2  U1

initial and final total kinetic energy of particles forming body

total work of internal and external forces acting on particles of body.

• Internal forces between particles A and B are equal and opposite.

• Therefore, the net work of internal forces is zero.

Work of Forces Acting on a Rigid Body

• Work of a force during a displacement of its point of application,

 





 

2

1 2

1

2 cos

1

s s A

A

ds F

r d F

U ^{ } 

• Consider the net work of two forces

forming a couple of moment during a displacement of their points of application.

F F 

and  M



 d

M

d Fr ds

F

r d F r

d F r

d F dU



















2

2 1

1

 

 

 





2 1

2

1





 













M d M U

if M is constant.

Work of Forces Acting on a Rigid Body

Do the pin forces at point A do work?

YES NO

Does the force P do work?

YES NO

Work of Forces Acting on a Rigid Body

Does the normal force N do work on the disk?

YES NO

Does the weight W do work?

YES NO

If the disk rolls without slip, does the friction force F do work?

YES NO

 



 F ds F v dt

dU _{C c}

Kinetic Energy of a Rigid Body in Plane Motion

• Consider a rigid body of mass m in plane motion consisting of individual particles i. The kinetic energy of the body can then be expressed as:

 

2 2

1 1

2 2

2 2 2

1 1

2 2

2 2

1 1

2 2

Δ Δ

i i

i i

T mv m v

mv r m

mv I





  

  

 





• Kinetic energy of a rigid body can be separated into:

- the kinetic energy associated with the motion of the mass center G and

- the kinetic energy associated with the rotation of the body about G.

2 2

1 1

2 2

T  mv  I 

Kinetic Energy of a Rigid Body in Plane Motion

• Consider a rigid body rotating about a fixed axis through O.

 

 

2 2 2

1 1 1

2 2 2

1 2 2

Δ _{i i} Δ _{i i i} Δ _i

O

T m v m r r m

I

 



  



  

• This is equivalent to using:

2 2

1 1

2 2

T  mv  I 

1 2 2 O

T  I 

• Remember to only use

when O is a fixed axis of rotation

Concept Quiz

The solid cylinder A and the pipe B have the same diameter and mass.

If they are both released from rest at the top of the hill, which will reach the bottom the fastest?

B A

a) A will reach the bottom first b) B will reach the bottom first c) They will reach the bottom

at the same time

Which will have the greatest

kinetic energy when it reaches the

bottom?

Systems of Rigid Bodies

• For problems involving systems consisting of several rigid bodies, the principle of work and energy can be applied to each body.

• We may also apply the principle of work and energy to the entire system,

2 2

1

1 U T

T  _  = arithmetic sum of the kinetic energies of all bodies forming the system

= work of all forces acting on the various bodies, whether these forces are internal or external to the system as a whole.

2 1,T T

12

U

T

T T

W= 120 g

Systems of Rigid Bodies

• For problems involving pin connected members, blocks and pulleys connected by inextensible cords, and meshed gears,

- internal forces occur in pairs of equal and opposite forces

- points of application of each pair move through equal distances - net work of the internal forces is zero

- work on the system reduces to the work of the external forces

Power

• Power = rate at which work is done

• For a body acted upon by force and moving with velocity ,F v v

dt F

dU  



 Power 

• For a rigid body rotating with an angular velocity and acted upon by a couple of moment parallel to the axis of rotation,

^ M

  dt M

d M dt

dU   Power 

Sample Problem 17.1

For the drum and flywheel,

The bearing friction is equivalent to a couple of At the instant shown, the block is moving downward at 2 m/s.

16kg m .2

= ×

I

90 N m.×

Determine the velocity of the block after it

SOLUTION:

• Consider the system of the flywheel and block. The work done by the internal forces exerted by the cable cancels.

• Apply the principle of work and kinetic energy to develop an expression for the final velocity.

• Note that the velocity of the block and the angular velocity of the drum and flywheel are related by

 r v 

Sample Problem 17.1

SOLUTION:

• Consider the system of the flywheel and block. The work done by the internal forces exerted by the cable cancels.

• Note that the velocity of the block and the angular velocity of the drum and flywheel are related by

1 2 2

1 2

2 m s

5rad s

0.4 m 0.4 m

w w w

= = v = = = v = v

v r

r r

• Apply the principle of work and kinetic energy to develop an expression for the final velocity.

