1. The document discusses center of mass and linear momentum. It defines center of mass as the point where the mass of an object appears concentrated. 2. Equations are provided to calculate the center of mass for point particles and rigid bodies by using integration and the type of mass distribution (linear, surface, volume). 3. Linear momentum is defined as the product of mass and velocity. Its rate of change gives the net force on an object. Momentum is conserved in the absence of external forces.

1. Center of Mass and Linear Momentum
Note: The center of mass of an object is a point where the mass of the
object appears to be concentrated.
• We classify objects onto point objects and rigid objects. For point
objects, the mass is assumed to be concentrated at a point, whereas
for a rigid object the mass is distributed throughout the object.
• For a system of point particles m1, m2,m3,.……..mn, each located at
(xi,yi,zi) the coordinates of the center of mass are obtained by using
the following equations.

2. By using summations we can write the center of mass equations as
follows:
where M is the total mass, i.e
similarly
and

3. For a rigid body we partition the mass into differential elements of
mass dm and then replace the summation by integration and write
where dm is a differential element of mass
Once we obtain each coordinate of the center of mass , then we write
the center of mass as:

4. In order to perform the integration we need to know the type of
distribution of the mass. There are three possible distributions of mass on
on a rigid. These are:
i) linear distribution
ii) surface distribution
iii) volume distribution
When mass is linearly distributed then dm is given by
Where λ is mass per unit length.
λ is called linear mass density. If the mass is uniformly distributed, then λ
is constant.
• When mass is distribution over an area then dm is given by
Where ς is surface charge density

5. Example: Find the center of mass of a thin uniform rod of mass M and total
length L.
• When mass is distribution over a volume then dm is given by
where ρ is volume mass density or simply density
Solution: Note. From our experience with physical objects we expect the
center of mass of a uniform rod should coincide with its geometric center.
Lets consider a differential element of mass dm distributed over a
differential element of length dx
The linear density λ =M/L or λ= dm/dx is constant
Let the rod lay on the x axis as shown below.

6. • We start from the following equation for center of mass
• Note: Since we have two variables under the integrand, we can not
perform the integration right away. We must have only one variable
• The way we can reduce the number of variables is by relating the
variables to each other
• In the present problem we use the linear density to relate dm and dx. i.e
Thus
• Upon integrating applying the limits we obtain
• Note: This was what we expecting at the beginning

7. 1. In the figure below two blocks are connected over a massless and
frictionless pulley. The mass of block A is 10kg and the coefficient of
kinetic friction between A and the incline is 0.2 and =300. If Block A
slides down at constant speed,
Review
Outline of the solution
i. Do a free body diagram for each object
ii. Find the net force on each block
iii. Apply Newton’s second law

8. 2. If a machine moves a package from its initial position (ri) to a new
position (rf) in 4.0 seconds by applying a force given by
to
i. Compute the amount of work done by the force.
ii. Compute the amount of work done by the force.

9. Momentum
• Definition: The momentum of an object is a physical quantity whose
rate of change gives us the net force on the object.
Note: Momentum denoted by the letter P
Thus we can write the above definition of the momentum mathematically
we obtain
The momentum is thus obtained by integration
Replacing F by
We get

10. Completing the integration yields the following expression for the
momentum of an object.
Lets revisit the following equation we had considered earlier
• If the net force F is zero, then
This means P is constant. This is a statement of conservation of
momentum.
• Which means
• conclusion.: In the absence of net external force momentum is
always conserved

11. • Note: Conservation of momentum has been a powerful tool in the
discovery of new particles. It has also been used solve collision
problems.
• There are two types of collisions. These are:
• i) Inelastic collision
• ii) Elastic collision
• In an inelastic collision momentum is conserved but kinetic energy
is not conserved. i.e
• In an elastic collision both momentum and kinetic energy are
conserved. i.e

12. Example: Two identical balls are on a frictionless, horizontal tabletop. Ball
X initially moves at 10 meters per second, as shown in figure on the left-
hand side. It then collides elastically with ball Y , which is initially at rest.
After the collision, ball X moves at 6 meters per second along a path at
530 to its original direction, as shown in on the right-hand side.
• Find the direction and speed of the second object. Since momentum
must be conserved, we can say the second object would scatter with
velocity v at an angle θ below the x- axis after the collision.
Solution:
Object x- component of Initial momentum y- component of initial momentum
m1=m 10m 0
m2=m 0 0

