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Engineering Mechanics: 

Dynamics in SI Units, 12e 

Copyright © 2010 Pearson Education South Asia Pte Ltd 

Chapter 19 

Planar Kinematics of a Rigid Body: Impulse and Momentum 

Chapter Objectives 

Develop formulations for the linear and angular 

momentum of a body 

Apply the principles of linear and angular impulse 

and momentum to solve rigid-body planar kinetic 

problems that involve force, velocity, and time 

Discuss application of the conservation of 

momentum 

Analyze the mechanics of eccentric impact 

Copyright © 2010 Pearson Education South Asia Pte Ltd

Chapter Outline 

1. Linear and Angular Momentum 

2. Principle of Impulse and Momentum 

3. Conservation of Momentum 

4. Eccentric Impact* 


19.1 Linear and Angular Momentum 

We will assume the body is symmetric with 

respect to an inertial x-y reference plane 

Linear Momentum 

The linear momentum of a rigid body is 

determined by L = m i v i 

m i v i = mv G , 

momentum is a vector quantity having a 

magnitude mv G , and a direction defined by v G 



Angular Momentum 

Consider the body subjected to general planar 

motion 

Arbitrary point P has a velocity v P , and body has 

an angular velocity 

The velocity of the ith particle is 

The angular momentum is 




Using Cartesian vectors, 

Letting m i dm and integrating over the entire 

mass m of the body, 




inertia computed about the z axis, I P = r 2 dm 

Equation can be reduced to a 

simpler form if point P coincides 

with the mass center G for the 

body where 




Thus 

It states that the angular momentum of the body 

computed about G is equal to the product of 

moment of inertia of the body about an axis 

velocity. 

Can also written as 



Translation 

When a rigid body of mass m is subjected to 

rectilinear or curvilinear translation, its mass 

center has a velocity of v G = v and = 0 

The linear momentum and the angular 

momentum computed about G is 

Since d is the moment arm, 

H A = (d)(mv G ) 



Rotation About a Fixed Axis 

When a rigid body is rotating about a fixed axis 

passing through point O, the linear momentum 

and the angular momentum about G are 

Noting that L (or v G ) is always 

perpendicular to r G , 



Rotation About a Fixed Axis 

Equation can be simplified by substituting 

v G = r G , and using parallel-axis theorem 

Hence, 

General Plane Motion 

When a rigid body is subjected 

to general plane motion, 



General Plane Motion 

When angular momentum is computed about a 

point A located either on or off the body, 


Example 19.1 

At a given instant the 5-kg slender bar has the 

motion as shown. Determine its angular momentum 

about point G and about the IC at this instant. 


Example 19.2 

Solution 

Bar 

The bar undergoes general plane motion, we have 

Thus, 


Example 19.2 

Solution 

Bar 

Adding I G w and the moment of mv G about the IC 

yields 


19.2 Principle of Impulse and Momentum 

Principle of Linear Impulse and Momentum 

The equation of translational motion for a rigid 

body can be written as F = ma G = m (dv G /dt) 

Since the mass of the body is constant, 

Multiplying both sides by dt and integrating from t 

= t 1 , v G = (v G ) 1 to t = t 2 , v G = (v G ) 2 yields 




This equation is referred to as the principle of 

linear impulse and momentum. 

It states that the sum of all impulses created by 

the external force system which acts on the body 

during the time interval t 1 to t 2 equal to the 

change in the linear momentum of the body 

during the time interval. 




+ = 



Principle of Angular Impulse and Momentum 

If the body has general plane motion we can 

write M G = I G = I G (d /dt) 

Since the moment of inertia is constant, 

Multiplying both sides by dt and integrating from t 

= t 1 , = 1 to t = t 2 , = 2 gives 




In a similar manner, for rotation about a fixed 

axis passing through point O, M O = I O when 

integrated becomes 

Both equations are referred to as the principle of 

angular impulse and momentum. 

They state that the sum of the angular impulses 

acting on the body during the time interval t 1 to t 2 

momentum during this time interval. 




To summarize the preceding concepts, 

m( 

v 

m( 

v 

I 

G 

Gx 

Gy 

) 

) 

1 

1 

1 

t 2 

t1 

t 2 

t1 

t 2 

t1 

F 

F 

M 

x 

y 

G 

dt 

dt 

dt 

m( 

v 

m( 

v 

I 

G 

Gx 

Gy 

2 

) 

) 

2 

2 




The resultant equations may be written in 

symbolic form as 

syst. 

linear 

syst. 

linear 

syst. 

linear 

momentum 

x1 

impulse 

x(1 

2) 

momentum 

x2 

syst. 

linear 

syst. 

linear 

syst. 

linear 

momentum 

y1 

impulse 

y(1 

2) 

momentum 

y2 

syst. 

angular 

syst. 

angular 

syst. 

angular 

momentum 

O1 

impulse 

O(1 

2) 

momentum 

O2 



Procedure for Analysis 

Free Body Diagram 

Establish x, y, z inertial frame of reference and 

draw the free-body diagram. 

Establish the direction and sense of the initial 

v G , 

Assume that the sense of its components is in 

the direction of the positive inertial coordinates 

Compute the moment of inertia I G or I O 




Principle of Impulse and Momentum 

Apply the three scalar equations of impulse and 

momentum. 

The angular momentum of a rigid body rotating 

about a fixed axis is the moment of mv G plus I G 

about the axis. 

diagram will create an impulse. 

-body 





Forces that are functions of time must be integrated to 

obtain the impulse. 

Principle of angular impulse and momentum is used to 

eliminate unknown impulsive forces that are parallel or 

pass through a common axis. 

