1
2
1.1 Linear Momentum and its Conservation
Momentum
Momentum can be defined as "mass in motion." The amount of momentum
that an object has is dependent upon two variables: how much stuff is
moving and how fast the stuff is moving. Momentum depends upon the
variables mass and velocity. The momentum of an object is equal to the
mass of the object times the velocity of the object.
Momentum = mass • velocity
p=m•v
where m is the mass and v is the velocity. The equation illustrates that
momentum is directly proportional to an object's mass and directly
proportional to the object's velocity.
Momentum is a vector quantity.
3
1.1 Linear Momentum and its Conservation
In 3 dimensions
Px = m vx
Py = m vy
Pz = m vz
Units of
momentum
(kg.m/s)
From Newton’s second law
F = ma
F = m(dv/dt)
F = (mdv/dt)
F = (dp/dt)
4
1.1 Linear Momentum and its Conservation
If two particles interact while isolated from their surroundings, the
particles exert forces on each other where Newton’s third law can be
applied.
F21 = dp1/dt
F12 = dp2/dt
F21 + F12 =0
dp1/dt + dp2/dt= 0
p1i + p2i = p1f + p2f
Momentum of an isolated
system is conserved
5
1.2 Impulse and Momentum
From Newton’s second law
F = (dp/dt)
dp = F dt
∆p = ∫ F dt
This is called the impulse which is the change in momentum.
I = ∆p = ∫ F dt
I = F ∆t
6
1.3 Collisions
The total momentum of a system before a collision is equal to the total
momentum of a system after a collision.
7
1.4 Elastics and Inelastic Collisions in one Dimension
A collision between two objects is considered elastic if the total kinetic
energy as well as the momentum before and after the collision is
conserved.
Where it is considered inelastic if the total kinetic energy is not conserved even if
the momentum is conserved before and after the collision.
8
1.4 Elastics and Inelastic Collisions in one Dimension
Elastic collisions
Momentum is conserved
m1v1i + m2v2i = m1v1f + m2v2f
Kinetic energy is conserved
(1/2)m1v1i2 + (1/2)m2v2i2 = (1/2)m1v1f2 + (1/2)m2v2f2
We can find that
v1f = ((m1-m2)/(m1+m2)) v1i
v2f = ((2m1)/(m1+m2)) v1i
9
1.5 Two Dimensional collisions
We use the same conservation laws of kinetic energy and momentum but
we analyze the equations in all dimensions.
You should remember how to find vector components using the angles !!
In order to solve a problem with collisions:
Choose the system. If it is complex, subsystems may be chosen where one
or more conservation laws apply.
Draw diagrams of the initial and final situations, with momentum vectors
labeled.
Define a coordinate system
Apply momentum conservation; there will be one equation for each
dimension. If the collision is elastic, apply conservation of kinetic energy
as well.
Solve.
10
1.6 The Center of Mass
The Center of Mass is the average position of the mass of an object. It is a
uniquely interesting point. We may consider an object as a hollow, mass
less shell with all its mass located at this Center of Mass. The general
motion of an object can be considered as the sum of the translational
motion of the CM, plus rotational, vibrational, or other forms of motion
about the CM.
In the first picture, the diver’s motion is pure translation; where in the
second picture it is translation plus rotation. There is one point that moves
in the same path a particle would take if subjected to the same force as the
diver. This point is called the center of mass
11
1.6 The Center of Mass
If the particles in the system have masses m1, m2, ...mN, with total mass
∑iN mi = m1+m2+···+mN ≡ M
and respective positions r1, r2, ...,rN, then the center of mass position rCM is:
rCM = (1/M) ∑iN mi ri
which means that the x, y and z coordinates of the center of mass are
xCM = (1/M) ∑iN mi xi
yCM = (1/M) ∑iN mi yi
zCM = (1/M) ∑iN mi zi
For an extended object rCM is given by an integral over the mass elements
of the object:
rCM = (1/M) ∫ r dm
which means that the x, y and z coordinates of the center of mass are
xCM = (1/M) ∫ x dm
yCM = (1/M) ∫ y dm
zCM = (1/M) ∫ z dm
12
1.7 Motion of System of Particles
The velocity of the center of mass of a system
vCM = drCM/dt = (1/M) ∑i mi (dri/dt) = (1/M) ∑i mi vi
Which gives the momentum of the system
MvCM = ∑i mi vi = ∑i pi = ptot
The acceleration of the center of mass
aCM = dvCM/dt = (1/M) ∑i mi (dvi/dt) = (1/M) ∑i mi ai
Which gives the force
MaCM = ∑i mi ai = ∑i Fi
which leads to
∑ Fext = MaCM = (d ptot / dt)
P is
constant
when the
external
force is zero
13
1.8 Rocket Propulsion
The operation of a rocket depends on the law of conservation of linear
momentum as applied to a system of particles where the system is the
rocket plus its ejected fuel.
Applying the law of conservation of momentum
(M+∆m)v = M(v+ ∆v) + ∆m(v-ve)
Where the velocity of the rocket changes after its ejects the fuel. The fuel is ejected
with speed ve relative to the rocket.
We get
M ∆v = ve∆m
Mdv = vedm = - vedM
Since dm = - dM
∫ dv = - ve ∫(dM/M)
vf – vi = ve ln (Mi/Mf)
We call the force exerted on the rocket by the ejected fuel as the thrust
Thrust = M (dv/dt)
14
Examples
 A car is stopped for a traffic signal. When the light turns green, the car
accelerates, increasing its speed from 0 to 5.2 m/s in 0.832 s. What linear impulse
and average force does a 70 kg passenger in the car experience?
Assume the car accelerates in the x-direction. The final momentum of the
passenger is pxf = (70 kg)(5.2 m/s) = 364 kgm/s. The initial momentum is zero.
∆px = pxf - pxi = 364 kgm/s = Favg∆t. Favg = (364 kgm/s)/(0.832 s) = 437.5 N in the xdirection.
 A 3 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60o with the
surface. It bounces off with the same speed and angle. If the ball is in contact with
the wall for 0.2 s, what is the average force exerted on the ball by the wall?
The balls initial momentum is
pi = pxii + pyij = (3 kg 10 m/s)sin60oi+(3 kg 10 m/s)cos60oj.
Its final momentum is
pf = pxfi + pyfj = -(3 kg 10 m/s)sin60oi+(3 kg 10 m/s)cos60oj.
∆p = pf - pi = -2(30 kgm/s)sin60oi = -(51.96 kgm/s)i.
∆p = Favg ∆t.
Favg = -(51.96 kgm/s)i/(0.2 s) = -259.8 Ni.
15
Examples
 A 0.1 kg ball is thrown straight up into the air with an initial speed of 15
m/s. Find the momentum of the ball
(a) at its maximum height and
(b) At half rise time.
(a) At the ball's maximum height its velocity is zero, and therefore its momentum is
zero.
(b) First we find maximum rise time.
Vf = vi-gt which gives that t=1.5s which means that half rise time is 0.75s.
Then Vf = 7.5 m/s and momentum is 1.125 kgm/s.
 A 10 g bullet is stopped in a block of wood (m = 5 kg). The speed of the bulletwood combination immediately after the collision is 0.6 m/s. What was the original
speed of the bullet?
m1v1 + m2v2 = (m1 + m2)vf.
m1 is the mass of the bullet, m2 is the mass of the block. Initially the block is at
rest, v2 = 0. Therefore
(0.01 kg)v1 = (5.01 kg)(0.6 m/s). v1 = 300.6 m/s.