13. Object x- component of final momentum y- component of final momentum
m1=m 6mcos53 6msin53
m2=m mvcosθ -mvsinθ
• The total initial momentum is thus given by
• Note: The initial momentum in the y-direction is zero.
• The total final momentum is thus given by
Now we can apply the rules of the conservation of momentum,.i.e
From the first equation we get

14. From the second equation we get
Dividing by m and rearranging the last two equations we get
and
Note: Since we have three unknowns, we can reduce the number of
unknowns by dividing the second equation by the first equation as follows.
Which results in
Hence θ=370
Once we obtain θ, we can find v by using any of the above equations
and obtain v=6m/s

15. Collision and Impulse
In the absence of a net external force momentum is conserved. However
when a nonzero net force acts on an object, the momentum of the
object changes. The change in momentum is called the impulse of the
object and it is denoted by the letter J, i.e
or
or
or
where J is the impulse of a collision

16. • In order to solve this problem with start from Newton’s second law, i.e
• Thus to complete the solution we need to find the change in momentum.
Example: A 5 kg steel ball strikes a wall with a speed of 20 m/s at an
angle of θ=530 with the normal to the wall. It bounces off with the same
speed and angle, as shown in the figure below.
If the ball is in contact with the wall for
0.250 s, what is the magnitude of the
average force exerted on the ball by the
wall?
Solution:

17. From the figure we see that
and
Thus the magnitude of the change in momentum is:
Hence
Practice different problems involving collsions
F= 2(5kg)(20m/s)cos53/0.25s)=480N

18. Chapter 10
Rotation
Now lets revisit circular motion and consider an object moving on a
circle with a radius r where the speed is not constant as shown below,
i.e. v1≠v2.
• Suppose the car moves from p1 to p2
θ is called angular displacement
How fast the angular displacement is called
angular velocity and it is denoted by ω, i.e
There are two types of motions
i. Translational motion
ii. Rotational motion

19. α is measured in rad/s2
Note: All the equations of motion we discussed in translational motion can
be applied for rotational motion by exchanging the translational variables
with the rotational variables.
• When θ is measured in radians, ω will be in radians/second
• Angular acceleration which is denoted by α is defined as the rate of
change of angular velocity, i.e.

20. Example1. A wheel rotating with a constant angular acceleration turns
through 20 revolutions during a 6 s time interval. Its angular velocity at
the end of this interval is 13 rad/s. What is the angular acceleration of the
wheel?
• Solution: We always start from the definition of what is being asked to find out.
Recall that angular acceleration is the rate of change of angular velocity, i.e
According to the above equation, we need to know the initial and final
angular velocities as well as the duration t.
(1)

21. • Solution: We start by listing the data given in the problem.
• Angular displacement θ=20 revolutions=20x2π radians=40π radians
• Time=t=6.0s • Final angular velocity=ω=13 rad/s
• Initial angular velocity ω0=? • Angular acceleration =α=?
• Thus we start by listing the data given in the problem.
• According to the information we have from the data there are two
unknowns ω0 and α. One has to look for ways to find out ω0 before
proceeding.
• Lets use the alternate definition of angular displacement θ, i.e
• In the new equation, one sees that only ω0 is unknown. Thus we
use this equation and solve for ω0 first and obtain
(2)

22. • Now we are in a position to solve for the angular acceleration, i.e
Example 2: The turntable of a record player rotates initially at 40.3
rev/min and takes 14.9 s to come to rest. i. What is the angular
acceleration of the turntable, assuming it is uniform? Ii. How many
rotations does the turntable make before coming to rest? Iii. If the
radius of the turntable is 0.109 m, what is the initial linear speed of a
bug riding on the rim?
• Note we solve example 2 in the same approach we solved example 1,
i.e

23. List all data:
• Angular displacement θ=?
• Time=t=14.9s• Final angular velocity=ω=0
• Initial angular velocity ω0=40.3 rev/min=4.22rad/s
• Angular acceleration =α=?
(2)

24. Acceleration
Consider a particle on the rim of circle that is rotating as shown
• Note: When the particle is moving on
the rim there will always be a nonzero
centripetal (radial) component of
acceleration.
• There may also be a tangential
component if the velocity changes in
magnitude as well, i.e if ω is not
constant
• These two components are given by the
following equation.