( )( ) ( ) ^{( )}

2 2

1 1

1 2 1 2 1

2 2 2

1 1

120 kg 2 m s 16 kg m 5rad s

2 2

440 J

w

= +

= + ×

=

T mv I

2 2

1 1

2 2 2 2 2

2

2 2 2

2 2

1 1

(120) (16) 110

2 2 0.4

w

= +

= + æ öçè ÷ø =

T mv I

v v v

Sample Problem 17.1

2 2

1 1

1 = 2 1 + 2 w1 = 440J

T mv I

2 2 2

1 1

2 = 2 2 + 2 w2 =110 2

T mv I v

• Note that the block displacement and pulley rotation are related by

2 2

1.25m

3.125rad

= s = 0.4m = q r

• Principle of work and energy:

1 1 2 2

2

440 J 1190 J 110 2

3.85m s + ® =

+ =

=

T U T

v

v ^{v =3.85m s}

( ) ( )

( ) ( ) ^{( )( )}

1 2 2 1 2 1

120 kg 9.81m/s2 (1.2 m) 90 N m 3.125rad 1190 J

U _® =W s - s - M q - q

= - ×

= Then,

Sample Problem 17.2

mm 80 kg

3

mm 200 kg

10









B B

A A

k m

k m

The system is at rest when a moment of is applied to gear B.

Neglecting friction, a) determine the number of revolutions of gear B before its angular velocity reaches 600 rpm, and b) tangential force exerted by gear B on gear A.

m N 6  M 

SOLUTION:

• Consider a system consisting of the two gears. Noting that the gear rotational speeds are related, evaluate the final kinetic energy of the system.

• Apply the principle of work and energy.

Calculate the number of revolutions required for the work of the applied

moment to equal the final kinetic energy of the system.

• Apply the principle of work and energy to a system consisting of gear A. With the final kinetic energy and number of

revolutions known, calculate the moment and tangential force required for the

indicated work.

Sample Problem 17.2

SOLUTION:

• Consider a system consisting of the two gears. Noting that the gear rotational speeds are related, evaluate the final kinetic energy of the system.

  

s rad 1 . 250 25

. 0

100 . 80 . 62

s rad 8 . min 62

s 60

rev rad 2

rpm 600











A B B A

B

r

 r



 

  

  

2

2 2 2

m kg 0192 .

0 m

080 . 0 kg 3

m kg 400 . 0 m

200 . 0 kg 10

















B B B

A A A

k m I

k m I

     

J 9 . 163

8 . 62 0192 .

0 1

. 25 400 .

0 ²

2 2 1 2

1

2 2

2 1 2

2 1









 I_{A A} I_{B B}

T  

Sample Problem 17.2

• Apply the principle of work and energy. Calculate the number of revolutions required for the work.

 

rad 32 . 27

163.9J J

6 0

2 2

1 1









 _

B

B

T U

T





rev 35 . 2 4

32 .

27 

 

_B

• Apply the principle of work and energy to a system consisting of gear A. Calculate the moment and tangential force required for the indicated work.







2 2 1 2

2  1I_{A A}  

T 

 

m N 52 . 11

J 0 . 26 1 rad 10.93 0

2 2

1 1













 _

F r M

M

T U

T

A A

A

rad 93 . 250 10

. 0

100 . 320 .

27 





A B B

A r

 r



N 2 . 250 46

. 0

52 .

11 

 F

Sample Problem 17.3

A sphere, cylinder, and hoop, each having the same mass and radius, are released from rest on an incline.

Determine the velocity of each body after it has rolled through a distance

corresponding to a change of elevation h.

SOLUTION:

• The work done by the weight of the bodies is the same. From the principle of work and energy, it follows that each body will have the same kinetic energy after the change of elevation.

• Because each of the bodies has a

different centroidal moment of inertia, the distribution of the total kinetic

energy between the linear and rotational components will be different as well.

Sample Problem 17.3

SOLUTION:

• The work done by the weight of the bodies is the same. From the principle of work and energy, it follows that each body will have the same kinetic energy after the change of elevation.

r

 v

 With

2 2 2

1

2 2

2 1 2 2 1 2 2 1 2 2 1

v r m I

r I v v

m I

v m T



 

 





 



 





 

2 2

2

2 2 2

1 2 2

1 1

1 2 2

0

mr I

gh r

I m v Wh

v r m I Wh

T U

T

 

 



 

 







 _

Sample Problem 17.3

2 2

1 2

mr I v gh

 

gh v

mr I

Hoop

gh v

mr I

Cylinder

gh v

mr I

Sphere

2 707 . 0 :

2 816 . 0 :

2 845 . 0 :

2 2 2 1

2 5 2













• Because each of the bodies has a different

centroidal moment of inertia, the distribution of the total kinetic energy between the linear and

rotational components will be different as well.