16
Examples
 A car of mass 1000 kg travelling west at 20 m s−1 crashes into the rear of a
stationary bus of mass 5000 kg. The vehicles lock together on impact. Assume that
road friction is negligible.
a What is their joint velocity immediately after the collision?
b What is the total kinetic energy of the system before the collision?
c What is the total kinetic energy of the system after the collision?
d Is this an elastic or inelastic collision? Explain.
a For this problem, west will be taken as the positive direction. Conservation of
momentum must be used to determine the final common velocity v' of the vehicles.
Σpi=Σpf
pi(car) +pi(bus) =pf(bus and car)
(1000 × 20) + 0 = (1000 + 5000) v'
v' = 20 000 ÷ 6000 = 3.3 m/s towards the west
b Since the bus is stationary before the collision, the initial kinetic energy of the
system is the initial kinetic energy of the car. It has a mass of 1000 kg and was
moving at 20 m/s before the collision.
KE= 12mv 2= 0.5 × 1000 × 202 = 2.0 × 105 J
c After the collision, the car and bus are locked together and moving at 3.3 m/s.
Their total kinetic energy is: KE= 12mv 2= 0.5 × (1000 + 5000) × 3.32= 3.3 × 104 J
d This is an inelastic collision.
17
A large amount of kinetic energy has been lost, so the collision is inelastic.
Questions
 An 8 N force acts on a 5 kg object for 3 sec. What impulse is given the object? What change in
momentum does this impulse cause? If the object’s initial velocity was 25 m/s. what is its final
velocity?
I = F ∆t = 8 (3) = 24 N.s
I = ∆p = 24 kg.m/s
∆p = mvf - mvi = 5 (vf – 25) = 24
vf = 29.8 m/s
 A 6 N force acts on a 3 kg object for 10 sec. What will be the final velocity of the object if its
initial velocity was 10 m/s?
∆p = F ∆t = 6 (10) = 60 kg.m/s
60 = 3 (vf - 10)
vf = 30 m/s
 A 20,000 N truck is acted upon by a force that decreases its speed from 10 m/s to 5 m/s in 5 sec.
What is the magnitude of the force?
m = 20000/9.8 = 2040.8 kg
F = ∆p/ ∆t = 2040.8(5-10)/5 = - 2040.8 N
18
Questions
A 2000 N force acts on a rocket of mass 1000 kg, increasing its speed from rest to 200 m/s. How
long did the force act?
F = ∆p/ ∆t
2000 = 1000(200-0)/t
t = 100 s
 A 700 kg car moving at 20 m/s collides with a stationary truck with mass 1400 kg. The two
vehicles interlock as a result of the collision. What is the velocity of the car/truck?
m1v1i + m2v2i = m1v1f + m2v2f
700(20) + 0 = (700+1400) vf
Vf = 6.67 m/s
 A rocket sits on a launch pad loaded with fuel. The igniter causes the rocket to rise off the pad,
oxidizing 100 g of fuel and ejecting it out the back of the rocket at -650 m/s. After launch, the
rocket has a mass of 3 kg. What is its speed after launch?
M∆v = m ve
3 ∆v = 0.1 (650)
∆v = 21.67 m/s
19
Questions
 A 0.4 kg ball is thrown straight up into the air with an initial speed of 15 m/s. Find the
momentum of the ball
(a) at its maximum height and (b) At half rise time. (c) At half maximum height.
(a) At max height
P=0
(b) First we find rise time
vf = vi – gt
0 = 15 – 9.8t
t = 1.53 s
t ½ = 0.77 s
Velocity at half rise time
vf = vi – gt = 15 – 9.8 (0.77) = 7.45 m/s
P = 2.98 kg.m/s
(c) The maximum height ∆y = vi t+ (1/2)gt2
∆y = 15 (1.53) + (1/2)(-9.8)(1.53)2 = 22.95 - 11.47 = 11.48 m
The half height ∆y1/2 = 5.74 m
vf2 = vi2 + 2g ∆y = (15)2 - 2 (9.8) (5.74) = 225 - 112.504 = 112.496
vf = 10.6 m/s
P = 4.24 kg.m/s
20
Questions
 A 0.2 kg ball is thrown straight up into the air with an initial speed of 30 m/s. Find the
momentum of the ball
(a) at its maximum height and (b) At half rise time. (c) At half maximum height.
(a) At max height
P=0
(b) First we find rise time
vf = vi – gt
0 = 30 – 9.8t
t = 3.06 s
t ½ = 1.53 s
Velocity at half rise time
vf = vi – gt = 30 – 9.8 (1.53) = 15 m/s
P = 3 kg.m/s
(c) The maximum height ∆y = vi t+ (1/2)gt2
∆y = 30(3.06) + (1/2)(-9.8)(3.06)2 = 91.8 - 45.88 = 45.92 m
The half height ∆y1/2 = 22.96 m
vf2 = vi2 + 2g ∆y = (30)2 - 2 (9.8) (22.96) = 449.984
vf = 21.21 m/s
P = 4.24 kg.m/s
21
Questions
 A 0.5 kg ball is thrown straight up into the air with an initial speed of 20 m/s. Find the
momentum of the ball
(a) at its maximum height and (b) At half rise time. (c) At half maximum height.
(a) At max height
P=0
(b) First we find rise time
vf = vi – gt
0 = 20 – 9.8t
t = 2.04 s
t ½ = 1.02 s
Velocity at half rise time
vf = vi – gt = 20 – 9.8 (1.02) = 10 m/s
P = 5 kg.m/s
(c) The maximum height ∆y = vi t+ (1/2)gt2
∆y = 20(2.04) + (1/2)(-9.8)(2.04)2 = 40.8 - 20.4= 20.4 m
The half height ∆y1/2 = 10.2 m
vf2 = vi2 + 2g ∆y = (20)2 - 2 (9.8) (10.2) = 200.08
vf = 14.14 m/s
P = 7.07 kg.m/s
22
Questions
 A 15 g pellet moving at 600 m/s hits a stationary 100 g block and becomes embedded in the
block. With what speed do they move off together?
m1v1i + m2v2i = m1v1f + m2v2f
0.015(600) + 0.1 (0) = (0.015 + 0.1) vf
vf = 78.26 m/s
 A 0.5 kg ball traveling at 6 m/s collides head-on with a 1 kg ball moving in the opposite
direction at 12 m/s. The 0.5 kg ball moves away at 14 m/s in the opposite direction after the
collision. Find the velocity of the 1 kg ball after the collision.
m1v1i + m2v2i = m1v1f + m2v2f
0.5 (6) + 1 (-12) = 0.5 (-14) + 1 v2f
v2f = 1.29 m/s
 A plastic ball of mass 0.2 kg moves at 0.6 m/s, east. It collides with a 0.1 kg ball moving at 0.8
m/s, west. After the collision, the velocity of the 0.1 kg ball is 0.6 m/s, east. What is the velocity of
the 0.2 kg ball after the collision?
0.2 (0.6) + 0.1 (-0.8) = 0.2 v1f + 0.1 (0.6)
v1f = -0.1 m/s
23
Questions
A machine gun fires 35.0g bullets at a speed of 750.0m/s. If the gun can fire 200bullets/min,
what is the average force the shooter must exert to keep the gun from moving?