25. • The the speed is calculated from
• The last step is an application of the product rule of differentiation
• If an object is moving a when a circular path the radius is constant
• This results in:
Suppose the particle moves a distance s on the rim from one point to
another point through an angle θ. If θ is measured in radians then one we
have the following relation.

26. Moment Inertia (I)
• The moment inertia of an object is a physical quantity that tells us
the degree of difficulty of the object to be rotated.
• The moment of an inertia of an object depends on the mass of the
object and the distance of the object from the axis of rotation.
• For a point object of mass m the moment of inertia is given by
• For a rigid (extended) object the moment of inertia is calculated by
• Note: Moment inertia is a scalar quantity
• Example: Calculate the moment of inertia of a uniform rod of mass
m and length L about an axis passing through its center.

27. • Solution: For the sake of convenience lets assume the rod is laying on
the x-axis as shown below.
• The problem is solved by partitioning the rod into differential (small)
elements of the mass dm distributed over a differential element of
length dx and located at distance x from the center of mass, cm.
• Note: x varies from x=-L/2 to x=L/2
x-axis
x=-L/2
cm
x
dx
dm
x=L/2
• Note: Since we have two variables under the integrand sign we have
to reduce that into one variable by relating the variables to each other
by using the following relations, i.e
dm=λdx r=x λ=mass density=M/L

28. • Note: The linear density is constant because the rod is uniform
The Parallel Axis Theorem
• This is a law that enables one to determine the moment inertia of
an object about an axis passing through a point p that is a distance
h away from the center provided Icm is known by using the equation
below.
• Completing the integration we get
Example: Find the moment of inertia of a thin uniform rod about an
axis passing through one of its ends.

29. Solution: Lets first list what data we have, i.e.
h=L/2
Hence
Ip=(1/12)ML2+M(L/2)2=(1/3)ML2
Kinetic energy of rotation
• An object rotating with an angular velocity ω posses a rotational
energy posses a rotational kinetic energy given by
• Note: The above equation is analogous to the translational kinetic
energy we discussed before, i.e

30. Torque (τ)
• Torque is power of a force to rotate an object. The power of a force
to rotate an object depends on the magnitude, direction and
location of the force from the axis of rotation.
• In general the torque is given as the cross product of the force and
the distance from the axis rotation, i.e.
where r is the vector from the axis of rotation to the point where the
force is applied. rsinθ is called the lever arm.
Consider the following situation and discuss qualitatively the torque
generated by each force.
F1 F2
F3
F4F5

31. • Suppose all the forces have the same magnitude of 20N
• F1 generates a clockwise torque
• F2 generates no torque because r=0
• F3 generates a counter clockwise torque
• F4 generates no torque because θ=0, because r and F have the same
direction
• F5 generates a clockwise torque
Newton’s Second Law for Rotation
For translational motion the net force on an object is related to the
acceleration by Newton’s second law, i.e

32. • Similarly for a rotating object, the net torque is given by
• where α is the angular acceleration
• Example: Consider a uniform disk with mass
M=5kg and R=20cm mounted on a fixed
horizontal axle. A block with mass m =1.5kg
hangs from a massless cord that is wrapped
around the rim of the disk. Find the
acceleration of the falling block, the tension
in the cord, the angular acceleration of the
disk.
Solution: One has to do the free body diagram of
each object as shown here.