Kinematics 

If more than 3 equations are needed, we relate the 

velocity using kinematics. 

Kinematic (velocity) diagrams are helpful in obtaining the 

necessary relation. 


Example 19.2 

The 100-N disk is assumed to be uniform and is pin 

supported at its center. If it is acted upon by a 

constant couple moment of 6 N.m and a force of 50 

N which is applied to a cord wrapped around its 

periphery, determine the angular velocity of the disk 

two seconds after starting from rest. Also, what are 

the force components of reaction at the pin 


Example 19.2 

Solution 

Free Body Diagrams 

The loading causes the disk to rotate clockwise. 

The moment of inertia of the disk about its fixed axis 

of rotation is 


Example 19.2 

Solution 


We have 

I 

m( 

v 

A 

m( 

v 

1 

Ax 

Ay 

) 

) 

1 

1 

0 

(2) 

m( 

v 

0 Ay(2) 

100(2) 50(2) 0 

t 

t 

1 

2 

M 

t 

A 

t 

t 

1 

t 

1 

A 

2 

0 6(2) [50(2)](0.25) 0.31855 

2 

x 

dt 

F 

F 

y 

x 

dt 

dt 

I 

A 

0 

m( 

v 

2 

Ax 

Ay 

) 

) 

2 

2 

2 


Example 19.2 

Solution 


Solving, 


Example 19.4 

The block has a mass of 6 kg. It is attached to a cord 

which is attached to a cord which is wrapped around 

the periphery of a 20-kg disk that has a moment of 

inertia IA = 0.40 kg.m2. If the block is initially moving 

downward with a speed of 2 m/s, determine its 

speed in 3 s. 


Example 19.4 

Solution 

Free Body Diagrams 

All the forces are constant since the weight of the 

block causes the motion. 

The downward motion of the block, v B , causes 

the disk to be clockwise. 

of 


Example 19.4 

Solution 


We can eliminate A x and A y from the analysis by 

applying the angular impulse and momentum about 

point A. 

Disk 


Example 19.4 

Solution 

Cylinder 

Kinematics 

Since = v B /r then 1 = 2/0.2 = 10 rad/s and 2 = 

(v B ) 2 /0.2 = 5(v B ) 2 

Substituting and solving the equations, 


19.3 Conservation of Momentum 

Conservation of Linear Momentum 

When the sum of all the linear impulses acting on 

the system of connected rigid is zero, the linear 

momentum of the system is conserved. 

This equation is the conservation of linear 

momentum. 



Conservation of Angular Momentum 

Angular momentum is conserved about the 

G when the sum of all 

the angular impulses is zero. 




Free-body Diagram 

Classify each of the applied forces as being 

- 

From FBD, the conservation of linear momentum 

applies when no external impulsive forces act on 

the body or system in that direction. 

The conservation of angular momentum applies 

at the mass center G when all external impulsive 

forces acting on the body or system create zero 

moment. 




Conservation of Momentum 

Apply the conservation of linear or angular 

momentum in the appropriate directions. 

Kinematics 

If the motion appears to be complicated, 

kinematics (velocity) diagrams may be helpful in 

obtaining the necessary kinematics relations. 


Example 19.6 

The 10 kg wheel has a moment of inertia I G = 0.156 

kg.m2. Assuming that the wheel does not slip or 

rebound, determine the minimum velocity v G it must 

have to just roll over the obstruction at A. 


Example 19.6 

Solution 

Impulse and Momentum Diagrams 

There is no slipping. 

We have the momentum of the wheel 

just before impact, the impulses given 

to the wheel during impact, and the 

momentum of the wheel just after 

impact. 


Example 19.6 

Solution 


Kinematics 

Since no slipping occurs, = v G /r = v G /0.2 = 5v G 

Substituting this into the above equation, 

(1) 


Example 19.6 

Solution 

Conservation of Energy 

In order to roll over the obstruction, the wheel must 

pass position 3. 

From conservation of energy equation, 

Substituting 2 = 5(v G ) 2 into Eq. , 


19.4 Eccentric Impact* 

Eccentric impact occurs when the line 

connecting the mass centers of the two bodies 

does not coincide with the line of impact. 

Occurs when one or both of the bodies are 

constrained to rotate about a fixed axis. 

A problem involving the impact of two bodies 

requires determining the two 

unknowns (v A ) 2 and (v B ) 2 , 

assuming (v A ) 1 and (v B ) 1 

are known or to be determined. 



To solve this problem, 2 equations are needed. 

The 1 st equation is the conservation of angular 

momentum. 

The 2 nd equation is using the definition of the 

coefficient of restitution, e, which is the ratio of 

the restitution impulse to the deformation 

impulse. 

(+ ) 

B 

A 

2 

1 

A 

B 

2 

1 




Example 19.8 

The 5-kg slender rod is suspended from the pin at A. 

If a 1-kg ball B is thrown at the rod and strikes its 

center with a horizontal velocity of 9 m/s, determine 

the angular velocity of the rod just after impact. The 

coefficient of restitution is e = 0.4 


Example 19.8 

Solution 


Angular momentum is conserved about point A since 

the impulsive force between the rod and the ball is 

internal. 

The weights of the ball and rod are non-impulsive. 


Example 19.8 

Solution 


We have 

A 

1 

A 

2 

B 

B 

1 

B 

B 

2 

R 

G 

2 

G 

2 

B 

2 

G 

2 

2 

2 

Since (v G ) 2 = 0.5 

2 , then 


Example 19.8 

Solution 

Coefficient of Restitution 

We have 

Solving

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