P = mv = 0.035 (750) = 26.25 kg.m/s
Time for one bullet t = 60/200 = 0.3 s
F = p/t = 26.25/0.3 = 86.5 N
 A 3.00kg particle is located on the x axis at x =−5.00m and a 4.00kg particle is on the x axis at
x =3.00m. Find the center of mass of this two–particle system.
The center of mss is at Xcm = (-15 + 12)/(3+4) = -0.43 m
 A 2.0kg particle has a velocity of v1 = (2.0i−3.0j) m/s, and a 3.0kg particle has a velocity
(1.0i+6.0j)m/s.
Find (a) the velocity of the center of mass and (b) the total momentum of the system.
Vcm = (2i-6j +3i+18j)/(5) = (5i +12j)/5 = (1i + 2.4j) m/s
24
25
26
2.1 Angular Position, velocity and acceleration
Rigid Object
A rigid object is one that is “non deformable” that is, the relative posed
remains constant.
Rotation
When a disc rotates about a fixed axis with an angle Ɵ, it moves along an
arc s where s = r Ɵ and Ɵ is measured by radians.
We can convert between degrees and radians using:
Ɵ (rad) = (π/180) Ɵ (degree)
27
2.1 Angular Position, velocity and acceleration
Angular Displacement
As a result of rotation the point at position p1 moves to position p2.
The angular displacement is expressed by
∆Ɵ = Ɵ2 – Ɵ1
Angular Speed
The angular speed w is the ratio of the angular displacement to the time
interval and has the units of (s-1).
w = ∆Ɵ/ ∆t
w = dƟ/dt
28
2.1 Angular Position, velocity and acceleration
Angular Acceleration
The angular acceleration α is the change of angular velocity in time.
α = ∆w/ ∆t
α = dw/ dt
29
2.2 Rotational Kinematics
The equations of rotational motion are
30
2.3 The Relation Between Linear and Rotational
There is a relation between linear and rotational quantities.
Angular velocity and Tangential velocity
Every point on the rotating object moves in the same direction and travels
the same distance s which means its velocity is
v = ds/dt
But s = r Ɵ
Which means that
v = d(rƟ)/dt = r dƟ/dt
v = rw
Which means that all points of the object have the same angular speed but different
tangential speed because they have different distances from the center of the object.
31
2.3 The Relation Between Linear and Rotational
Angular acceleration and Linear Acceleration
The tangential acceleration of any point is resulted from change in
tangential velocity.
at = dv/dt = d(rw)/dt = r dw/dt
at = rα
Where the rotational acceleration is the result from changing the direction of the
velocity with time.
ar = v2/r = r2w2/r
ar = rw2
The total acceleration
a = (at2 + ar2)1/2 = (r 2 α2 + r 2 w4 )1/2 = r(α2 + w4 )1/2
32
2.4 Rotational Kinetic Energy
The kinetic energy of the ith particle of the rigid object
KEi = (1/2)mivi2
The total kinetic energy of the object
KE = ∑iKEi
The moment of inertia I is
I = ∑imiri2
Which means
KE = ∑iKEi = ∑i (1/2)mivi2 = ∑i (1/2) miri2wi2 = ∑i (1/2)Iwi2
33
2.5 Moment of Inertia
The moment of inertia of an object can be found by dividing the object into
small elements each of mass ∆m then taking the limit where ∆m
approaches zero.
I = lim∆m0∑imiri2 = ∫r2dm
Uniform Thin Ring
The ring is thin we could neglect the thickness which means that all
elements have the same distance from the axis.
I = ∫r2dm = R2 ∫dm = MR2
34
2.5 Moment of Inertia
Long Thin Rod Rotating about an axis passing through the center
The mass dm is related to the mass per unit length λ
dm = λ dx = (M/L)dx
I = ∫r2dm = (1/12)ML2
Long Thin Rod Rotating about an axis passing through the end
I = ∫r2dm = (1/3)ML2
35
2.5 Moment of Inertia
36
2.6 Torque
Torque, also called moment or moment of force is the tendency of a force to
rotate an object about an axis. Just as a force is a push or a pull, a torque
can be thought of as a twist.
A torque is a force exerted at a distance from the axis of rotation; the
easiest way to think of torque is to consider a door. When you open a door,
where do you push?
If you exert a force at the hinge, the door will not move; the easiest way to
open a door is to exert a force on the side of the door opposite the hinge,
and to push or pull with a force perpendicular to the door. This maximizes
the torque you exert.
37
2.6 Torque
Torque is the product of the distance from the point of rotation to where
the force is applied multiplied by the force multiplied by the sine of the
angle between the line you measure distance along and the line of the
force.
τ = r F sinƟ
The unit of torque is N.m.
If an object is under multiple torques then the net torque is the summation
of all torques.
τ = ∑iτ i
38
2.6 Rigid Object under Torque
If a particle of mass m rotating in a circle of radius r under the influence of a
tangential force Ft and a radial force Fr. The tangential force provides a tangential
acceleration a
Ft = mat
The torque about the center of the circle
τ = rFt = rmat = rmrα = mr2α = I α
39
2.7 Work and Energy in Rotational Motion
The work done on an object that is under a translational and rotational motion is
given by the change in both translation kinetic energy and the rotational kinetic
energy.
W = ∆KEt + ∆KEr = [(1/2)mvf2 – (1/2)mvi2] + [(1/2)Iwf2 – (1/2)Iwi2]
40
Examples
-A disc rotates at 1200 rev/min about its central axis. What is the linear speed of a
point 3.0 cm from its center?
∆θ = 1200 x 2π = 7536 rad
ω = 7536 / 60 = 125.6 rad/s
vt = 3.768 m/s
-During a certain period of time, the angular position of a swinging door is described
by θ = 5.00 + 10.0 t + 2.00 t², where θ is in radians and t is in seconds. Determine
the angular position, angular speed, and angular acceleration of the door (a) at t = 0
and (b) t = 3.00 s.
ω = 10 + 4 t
α=4
At t = 0 then θ = 5 , ω = 10 , α = 4
At t = 3 then θ = 53 , ω = 22 , α = 4
-A racing car travels on a circular track of radius 250 m.
If the car moves with a constant linear speed of 45.0 m/s,
find (a) its angular speed and (b) the magnitude of its acceleration.
ω = 0.18 rad/s , at = 0 , ar = 8.1
Total acceleration a = 8.1
41
Examples
-A bicycle travels 141 m along a circular track of radius 15 m. What is the angular
displacement in radians of the bicycle from its starting position?
∆θ = s / r = 9.4 rad
- A grindstone, originally rotating at 126 rad/s undergoes a constant angular
acceleration so that it makes 20.0 rev in the first 8.00 s. What is its angular
acceleration?
∆θ = 125.6 rad , ∆θ = ωit + (1/2) αt2
α = -27.6 rad/ s2
- A ball attached to a string starts at rest and undergoes a constant angular
acceleration as it travels in a horizontal circle of radius 0.30 m. After 0.65 sec, the
angular speed of the ball is 9.7 rad/s, what is the tangential acceleration of the ball?
ωi = 0 , ωf = 9.7 rad/s , t = 0.65 s , ωf = ωi + αt , α = 14.9 rad/ s2
a = 4.97 m/s2
- Two motorcycles are riding around a circular track at the same angular velocity.
One motorcycle is at a radius of 15 m; and the second is at a radius of 18 m. What is
the ratio of their linear speeds, v2/v1?