33. Object= block
Fnet=T-mg=-ma (1)
Object =disk
τnet==Iα=TR (2)
For a disk =(1/2)MR2
To reduce the number of variables use a=αR
Solving for T from equation (1) and substituting that into equation (2)
one obtains

34. Angular Momentum
The angular momentum is a physical quantity of rotating
object whose rate of change gives the net torque causing
the rotation. Angular momentum is denoted by L
substituting
and writing the force as

35. Note: Angular momentum is sometimes called the momentum of
momentum .
The magnitude of the angular momentum
Since the velocity is tangential, the angle θ=900
Hence
but
Thus the angular momentum can be written as
But mr2=I, the moment inertia of the rotating object
And hence

36. Conservation of Angular Momentum
Question: What is the condition for the conservation of angular
momentum?
To answer this question we need to revisit the definition of angular
momentum earlier, i.e
If angular momentum of a system of particles is conserved then its
value does not change, hence its derivative is zero. This will happen
provided the net torque is zero. Thus the condition for angular
momentum conservation is the absence of net external torque.
When angular momentum is conserved we have: Lf=Li, or written in a
different form

37. Summary
Translational motion Rotational Motion
1. Displacement, x 1. Angular displacement, θ
2. Velocity, v=dx/dt 2. Angular velocity, ω=dθ/dt
3. Acceleration, a=dv/dt 3. angular acceleration, α=dω/dt
4. Mass, m 4. Inertia, I
5. Momentum, P=mv 5. Angular momentum, L=Iω
6. Force, F=ma 6. Torque, τ=Iα
7. Kinetic energy Kt=(1/2)mv2 7. Rotational KE=Kr= (1/2)Iω2
Example: A man stands on a platform that is rotating (without friction)
with the angular speed of 1.5 rev/s, his arms are outstretched and he holds
a brick in each hand. The rotational inertia of the system consisting of the
man, bricks and platform about the central vertical axis of the platform is
16.0kg.m2. If by moving the bricks the man decreases the rotational inertia
of the system to 4.0 kg.m2, a) what is the physical basis of the situation?, b)
what is the angular speed of the platform?, c) what is the ratio of the new
kinetic energy of the system to the original kinetic energy?, d) what
provided the added kinetic energy?

38. Solution:
a) Since no external torques are involved, this is an internal interaction
which conserves angular momentum.
b. We solve for the new angular velocity by applying the conservation
of angular momentum, i.e Lf=Li
, where
and
From the problem we know that, Ii=16.0kg.m2, ωi=1.5 rev/s, and
If=4kg.m2, hence
Which results in

39. Example: Consider two spheres of the same mass 2kg and radius 10cm on
an inclined planes of different surfaces. If both spheres starts from rest at
the top of the incline of height 10.0m, one sliding without friction while
the other rolling down how fast will each be travelling at the bottom of
the incline?
Solution: This problem is best solved by applying the conservation of
mechanical energy in both situations, i.e.
Recall that
Which translates to
Object= sliding sphere: In this case the sphere is executing only
translational motion. Lets also take our reference for the potential
energy at the bottom, which means Uf=0 and Ui=Mgh for both spheres.
Lets consider each object separately

40. This results in
Hence, the speed
Object=rolling sphere:
In this case there are two parts to the kinetic energy, i.e, translational
part and rotational part thus,
Recall the relation between angular and translational velocity
We also use I=(2/5)mr2 for a sphere
Thus:

41. Equating the initial potential energy with the final kinetic energy yields
Hence

42. Equilibrium
• a. An object is said to be in translational equilibrium if the net force
(vector sum of all forces) on the object is there. In other words this
means the forces are balanced. Mathematically this is written as
follows.
• b. An object is said to be in rotational equilibrium if the net torque
(vector sum of all torques) on the object is zero. In other words this
means the total clockwise torques are balanced by the total counter
clockwise torques. Mathematically this is written as follows.
• When both conditions (a) and (b) are satisfied the object is said to be
in complete equilibrium.

43. • Question: Can a moving object be in equilibrium?
• Answer: Yes, but it is moving with no acceleration.
Note: In order to calculate torque one has to specify the axis of rotation.
Any force located at the axis rotation does not generate any torque.
Example: The figure below shows a safe (mass M=430kg) hanging by a
rope (negligible mass) from a boom (a=1.9m, and b=2.5m) that consists
of a uniform hinged beam (m=85kg) and horizontal cable (negligible
mass)
a) Find the tension Tc in the cable
b) Find the magnitude of the net
force on the beam from the hinge?
Solution: the first thing to do is to do a
free body diagram on the beam