• The velocity of the body is independent of its mass and radius.

NOTE:

• For a frictionless block sliding through the same distance,  0, v  2gh

• The velocity of the body does depend on k2

I 

Sample Problem 17.4

A 15-kg slender rod pivots about the point O. The other end is pressed against a spring (k = 300 kN/m) until the spring is compressed 40 mm and the rod is in a horizontal position.

If the rod is released from this position, determine its angular velocity and the reaction at the pivot as the rod passes through a vertical position.

SOLUTION:

• The weight and spring forces are

conservative. The principle of work and energy can be expressed as

2 2

1

1 V T V

T   

• Evaluate the initial and final potential energy.

• Express the final kinetic energy in terms of the final angular velocity of the rod.

• Based on the free-body-diagram

equation, solve for the reactions at the pivot.

Sample Problem 17.4

SOLUTION:

• The weight and spring forces are conservative. The principle of work and energy can be expressed as

2 2

1

1 V T V

T   

• Evaluate the initial and final potential energy.

( )( )

1 2 1

1 0 2 1 2 300,000 N m 0.04 m

240 J

= + = + =

=

g e

V V V kx

( )( )

2 0 147.15 N 0.75m

110.4 J

= + = + =

=

g e

V V V Wh

• Express the final kinetic energy in terms of the angular velocity of the rod.

( )( )

1 2 12

2

2

1 15kg 2.5m 12

7.81kg m

=

=

= ×

I ml

( )

( ) ( ) ( )

2 2 2 2

1 1 1 1

2 2 2 2 2 2 2 2 2

2 1 2 2

2 2 2 2

1 15 0.75 7.81 8.12

2

w w w

w w w

= + = +

= + =

T mv I m r I

Sample Problem 17.4

1 1 2 2

2

0 240 J 8.12 2 110.4 J

T V T V

w + = +

+ = +

From the principle of work and energy,

• Based on the free-body-diagram equation, solve for the reactions at the pivot.

( )( )

2 2

2 0.75m 3.995rad s 11.97 m s w

a

= = =

=

n t

a r a r

11.97 m s2

a

=

=

n t

a a r

 



MO ^ MO _eff 0  I m

 



 

Fx ^ Fx _eff R_x  m

 



 

Fy ^ Fy _eff

( ) (

)

147.15 N

15kg 11.97 m s 32.4 N

- = -

= -

= -

y n

y

R ma

R

32.4 N

= R

2 3.995rad s w =

Sample Problem 17.5

Each of the two slender rods has a mass of 6 kg. The system is released from rest with b = 60^o.

Determine a) the angular velocity of rod AB when b = 20^o, and b) the velocity of the point D at the same instant.

SOLUTION:

• Consider a system consisting of the two rods. With the conservative weight force,

2 2

1

1 V T V

T   

• Express the final kinetic energy of the

system in terms of the angular velocities of the rods.

• Evaluate the initial and final potential energy.

• Solve the energy equation for the angular velocity, then evaluate the velocity of the point D.

Sample Problem 17.5

• Evaluate the initial and final potential energy.

  

J 26 . 38

m 325 . 0 N 86 . 58 2 2 ₁

1





 Wy V

  

J 10 . 15

m 1283 .

0 N 86 . 58 2

2 ₂

2





 Wy V

SOLUTION:

• Consider a system consisting of the two rods. With the conservative weight force,

2 2

1

1 V T V

T   

   

N 86 . 58

s m 81 . 9 kg

6 ²





 mg W

Sample Problem 17.5

Since is perpendicular to AB and is horizontal, the instantaneous center of rotation for rod BD is C.

m 75 .

 0

BC CD  2





vB

vD

• Express the final kinetic energy of the system in terms of the angular velocities of the rods.

^{0.375 m}

vAB  

   

B AB BC

v     _BD  Consider the velocity of point B





BD  v

  

12 2 1 12

1  6kg 0.75m  0.281kg m



 I ml

I_{AB BD}

For the final kinetic energy,

  

 

  

 

12 1

2 2

2 1 12 2 1

2 2 1

12 2 1

281 . 0 522

. 0 6 281

. 0 375

. 0

6    



















 mv_AB I_{AB AB} mv_BD I_{BD BD} T

Sample Problem 17.5

s rad 3.90

J 10 . 15 1.520

J 26 . 38

0 ²

2 2

1 1

















 V T

V T

• Solve the energy equation for the angular velocity, then evaluate the velocity of the point D.

s rad 90 .

3

^AB

 

  

s m 00 . 2

s rad 90 . 3 m 513 . 0





 CD  v_D

s m 00 .