ω1 = ω2 , v1 r1 = v2 r2
42
v1 15 = v2 18 , v2 /v1 = 1.2
Examples
- A 100 cm meter rule is pivoted at its middle point (that is, at the 50 cm point). If a
weight of 10 N is hanged from the 30 cm mark and a weight of 20 N is hanged from
its 60 cm mark, find out whether the meter rule will remain balanced about its pivot
or not?
τ1 = 10 (0.2) = 2 N.m
τ2 = 20(0.1) (-) = - 2 N.m
The meter rule will balance
-The net torque on the pulley about the axle is the
torque due to the 30 N force plus the torque due to
the 20 N force
τ1 = 20(r)sin(90) = 20r N.m
τ2 = 30(r)sin(-90) =-30r N.m
τ = -10r
43
Examples
-Find the total torque at point A then B
Around A:
τ = 0 + 80(0.5 L)sin(37) + 70(L)sin(120) + 0 + 60(0.5L) sin(-90)
τ = 0.3L + 60.6L – 30L = 60.9 L mN
Around B:
τ = 90Lsin(-140) + 80(0.5 L)sin(-143) + 0 + 0 + 60(0.5L) sin(90)
τ = -57.9L - 24.1L + 30L = -52L mN
44
Questions
 A propeller (arm length 1.2 m) starts from rest and begins to rotate
counterclockwise with a constant angular acceleration of size 2.7 rad/s2.
a. How long does it take for the propeller's angular speed to reach 5.7 rad/s?
b. How many revolutions does it take for the propeller's angular speed to reach 5.7
rad/s?
c. What is the linear speed of the tip of the propeller at 5.7 rad/s?
d. What is the linear, rotational and total acceleration of the tip of the propeller at
this point?
wi = 0 α = 2.7 rad/s2 wf = 5.7 rad/s r = 1.2 m
wf = wi + αt 
5.7 = 2.7 t 
t = 2.11 s
∆θ = wi t + (1/2) αt2  ∆θ = 6.01 rad
Number of revolutions = 0.96
v = rw  v = 6.84 m/s
a = rα  a = 3.24
The total acceleration
a = 39.12 m/s2
ar = rw2  ar = 38.99
45
Questions
An
airplane
propeller
is
rotating
at
1900
rev/min.
a. Compute the propeller's angular velocity in rad/s.
b. How long in seconds does it take for the propeller to turn through 30 degrees?
w = [1900 (2)(3.14)]/60 = 198.87 rad/s
∆θ = 30(2)(3.14)/180 = 1.05 rad
t = 0.0053 s
 A 500 N rock rests on an 8.0-m-long board that weighs 100 N. The board is
supported at each end. The support force at the right end is 3 times the support
force at the left end. How far from the right end is the rock standing?
F1 + F2 – 100 – 500 = 0  F1 + F2 = 600
4 F2 = 600  F2 = 150 N
The pivot point at the center
∑τ = 0
4F2sin(-90) + (4-x)(500)sin(-90) + F1(4)sin(90) = 0
-4F2 - 2000 + 500x + 12 F2 = 0
X = 1.6 m
46
Questions
 A uniform horizontal beam 6.0 m long and weighing 4 kg is supported at points p
and q each 1.0 m from opposite ends of the beam. Masses of 10 kg and 8 kg
are placed near p and q respectively, one on each end of the beam.
Calculate the reaction forces at p and q.
Fp + Fq – (8)(9.8) – (10)(9.8) – (4)(9.8) = 0
 Fp + Fq = 215.6
The pivot point is q
∑τ = 0
(1)(8)(9.8) – 2(4)(9.8) + 2Fp – 3(10)(9.8) = 0
78.4 – 78.4 + 2Fp – 294 = 0
Fp = 147 N
Fq = 68.6 N
47
Questions
A spot of paint on a tire moves in a circular path of radius 0.42 m. Through what
angle must the tire rotate for the spot to travel 1.48 m?
s = rθ
θ = 3.523 rad
 A disk with a 1.0-m radius reaches a maximum angular speed of 18 rad/s before it
stops 35 revolutions (220 rad) after attaining the maximum speed. how long did it
take the disk to stop?
t = 24.44 s
 How many revolutions does a disc make in in 5.44 seconds accelerating from 0 to
7.598 thousand rpm?
Number of revolutions = 688.9
 Two 0.51 kg cartons are at one end of a bin that is 0.69 m long. Where should a
1.8 kg carton be placed so that the center of mass of the three cartons is at the
center of the bin?
48
Should be at 0.098 m
Questions
An Oxygen molecule rotating in the xy plane about an axis passing through the
center of the molecule perpendicular to its length. The mass of each atom is
2.66x10-26 kg and at room temperature the average seperation between the two
atoms is 1.21x10-10 m. Calculate the moment of inertia of the molecule about the z
axis. If the angular speed of the molecule about the z axis is 4.6x1012 rad/s. what is
its kinetic energy?
I = ∑mr2 = m(d/2)2 + m(d/2)2 = 1.95x10-46 kgm2
KE =(1/2)Iw2
KE =(1/2)(1.95x10-46)(4.6x1012) = 2.06x10-21 J
49
Questions
 If A = 3N, determine the weight of B,C,D,E, and F
50
51
2
3.1 Vector Product
vector product is the product of two vectors in three-dimensional space. It results
in a vector which is perpendicular to both of the vectors being multiplied and
normal to the plane containing them.
Consider two vectors A and B in the standard unit-vector notation
The cross product A x B is a vector perpendicular on both vectors A and B where its
magnitude could be calculated by
A x B = |A | | B | sinθ
Where θ is the angle between the two vectors.
The vector product could be found by finding the determinant
53
3.1 Vector Product
The direction of the resulted vector could be found using the right hand rule.
The unit vectors i, j, and k from the given orthogonal coordinate system satisfy the
following equalities:
i×i=j×j=k×k=0
i×j=k
j×k=i
j × i = −k
k × j = −i
k×i=j
i × k = −j
54
3.2 Angular Momentum
Angular momentum (L) is a measure of an object's tendency to keep
rotating and to maintain its orientation. Mathematically it depends on the
object's mass, m, radius, r, and rotational velocity, v, and is proportional to
mvr.
L=rxp
L = r m v sinθ
Where θ is the angel between r and p
The total torque of a particle is
∑τ = r x ∑F = r x (dp/dt) = d(rxp)/dt = dL/dt
55
3.3 Angular Momentum of a Rotating Rigid Object
The angular momentum of a rotating object is
L = m r2 w = I w
Where I is the moment of inertia of the object
The total external torque is the change of angular momentum in time is
∑τext = dL/dt = d(Iw)/dt = Iα
56
3.4 Conservation of Angular Momentum
If the total external torque acting on a system equals zero then the angular
momentum (magnitude and direction) is conserved.
∑τext = 0  Li = Lf
57
3.5 The Motion of Gyroscopes and Tops
A gyroscope is a fascinating mechanical device with a lot of technological
applications. You can easily make a gyroscope from a wheel with a handle
put through the middle.
If you dangle the end of the handle from a string, the wheel doesn't flop
over but stays up straight! This might seem a bit odd because the weight is
pointing straight down. Instead it slowly moves around in a circle, that is
called precess. The angular momentum of the wheel points along its axis of
rotation.
58
3.5 The Motion of Gyroscopes and Tops
We know that
Then
τ = dL/dt
dL = L(dθ/dt)
τ = L(dθ/dt) = Iw(dθ/dt)
The torque due to gravity is mgh
mgh = Iw(dθ/dt)
(dθ/dt) = mgh/Iw
wp = mgh/Iw
Where h is the distance between the pivot point and the center of mass and
(dθ/dt) is called the precessional frequency wp
This is valid when w<<wp
59
3.7 Angular Momentum as a Fundamental Quantity
The angular momentum is an intrinsic property of atoms, molecules, and
their constituents, a property that is a part of their very nature.