44. • As we see in the free body diagram
there are four forces acting on the
beam
• The beam is in equilibrium
under the action of all these
forces
• Lets resolve these forces into
vertical and horizontal
components.
Force horizontal component Vertical component
Tc -Tc 0
F Fh Fv
mg 0 -mg
Mg 0 -Mg
• Note: Since the beam is in equilibrium net force on the beam is zero.
Fh
Fv

45. • Which means
and
• Examining equation (1) one sees that there are two unknowns,
namely, Fh, Tc.
• This means one needs to obtain an additional equation. To do so
one applies the second condition of equilibrium, i.e, the fact that
the net torque is zero.
• To calculate the net torque one has to choose an axis of rotation. The axis
of rotation can be chosen anywhere. However it is recommended that one
choose where it is convenient and advantageous.
• In the present problem the best choice is to consider rotation about the hinge. In
this case both Fh and and Fv do not generate torque, while Tc generates a counter
clock wise torque, where as mg and Mg generate a clockwise torque.
• From equation (2) one can solve for Fv, where

46. Counter clockwise torque is:
Clockwise torque is:
Equating equations (3) and (4) and solving for Tc one gets
• Once Tc is found we can use equations (1) to solve for Fh and Fv , i.e,
And hence

47. Gravitation
• Newton’s law of gravitation states that any two masses of mass m1
and m2 separated by a distance r exert gravitational force on each
other.
• By doing an experiment one can find out that the force is directly
proportional to the masses of the two objects and inversely
proportional to the square of the separation.
• The gravitational force between these two objects is always attractive
• The magnitude of the gravitational force is given by
• Where G is called universal gravitational constant=6.67x10-11 Nm2/kg2
Equation (1) is called Newton’s law of gravitation.

48. • Note: The gravitational force between the Sun and the Earth is
responsible for keeping the Earth on its orbit. Similarly the
gravitational force between the Earth and the Moon keeps the
moon moving around the Earth.
• Note: Each object exerts gravitational force on another object
independent of the presence of other objects. This is called the
superposition principle.
Gravitational Potential Energy
• It takes energy to assemble objects in a certain order. This energy is called
the gravitational energy of the system of particles. For two particles of
mass M and m that are separated by a distance r is given by

49. Escape Speed
• Note: If an object is fired straight up with an initial velocity, the
force of gravity exerted by the Earth slows it down to a stop and
the object returns to the Earth with the same initial velocity.
• However when the initial speed is increased gradually the object
would rise higher and higher before it returns to the Earth. When
the initial speed reaches a certain velocity the object would reach
high enough and the influence of the force gravity will be
diminished to zero and the object will not return to the Earth
anymore (i.e it escapes from the Earth).
• The initial velocity that resulted in the escape of the object is called
escape speed. One can obtain the escape speed by applying the
conservation of mechanical energy.
• Let Ei be the mechanical energy of the object where,

50. where
and
Note: R is the radius of the Earth= 6.37x106m, M is mass of the
Earth=5.98x1024kg, and G=6.67x10-11 Nm2/kg2
Hence the initial energy is written as:
Similarly the final energy is written as:
Applying conservation of energy results in:

51. • As the objects leaves the influence of the Earth r approaches
infinity, which means Uf approaches zero. At minimum v0, the final
velocity is also zero. Thus after rearranging we get the following
equation for the escape speed.
Note. Thus if an object is fired at 11.2x103m/s or greater the object will
be gone forever.

52. Kepler’s Laws
• Kepler’s laws deal with the motion of planets or satellites. These
are the law of orbits, the law of areas and the law of periods.
1. The law of orbits states that all planets move in elliptical orbits with
the sun at one focus.
2. The law of areas states that a line joining any planet to the Sun
sweeps out equal areas in equal time intervals. (This statement is
equivalent to the conservation of angular momentum).
3. The law of period states that the square of the period T of any
planet is proportional to the cube of the semimajor axis a of its orbit.
For circular orbits with radius r, this states:
Where M is the mass of the attracting body-the Sun in the case of the
solar system.

53. Energy In Planetary Motion
When a planet or satellite with mass m moves in a circular orbit o
radius r, around the Earth or a planet with mass M, The gravitationa
force is responsible for keeping the satellite in its orbit, i.e.
• Since F is a centripetal force one can also write:
Equating these two equation yields
Hence the kinetic energy of the satellite is

54. Thus the total energy will be given by