 2 vD

Team Problem Solving

A slender 4-kg rod can rotate in a vertical plane about a pivot at B. A spring of

constant k = 400 N/m and of unstretched length 150 mm is attached to the rod as shown. Knowing that the rod is released from rest in the position shown,

SOLUTION:

• Because the problem deals with positions and velocities, you should apply the principle of work energy.

• Draw out the system at position 1 and position 2 and define your datum

• Use the work-energy equation to determine the angular

velocity at position 2

Draw your diagrams, set your datum and apply the work energy equation

1 1 1 2 2 2

T   V U

 T  V

Are any of the terms zero?

1 1 1 2 2 2

T  V  U

 T  V

Team Problem Solving

Determine the spring energy at position 1

Unstretched Length 1

2 2

1

(150 mm ) 370 150 220 mm 0.22 m

1 1

(400 N/m)(0.22 m) 9.68 J

2 2

e

x CD V kx

     

  

2

1 (4 kg)(9.81 m/s )( 0.22 m) 7.063 J Vg Wh  mgh   

Determine the potential energy due to gravity at position 1

Determine the spring energy at position 2

2

2 2

2 2

230 mm 150 mm 80 mm 0.08 m

1 1

(400 N/m)(0.08 m) 1.28 J

2 2

e

x

V kx

   

  

Determine the potential energy due to _{V  0}

Team Problem Solving

2 2

2 2 2

1 1

2 2

T  mv  I

2 2 (0.18 m) 2

v  r  

Determine an expression for T₂

Can you relate v₂ and ₂?

2 2 2

2 2 2 2 2

2 2 2 2 2 2

1 1

(4 kg)(0.6 m) 0.12 kg m

12 12

1 1 1 1

(4 kg)(0.18 ) (0.12) 0.1248

2 2 2 2

I mL

T mv I   

   

    

Find I and substitute in to T₂

2 2 2

2

9.68 7.063 0.1248 1.28 J 10.713





  

 

 3.273 rad/s

Substitute into T₁ + V₁ = T₂ + V₂

Concept Question

For the previous problem, how would you determine the reaction forces at B when the bar is horizontal?

a) Apply linear-momentum to solve for B

Dt and B

Dt b) Use work-energy to determine the work done by the

moment at C

c) Use sum of forces and sum of moments equations when

the bar is horizontal

Angular Impulse Momentum

When two rigid bodies collide, we typically use principles

of angular impulse momentum. We often also use linear

impulse momentum (like we did for particles).

Introduction

Forces and Accelerations

Velocities and Displacements

Velocities and Time

Approaches to Rigid Body Kinetics Problems

Newton’s Second Law (last chapter)

Work-Energy Impulse-

Momentum

2 2

1

1 U T

T  _ 

F maG

M H









2

1

1 2

t

mv 



t2

I  



Principle of Impulse and Momentum

• Method of impulse and momentum:

- well suited to the solution of problems involving time and velocity - the only practicable method for problems involving impulsive

motion and impact.

Sys Momenta₁ + Sys Ext Imp_1-2 = Sys Momenta₂

Principle of Impulse and Momentum

v m m

v

L _{i i} 







• The momenta of the particles of a system may be reduced to a vector attached to the mass center equal to their sum,

i i i

G r v m

H   Δ







and a couple equal to the sum of their moments about the mass center,

 I H_G 

• For the plane motion of a rigid slab or of a rigid body symmetrical with respect to the reference plane,

Principle of Impulse and Momentum

• For plane motion problems, draw out an impulse-momentum diagram, (similar to a free-body diagram)

• This leads to three equations of motion:

- summing and equating momenta and impulses in the x and y directions

- summing and equating the moments of the momenta and impulses with respect to any given point (often choose G)

Impulse Momentum Diagrams

A sphere S hits a stationary bar AB and sticks to it. Draw the impulse-momentum diagram for the ball and bar separately;

time 1 is immediately before

the impact and time 2 is

immediately after the impact.

Impulse Momentum Diagrams

F_impDt

Momentum of the ball before impact

Impulse on ball

Momentum of the ball after impact

F Dt

Momentum of the bar before impact

Impulse on bar

Momentum of the

bar after impact

Principle of Impulse and Momentum

• Fixed axis rotation:

- The angular momentum about O

 

 









r2

m I

r r m I

r v m I

I_O













- Equating the moments of the momenta and impulses about O,

2 1

2

1



 ^t _O

t

O

O M dt I

I 

 

The pin forces at point O now contribute no moment to the equation

Systems of Rigid Bodies

• Motion of several rigid bodies can be analyzed by applying the principle of impulse and momentum to each body

separately.