The angular momentum of these systems have discrete values which are
multiples of the fundamental unit of angular momentum ћ = h/2π, where h
is called Planck’s constant.
Fundamental unit of angular momentum = ћ=1.054 × 10-34 kg.m2/s.
60
Examples
A light rigid rod 1.00 m in length rotates in the xy plane about a pivot
through the rod's center. Two particles of masses 4.00 kg and 3.00 kg are
connected to its ends. Determine the angular momentum of the system
about the origin at the time the speed of each particle is 5.00 m/s.
LTot = Lrod + L1 + L2
We are told the rod is a "lightweight" rod, which is another way of saying
its mass and its moment of inertia are small enough they may be ignored.
LTot = L1 + L2
The angular momentum for a "point particle" is
L = m r vt
L1 = (3.0 kg) (0.5 m) (5.0 m/s) = 7.5 kg m2/s
L2 = (4.0 kg) (0.5 m) (5.0 m/s) = 10.0 kg m2/s
61
LTot = L1 + L2 = 17.5 kg m2/s
Examples
A playground merry-go-round of radius R = 2.0 m has a moment of
inertia I = 250 kg m2 and is rotating at 10 rev/min. A 25-kg child jumps
onto the edge of the merry-go-round. What is the new angular speed of the
merry-go-round?
Lf = Li
Li = Iiwi
Ii = 250 kg m2
wi = 10 rev/min
Li = (250 kg m2) ( 10 rev/min) = 2 500 (kg m2 rev / min)
Lf = If wf
If = IMgR + Ichild
Ichild = m r2 = (25 kg) (2.0 m)2 = 100 kg m2
If = (250 + 100) kg m2 = 350 kg m2
62
wf = [ 2 500 kg m2 rev / min ] / [ 350 kg m2 ] = 7.14 rev / min
Examples
A uniform solid disk of mass 3.00 kg and radius 0.200 m rotates about a
fixed axis perpendicular to its face. If the angular frequency of rotation is
6.00 rad/s, calculate the angular momentum of the disk when the axis of
rotation passes through its center of mass
ICM = (1/2) M R2
ICM = (1/2) (3.00 kg) (0.200 m)2
ICM = 0.06 kg m2
L = I = ( 0.06 kg m2) (6.0 1/s)
L = 0.54 kg m2 / s
63
Examples
A 1000 kg car has four 10 kg wheels. When the car is moving, what
fraction of its total kinetic energy is due to rotation of the wheels about
their axles? Assume that the wheels have the same rotational inertial as
uniform disks of the same mass and size. Why do you not need to know the
radius of the wheels?
64
Examples
A solid cylinder of radius 10 cm and mass 12 kg starts from rest and rolls
without slipping a distance L=6 m down a roof that is inclined at angle
θ=30°. What is the angular speed of the cylinder about its center as it
leaves the roof?
65
Examples
Two disks of moments of inertia I1 and I2 and having angular velocities
ω 1 and ω 2 respectively are brought in contact with each other face to
face such that their axis of rotation coincides. The situation is as shown in
the figure just before the contact. What is the angular velocity of the
combined system of two rotating disks, if they acquire a common angular
velocity?
66
Questions

67
Questions
 Li = Lf
I1w1i + I2w2i = I1w1f + I2w2f
I1w1i + 0 = (I1+ I2)wf
wf = I1w1i/(I1+ I2)
KEi= (1/2) I1w1i2
KEf= (1/2) (I1+ I2) wf2 = (1/2) (I1+ I2) [I1w1i/(I1+ I2)]2
KEf= (1/2) (I1w1i) 2 /(I1+ I2)= KEi (I1 /(I1+ I2))
Which means tha the kinetic energy decreased
68
Questions

69
Questions
 Li = Lf
Li = mvl
Lf = (m+M)vfl
mvl = (m+M)vfl
vf=mv/(m+M)
KEi = (1/2)mv2
KEf = (1/2)(m+M)vf2
Fraction lost of kinetic energy
=[(1/2)mv2- (1/2)(m+M)vf2]/ (1/2)mv2
=[mv2- (m+M)vf2]/ mv2
=[mv2- (m+M) m2v2/(m+M)2]/ mv2
=M/(m+M)
70
Questions

v=dr/dt
v=5j
P=mv
P=10j
L=rxP
L=(6i+5tj)x(10j)=60k
71
Questions

Lgyro=Lspace
Igyrowgyro=Ispacewspace
20(100)=5x105(dθ/dt)
20(100)=5x105(30/dt)
dt=131 s
72
73
74
4.1 Equilibrium Conditions
Equilibrium means that the object is at rest or that the center of mass is moving
with constant velocity relative to the observer.
We will deal here with objects in Static Equilibrium where the object is at rest.
Static Equilibrium Conditions
An object is in static equilibrium (it is not moving) IF:
1) it is not translating (not moving up, down, left, or right) AND
2) it is not rotating (not spinning clockwise or counter clockwise)
(We are talking about motion in a 2D plane (x,y) here.)
75
4.1 Equilibrium Conditions
If a stationary mass is acted on by several forces, then in order to NOT translate,
the net force must be zero.
Fnet = 0  ∑Fx = 0 and ∑Fy = 0
This is not enough!!
Look at the figure below, the two forces are equal and the net force is zero but the
object will rotate. Which means we need to add another condition for static
equilibrium.
The net torque must be zero as well.
∑τ = 0
76
4.2 Center of Gravity
The center of gravity is located at the center of mass of an object as long as the
gravitational acceleration is uniform along the entire object.
77
4.3 Elastic Properties of solids
Deformation of objects happens when acted on by external forces. The
deformation could be change in shape, size or both.
Stress is the net force acting on an object per unit area.
Strain is the measure of degree of deformation.
Modulus describes the behavior of an object under stress or how easily
does it deforms
78
4.3 Elastic Properties of solids
Young Modulus is the length deformation modulus.
The stress F/A
The strain ∆L/L0
Y = stress / strain
79
4.3 Elastic Properties of solids
Shear Modulus is the shape deformation modulus. It measures the
resistance to motion in parallel planes within a solid When an object is
subjected to a force parallel to one of its faces while the other face is held
fixed by another force, the object is called deformed in shape.
The stress F/A
The strain ∆x/h
S = stress / strain
80
4.3 Elastic Properties of solids
Bulk Modulus is the volume deformation modulus. It measures the
resistance of solids and liquids to change their volumes. The object
undergoes a change in volume when it is subjected to equal forces applied
perpendicular over its entire surface.
The stress F/A
The strain ∆V/Vo
B = stress / strain
81
Examples
The system is at static equilibrium, find T1, T2, T3 and θ
The total force = zero
T1cos35 - 40=0  T1 = 40/cos35 = 40/0.82 = 48.78 N
T2 - T1sin35=0  T2 = 28 N
T3cosθ - 50=0
T3sinθ - T2 =0  T3sinθ - 28 =0
tan θ = 28/50  θ = 29.3
T3 = 57.3 N
82
Examples
 At what minimum angle can the ladder lean without slipping? The wall
is frictionless and there is friction between the floor and the ladder.
Forces:
1. Normal force at bottom of ladder
2. Friction force at bottom of ladder
3. Ladder’s weight
4. Normal force at top of ladder
Pivot: Choose bottom of ladder, Why?
Torques:
1. Due to ladder’s weight
2. Due to normal force at top of
ladder
m n1 – n2 = 0 Force, y:
n1 – mg = 0
Force, x:
Torque:
Ln2sinf – (L/2) mg cosf = 0
From the force equations we get n2= m mg.
Therefore, msinf – (1/2)cosf = 0
and so,
tanf = 1/(2m)
83
Examples
 What is the force on the elbow?
m = 6 kg
Assume biceps force acts
3.4 cm from pivot point O. The force we know least
about is the force on the elbow. So, let’s
take the elbow (O) as the pivot.
Net torque:
1
 L
Fm   mh  m  g
2
 d
Fua , x  0  0  0  0
L
mh  Fmd  mgL  0
2
Fua , y  Fm  mh g  mg  0 Fua , y  (m  mh ) g  Fm
84
Examples
A steel wire of 2 mm in a diameter is stretched by applying a force of
72N. Find the stress in the wire.
Diameter = 2 mm; radius r = diameter / 2 = 2/2 = 1 mm = 1 x 10 -3 m
; F = 72 N
The Stress = F/A = F/πr2 = 2.292 x 107 N m-2
A brass wire of length 5m and cross section 1mm2 is hung from a rigid
support, with a brass weight of volume 1000cm3 hanging from the other
end. Find the decrease in the length of the wire, when the brass weight is
completely immersed in water. (Ybrass = 1011 NM-2 ; G = 9.8 ms2 ; water = 1 g
cm-3 . )
When the weight is hung in water, it loses weight which is equal to the
weight of the liquid displaced, according to Archimedes, principle .
Weight of the body hung in a liquid or water = weight of the body in air weight of the liquid displaced .
The decrease in length = 0.49 mm
85
Examples
A rectangular plastic block of length 600 mm, width 20 mm and height
400 mm has its lower face fixed to a bench and a shearing force of 200 N is
applied to the upper face. If the upper face is displaced by 20 mm, find the
shear stress, the shear strain and the shear modulus.
∆x=20 mm = 0.02 m
h=400mm = 0.4 m
A= 20x600 = 12000 mm2 = 0.012 m2
Stress = F/A = 200/0.012 = 16666.7 N/ m2
Strain = ∆x/h = 0.02/0.4 = 0.05
S = stress/strain = 16666.7/0.05 = 333333.3 N/ m2
86
Questions
 Copper has a tensile strength of about 3.0 × 108 N/m2. (a) What is the
maximum load that can be hung from a copper wire of diameter 0.42 mm?
(b) If half this maximum load is hung from the copper wire, by what
percentage of its length will it stretch?
F/A = 3x108 N/m2
r = 0.21 x 10-3 m
The maximum load Fmax = 41.5422 N
F = 20.77 N
Y=(F/A)/(∆L/L )
Y=12x1010
∆L/L = 0.00125 = 0.125%
87
Questions
An elevator cable is to be made of a new type of composite developed by
Acme Laboratories. In the lab, a sample of the cable that is 2.00 m long
and has a cross-sectional area of 0.200 mm2 fails under a load of 1000 N.
The actual cable used to support the elevator will be 20.0 m long and have
a cross-sectional area of 1.20 mm2. It will need to support a load of 20,000
N safely. Will it?
Stress = F/A = 1000/(0.2) = 500 N/mm2
Stress = F/A = 20000/(1.2) = 16666.7 N/mm2
It will fail to support the load, the stress is larger
88
Questions
A ladder with length L and mass m rests against
a wall. Its upper end is a distance h above the
ground. The center of gravity of the ladder is
one-third of the way up the ladder.
A firefighter with mass M climbs halfway up the
ladder. Assume that the wall, but not the ground, is frictionless.
What is the force exerted on the ladder by the wall and by the ground ?
89
Questions
∑F=0
Fw – Fgx = 0
Fw = Fgx
Fgy – mg – Mg = 0
Fgy = (m+M)g
∑ τ=0
mg (a/3) + Mg (a/2) – Fwh = 0
Fwh = ag ((m/3)+(M/2))
Fw = (g/h) ((m/3)+(M/2))
90
91
92
5.1 Newton’s Law of Universal Gravitation
Before Newton, scientists studied the motion of planets but did not have any idea
about the forces between them.
Then Newton discovered gravity when an apple fell on his head and he suggested
that there are forces between planets that keep them together and moving in the
same path.
Newton’s Law of Universal Gravitation states that every point mass in the
universe attracts every other point mass with a force that is directly proportional to
the product of their masses and inversely proportional to the square of the distance
between them.
where
93
5.2 Free Fall Acceleration
In the same way, Earth attracts objects with
a gravitational force.
When the object is on the surface of the Earth
Fg 
mg 
g
GMe m
re2
GMe m
re2
GMe
re2
If the object is not on the surface.
g
GMe
re  h 2
94
5.3 Kepler’s Laws and Motion of Planets
Kepler's first law
The path of each planet around the sun is an ellipse
with the sun at one focus. If F1 and F2 are the two foci,
P is the planet and S is the sun:
F1P + F2P = constant.
Kepler's second law
A line joining the sun to the planet sweeps out equal
areas in equal intervals of time. If the time taken by
a planet to travel from P1 to Q1 is equal to the time
taken to travel from P2 to Q2, the areas covered
are equal (shaded region).
Kepler's third law
A planet moves around the sun in such a way that the square of its time period is
proportional to the cube of the semi-major axis of its elliptical orbit. If T is the time
period of revolution and 'a' the semi-major axis then, for circular orbits
95
2
3
a = r (radius) , T =Kr
5.3 Kepler’s Laws and Motion of Planets
Deriving Kepler's third law from Newton’s law
Assuming the path of the planet around the sun is circular (in real life they are
almost circular except for Mercury and Pluto)
We know that ac=v2/r
And Newton’s law is F= ma
Where Ms is the mass of the sun
and Mp is the mass of the planet.
The orbital speed is 2лr/T
Where T is the period or the time to
complete a whole revolution
F
GM s M p
M pac 
r2
GM s M p
v2 GM s
 2
r
r
r2
GM s
 2r 

 
r
 T 
4 2 3
2
T 
r
GM s
2
T  Kr
2
3
96
5.4 Gravitational Field
The field lines are directed radically inwards, because at any point in the
Earth’s field, a body will feel a force directed toward the centre of the
Earth. The field lines become more spread out as the distance from the
Earth increases.
GMe
g  2  9.8 N / kg
re
97
5.5 Gravitational Potential Energy
The general expression for gravitational potential energy arises from the
law of gravity and is equal to the work done against gravity to bring a
mass to a given point in space. happens when acted on by external forces.