• For problems involving no more than three unknowns, it may be convenient to apply the principle of impulse and

momentum to the system as a whole.

• For each moving part of the system, the diagrams of momenta should include a momentum vector and/or a momentum couple.

• Internal forces occur in equal and opposite pairs of vectors and generate impulses that cancel out.

Practice

F_impDt

From the previous problem, notice that the impulse acting on the sphere is equal and opposite to the impulse acting on the bar. We can take advantage of this by drawing the impulse-momentum diagram of the entire system, as shown on the next slide.

Practice – Diagram for combined system

F_impDt

F_impDt

Conservation of Angular Momentum

I   I 

The moments acting through the skater’s center of gravity are negligible, so his angular momentum remains constant. He can adjust his spin rate by changing his moment of inertia.

2

1

1 2

t

G G G

t

I  



Conservation of Angular Momentum

• When the sum of the angular impulses pass through O, the linear momentum may not be conserved, yet the angular momentum about O is conserved,

   

• Two additional equations may be written by summing x and y components of momenta and may be used to determine two unknown linear impulses, such as the impulses of the reaction components at a fixed point.

• When no external force acts on a rigid body or a system of rigid bodies, the system of momenta at t₁ is equipollent to the system at t₂. The total linear momentum and angular momentum about any point are conserved,

   

2

1 L

L 



Concept Question

For the problem we looked at previously, is the

angular momentum about G conserved?

YES NO YES NO

For the problem we looked at previously, is the angular momentum about point A conserved?

For the problem we looked at previously, is the

YES NO

Sample Problem 17.6

The system is at rest when a moment of is applied to gear B.

Neglecting friction, a) determine the time required for gear B to reach an angular velocity of 600 rpm, and b) the tangential force exerted by gear B on gear A.

m N 6 

 M

mm 80 kg

3

mm 200 kg

10









B B

A A

k m

k m

SOLUTION:

• Considering each gear separately, apply the method of impulse and momentum.

• Solve the angular momentum equations for the two gears simultaneously for the unknown time and tangential force.

Sample Problem 17.6

SOLUTION:

• Considering each gear separately, apply the method of impulse and momentum.

 

    

s N 2 . 40

s rad 1 . 25 m kg 400 . 0 m

250 . 0

0 ₂













 Ft Ft

I

Ftr_{A A} _A moments about A:

moments about B:

 

   



 ^

^

m N 6 0

2 2















Ft t

I Ftr

Mt _{B B} _B

• Solve the angular momentum equations for the two gears simultaneously for the unknown time and tangential force.





Sample Problem 17.7

Uniform sphere of mass m and radius r is projected along a rough horizontal surface with a linear

velocity and no angular velocity.

The coefficient of kinetic friction is

Determine a) the time t₂ at which the sphere will start rolling without sliding and b) the linear and angular velocities of the sphere at time t .

k.



v1

SOLUTION:

• Apply principle of impulse and momentum to find variation of linear and angular

velocities with time.

• Relate the linear and angular velocities when the sphere stops sliding by noting that the velocity of the point of contact is zero at that instant.

• Substitute for the linear and angular

velocities and solve for the time at which sliding stops.

• Evaluate the linear and angular velocities at that instant.

Sample Problem 17.7

SOLUTION:

• Apply principle of impulse and momentum to find variation of linear and angular

velocities with time.

 0

Wt Nt

y components:

x components:

2 1

2 1

v m mgt v

m

v m Ft

v m

k 







 v₂  v₁ _kgt mg W

N  

moments about G:





 

Sys Momenta₁ + Sys Ext Imp_1-2 = Sys Momenta₂

• Relate linear and angular velocities when sphere stops sliding by noting that velocity of point of contact is zero at that instant.



 



 





r t r g

gt v

r v

k k

 

 2 5

1

2 2

• Substitute for the linear and angular

velocities and solve for the time at which sliding stops.

g t v

k¹

7

 2

Sample Problem 17.7

x components: ^v₂  ^v₁ _k^gt y components: ^{N W  mg}

moments about G: ^t

r

kg

 

2 5

2 

Sys Momenta₁ + Sys Ext Imp_1-2 = Sys Momenta₂



 



 





r t r g

gt v

r v

k k

 

 2 5

1

2 2

g t v

k¹

7

 2

• Evaluate the linear and angular velocities at that instant.



 



 

 g

g v v

v

k

k 

 ¹

1

2 7

2



 



 

g v r

g

k

k 

₂  ¹

7 2 2

5

1

2 7

5v v 

r v₁

2 7

 5