U  U f  U i   F (r )dr
rf
ri
U f Ui 
 GMe m
r r 2 dr
i
rf
U f  U i  GMe m 
rf
ri
When ri =infinity then Ui=0
Uf 
dr
r2
 GMe m
r
In the case of multiple objects U is the work needed to separate
those particles, for example for 3 particles: Utot=U12+U13+U23
98
5.6 Energy Considerations in Planetary and Satellite Motion
E  K U
The total energy of a planet orbiting is
Where
GMm
mv2
Fg  2  ma 
r
r
Which leads to
E
Then
1 2 GMm
mv 
2
r
K
GMm
2r
E  K U
E
GMm GMm

r
2r
 GMm
E
2r
In case of elliptical orbits where a is the semi major axis
E
 GMm
2a
99
Examples
The free-fall acceleration on the surface of the Moon is about one-sixth
that on the surface of the Earth. If the radius of the Moon is about 0.250
RE, find the ratio of their average densities.
gmoon/gearth = 1/6
6GMmoon/rmoon2= GMearth/rearth2
M=Vρ
ρmoon/ ρearth = 2/3
100
Examples
Plaskett’s binary system consists of two stars that revolve in a circular
orbit about a center of mass midway between them. This means that the
masses of the two stars are equal . Assume the orbital speed of each star
is 220 km/s and the orbital period of each is 14.4 days. Find the mass M of
each star. (For comparison, the mass of our Sun is 1.99  1030 kg.)
V=s/T=2πr/T
r=4.359x1010 m
V2/r=GM/(2r)2
M=1.265x1032 kg
101
Examples
Three uniform spheres of mass 2.00 kg, 4.00 kg, and 6.00 kg are placed
at the corners of a right triangle. Calculate the resultant gravitational
force on the 4.00-kg object, assuming the spheres are isolated from the rest
of the Universe.
F64=Gm6m4/r264
F64=G(6)(4)/16
F64=10.005x10-11N
F24=Gm2m4/r224
F24=G(2)(4)/9
F24=5.93x10-11N
Ftot=-10.005x10-11i+5.93x10-11j
102
Examples
Three objects of equal mass are located at three corners of a square of
edge length. Find the gravitational field at the fourth corner due to these
objects.
gx=(Gm/l2)+(Gmcos45/(2l2))
gx=(Gm/l2)+ (0.707Gm/2l2)
gx=(Gm/l2)+0.354 (Gm/l2)=9.03x10-11(m/l2)
gy=(Gm/l2)+(Gmcos45/(2l2))
gy=(Gm/l2)+(0.707Gm/2l2)
gy =(Gm/l2)+0.354 (Gm/l2)=9.03x10-11(m/l2)
103
Questions
If the mass of Mars is 0.107 ME and its radius is 0.53 RE, estimate the
gravitational field g at the surface of Mars.
gmars = GMmars/rmars2
gmars = G0.107Mearth/(0.53rearth)2
gmars = G0.107Mearth/(0.53rearth)2
gmars = 0.381GMearth/rearth2
gmars = 0.381gearth
gmars = 0.381(9.8)=3.733 N/kg
104
Questions
During a solar eclipse, the Moon, Earth, and Sun all lie on the same line,
with the Moon between Earth and the Sun. (a) What force is exerted on the
Moon by the Sun? (b) What force is exerted on the Moon by Earth? (c)
What force is exerted on Earth by the Sun?
Where mass of sun =1.991x1030 kg, mass of moon =7.36x1022 kg, mass of
earth =5.98x1024 kg, Distance between moon and sun = 1.496x1011 m
Distance between moon and earth = 3.81x1010 m
Fmoon-sun=Gmmms/rms2
Fmoon-sun=G(7.36x1022)(1.991x1030)/(1.496x1011)2
Fmoon-sun=43.67x1019N
Fmoon-earth=20.22x1015N
Fearth-sun=22.54x1023N
105
Questions
A satellite of Earth has a mass of 100 kg and is at an altitude of 2.0 x 106
m. (a) What is the potential energy of the satellite-Earth system? (b) What
is the magnitude of the gravitational force exerted by Earth on the
satellite?
U=GMm/r
U=GME (100)/(RE+(2.0 x 106 ))
U=G(5.98x1024 (100)/(6371000+(2.0 x 106 ))
U=4.76x109 J
F=GMm/r2
F=568.6 N
106
107
108
6.1 Periodic and Simple Harmonic Motion
Periodic motion is the motion which is repeated in equal intervals of time.
Simple harmonic motion is a special type of periodic motion where a force acting
on the object always acting towards the position of equilibrium.
109
6.2 Motion of an Object Attached to a Spring
When a pulling force is applied to a spring it stretches and undergoes a
displacement of x from its original, or “unstrained”. Also when a pushing force is
applied to the spring, and it again undergoes a displacement from its unstrained
length.
Experiment reveals that for relatively small displacements, the applied force
required to stretch or compress a spring is directly proportional to the displacement
x.
F=-kx
Where the constant k is called the spring constant.
110
6.2 Motion of an Object Attached to a Spring
When the restoring force has the mathematical form given by F=–kx, the type of
friction-free motion illustrated in the Figure below and is called as “simple harmonic
motion.” By attaching a pen to the object and moving a strip of paper past it at a
steady rate, we can record the position of the vibrating object as time passes.
F = ma
F = - kx
ma = - kx
ax = - (k/m) x
111
6.3 The Particle in Simple Harmonic Motion
Simple harmonic motion, like any motion, can be described in terms of
displacement, velocity, and acceleration.
Using a reference circle (radius=A) that indicates how to determine the
displacement.
The ball starts on the x axis at x=+A and moves through the angle θ in a time t.
Since the circular motion is uniform, the ball moves with a constant angular speed
w (in rad/s). Therefore, the angle has a value (in rad) of θ =wt. The displacement x
can be written by:
x = A cos θt = A cos wt
If the object doesn’t start at zero angle we add the phase φ
x = A cos (wt+ φ)
As time passes, the shadow of the ball oscillates
between the values of x=+A and x=–A,
corresponding to the limiting values of +1 and –1
for the cosine of an angle. The radius A of the
reference circle, then, is the amplitude of the simple harmonic motion.
112
6.3 The Particle in Simple Harmonic Motion
As the ball moves one revolution or cycle around the reference circle, its shadow
executes one cycle of back-and-forth motion. For any object in simple harmonic
motion, the time required to complete one cycle is the period T.
w = 2л/T
Often, instead of the period, it is more convenient to speak of the frequency f of the
motion, the frequency being just the number of cycles of the motion per second.
f = 1/T
The velocity can be found from the displacement
v = dx/dt
v = -Aw sin (wt+ φ)
vmax = Aw
The acceleration can be found from the velocity
a=
a = dv/dt
cos(wt+ φ)= w2 x
amax = Aw2
Aw2
113
6.4 Energy in Simple Harmonic Motion
The object's kinetic energy is
K = (1/2)mv2 = (1/2)mω2A2sin2(ωt + φ)
Its potential energy is elastic potential energy. The elastic potential energy stored
in a spring displaced a distance x from its equilibrium position is U=(1/2)kx2. The
object's potential energy therefore is
U = (1/2)kx2 = (1/2)mω2x2 = (1/2)mω2A2cos2(ωt + φ)
The total mechanical energy of the object is
E = K+U = (1/2)mω2A2(sin2(ωt + φ)+cos2(ωt + φ)) = (1/2)mω2A2
The energy E in the system is proportional to the square of the amplitude
E = (1/2)kA2
It is a continuously changing mixture of kinetic energy and potential energy.
114
6.5 The Pendulum
An ideal simple pendulum consists of a point mass m suspended from a
support by a massless string of length L. If the mass is displaced from its
equilibrium position while keeping the string taut, it exhibits periodic
motion, moving in a vertical plane along a circular arc.
F = -mg (sinθ) = ma = md2s/dt2
And for small angles sin θ = θ ,
s=Lθ
Then
d2 θ /dt2 = -g θ / L
Which is equivalent to the simple harmonic motion equation
θ(t) = θmaxcos(ωt + φ)
Where
w2=g/L
115
6.5 The Pendulum
A physical pendulum is an object suspended in a uniform gravitational
field from a point other than its center of mass. The object can rotate
about an axis through the suspension point. When the CM is displaced
from its stable equilibrium point under the support, the gravitational force
exerts a torque about the support, resulting in angular acceleration. The
CM accelerates towards its equilibrium position. The object exhibits
periodic motion. For small displacements, when sinθ ~ θ, the motion is
simple harmonic, θ(t) = θmaxcos(ωt + φ), with ω2 = (mgd)/I. Here I is the
moment of inertia of the object about the axis of rotation through the
support and d is the perpendicular distance of the CM from the axis of
rotation through the support.
116
Examples
 A particle oscillates with simple harmonic motion, so that its
displacement varies according to the expression x = (5 cm)cos(2t + π/6)
where x is in centimeters and t is in seconds.
At t = 0 find
(a) the displacement of the particle,(b) its velocity, and(c) its acceleration.
(d) Find the period and amplitude of the motion.
(a) The displacement as a function of time is
x(t) = Acos(ωt + φ).
Here ω = 2/s, φ = π/6, and A = 5 cm.
The displacement at t = 0 is x(0) = (5 cm)cos(π/6) = 4.33 cm.
(b) The velocity at t = 0 is v(0) = -ω(5 cm)sin(π/6) = -5 cm/s.
(c) The acceleration at t = 0 is a(0) = -ω2(5 cm)cos(π/6) = -17.3 cm/s2.
(d) The period of the motion is T = πs, and the
amplitude is 5 cm.
117
Examples
 A 20 g particle moves in simple harmonic motion with a frequency of 3
oscillations
per
second
and
an
amplitude
of
5cm.
(a) Through what total distance does the particle move during one cycle of
its motion? (b) What is its maximum speed? Where does that occur?
(c) Find the maximum acceleration of the particle. Where in the motion
does the maximum acceleration occur?
(a) The total distance d the particle moves during one cycle is from x = -A to x = +A
and back to x = -A, so d = 4A = 20 cm.
(b)
The
maximum
speed
of
the
particle
is
vmax = ωA = 2πfA = 2π 15 cm/s = 0.94 m/s. The particle has maximum speed when it
passes through the equilibrium position.
(c)
The
maximum
acceleration
of
the
particle
is
amax = ω2A = (2πf)2A = 17.8 m/s2. The particle has maximum acceleration at the turning
points, where it has maximum displacement.
118
Examples
 A 1kg mass attached to a spring of force constant 25N/m oscillates on a
horizontal frictionless track. At t = 0 the mass is released from rest at x = 3cm, that is the spring is compressed by 3cm. Neglect the mass of the
spring. Find (a) The period of its motion, (b) the maximum value of its
speed and acceleration, and (c) the displacement, velocity and acceleration
as a function of time.
(a) The period is T = 2πSQRT(m/k) = 2πSQRT(1 s2/25) = 1.26 s.
(b) The angular acceleration is ω = SQRT(k/m) = 5/s.
The maximum speed is vmax = ωA = 15 cm/s.
The maximum acceleration of the particle is amax = ω2A = 0.75 m/s2.
(c) x(t) = Acos(ωt + φ) = (3 cm)cos((5/s)t + π) = -(3 cm)cos((5/s)t)
v(t) = -ωAsin(ωt + φ) = (15 cm/s)sin((5/s)t)
a(t) = -ω2Acos(ωt + φ) = (0.75 m/s2)cos((5/s)t)
119
Examples
 A child swings on a playground swing with a 2.5 m long chain.
a) What is the period of the child’s motion?
b) What is the frequency of the vibration?
a) for a pendulum T = 2*π*sqrt (l/g)
= 2*3.14*sqrt(2.5/9.8)
T=3.1735 sec/swing
b) f = 1/T = 1/3.17=.315 swings/sec=.315 Hz
 By what factor should the length of a simple pendulum be changed if the period of
vibration were to be tripled?
T2=3T1
2π (l2/g)(1/2)=(3) 2π (l1/g)(1/2)
(l2)(1/2)=(3) (l1)(1/2)
(l2)=(9) (l1)
120
Questions
 The position of a
particle is given by
x = 2.5 cos πt,
where x is in meters
and t is in seconds.
(a) Find the maximum
speed and maximum
acceleration of the particle.
(b) Find the speed and
acceleration of the
particle when x = 1.5 m.
121
Questions
A particle moves in a circle of radius 15 cm, making 1 revolution every 3 s. (a) What is the
speed of the particle? (b) What is its angular velocity ω? (c) Write an equation for the x
component of the position of the particle as a function of time t, assuming that the particle is on
the positive x axis at time t = 0.
A 2.4-kg object is attached to a horizontal spring of force constant k = 4.5 kN/m. The spring is
stretched 10 cm from equilibrium and released. Find its total energy.
122
Questions
 Find the total energy of a 3-kg object oscillating on a horizontal spring with an amplitude of
10 cm and a frequency of 2.4 Hz.
A 3-kg object oscillating on a spring of force constant 2 kN/m has a total energy of 0.9 J. (a)
What is the amplitude of the motion? (b) What is the maximum speed?
A 2.4-kg object is attached to a horizontal spring of force constant k = 4.5 kN/m. The spring is
stretched 10 cm from equilibrium and released. Find (a) the frequency of the motion, (b) the
period, (c) the amplitude, (d) the maximum speed, and (e) the maximum acceleration. (f) When
does the object first reach its equilibrium position? What is its acceleration at this time?
123
Questions
A 3-kg object attached to a horizontal spring oscillates with an amplitude A = 10 cm and a
frequency f = 2.4 Hz. (a) What is the force constant of the spring? (b) What is the period of the
motion? (c) What is the maximum speed of the object? (d) What is the maximum acceleration of
the object?
 A simple pendulum of length L is released from rest from an angle φ0. (a) Assuming that the
pendulum undergoes simple harmonic motion, find its speed as it passes through φ = 0. (b)
Using the conservation of energy, find this speed exactly. (c) Show that your results for (a) and
(b) are the same when φ0 is small. (d) Find the difference in your results for φ0 = 0.20 rad and L
= 1 m.
124
Questions
A 3-kg plane figure is suspended at a point 10 cm from its center of mass. When it is
oscillating with small amplitude, the period of oscillation is 2.6 s. Find the moment of inertia I
about an axis perpendicular to the plane of the figure through the pivot point.
The displacement of a particle at t = 0.25 s is given by the expression
x = (4.0 m) cos (3.0 p t + p), where x is in meters and t is in seconds. Determine
(a) the frequency and period of the motion,
(b) the amplitude of the motion,
(c) the phase constant, and
(d) the displacement of the particle at t = 0.25 s
x=Acos(wt+ φ)
w=3π
T=0.66 s
f=1.5 Hz
A=4 m
φ= π
x at t=0.25
x=2.83 m
125
Questions
 A block of unknown mass is attached to a spring of spring constant 6.50 N/m and undergoes
simple harmonic motion with an amplitude of 10.0 cm. When the mass is halfway between its
equilibrium position and the endpoint, its speed is measured to be + 30.0 cm/s. Calculate
(a) the mass of the block,
(b) the period of the motion, and
(c) the maximum acceleration of the block.
When x = A/2 = 5cm = 0.05 m then v=0.3 m/s
x = Acos(wt) then cos(wt) = 0.5
w=(k/m)(1/2)
m=0.5 kg
T=1.8 s
amax=Aw2
126
127