Chapter 12: Kinematics of a Particle

[12.4, 12.5, 12.6]: Curvilinear motion: Rectangular Components & Projectile Motion

Rectangular Components fixed frame of reference

The position of the particle can be defined at any instant by the

position vector:

r = x i + y j + z k

The x, y, z components may all be functions of time, i.e.,

x = x(t), y = y(t), and z = z(t) .

The magnitude of the position vector is: r = (x2 + y

2 + z

2)

0.5

The direction of r is defined by the unit vector: ur = (1/r)r

The velocity vector is the time derivative of the position vector:

v = dr/dt = d(xi)/dt + d(yj)/dt + d(zk)/dt

Since the unit vectors i, j, k are constant in magnitude and

direction, this equation reduces to v = vx i + v

y j + v

z k

where:

vx = = dx/dt

vy = y = dy/dt

vz = z = dz/dt

x •

•

•

•

The magnitude of the velocity vector is:

v = [(vx)

2 + (v

y)

2 + (v

z)

2]

0.5

The direction of v is tangent to the path of motion.

The acceleration vector is the time derivative of the

velocity vector (second derivative of the position vector):

•

•

a = dv/dt = d2r/dt

2 = a

x i + a

y j + a

z k

ax = v

x = x = dv

x /dt

ay = v

y = y = dv

y /dt

az = v

z= z = dv

z /dt

•

•

•

••

••

••

The magnitude of the acceleration vector is:

a = [(ax)

2 + (a

y)

2 + (a

z)

2 ]

0.5

The direction of a is usually not tangent to the path of the

particle.

Projectile Motion:

KINEMATIC EQUATIONS: HORIZONTAL MOTION:

Since ax = 0, the velocity in the horizontal direction remains constant (vx = vox)

and the position in the x direction can be determined by:

x = xo + (vox) t

KINEMATIC EQUATIONS: VERTICAL MOTION:

Since the positive y-axis is directed upward, ay = – g. Application of the

constant acceleration equations yields:

vy = voy – g t

y = yo + (voy) t – ½ g t2

vy2 = voy

2 – 2 g (y – yo)

[12.7]: Curvilinear motion using n-t coordinate system

ACCELERATION IN THE n-t COORDINATE SYSTEM:

The magnitude is determined by taking the time derivative of the path

function, s(t):

v = v ut where v = s = ds/dt

Here v defines the magnitude of the velocity (speed) and

ut defines the direction of the velocity vector.

VELOCITY IN THE n-t COORDINATE SYSTEM:

The velocity vector is always tangent to the path of motion (t-direction)

Acceleration is the time rate of change of velocity:

a = dv/dt = d(vut)/dt = vu

t + vu

t

. .

a = v ut + (v

2/) u

n = a

t u

t + a

n u

n, a = at u

t + a

n u

n

•

.

CASES OF MOTION:

• The tangential component is tangent to the curve and in the direction of

increasing or decreasing velocity:

at = v = dv/dt or a

t ds = v dv

• The normal or centripetal component is always directed toward the

center of curvature of the curve. an = v

2/

• The magnitude of the acceleration vector is

a = [(at)

2 + (a

n)

2]

0.5

1) The particle moves along a straight line.

=> an = v

2/ = = a = a

t = v

The tangential component represents the time rate of change in the

magnitude of the velocity.

2) The particle moves along a curve at constant speed.

at = v = 0 => a = a

n = v

2/

.

.

.

3) The tangential component of acceleration is constant, at = (a

t)

c.

In this case:

s = so + v

o t + (1/2) (a

t)

c t2

v = vo + (a

t)

c t

v2 = (v

o)

2 + 2 (a

t)

c (s – s

o)

As before, so and v

o are the initial position and velocity of the particle at t = 0.

The normal component represents the time rate of change in the

direction of the velocity.

4) The particle moves along a path expressed as y = f(x).

The radius of curvature, at any point on the path can be

calculated from

[12.8]: Curvilinear motion using cylindrical components

We can express the location of P in polar coordinates as r = r ur

Note that the radial direction, r, extends outward from the fixed

origin, O, and the transverse coordinate, is measured counter-

clockwise (CCW) from the horizontal.

VELOCITY in POLAR COORDINATES

The instantaneous velocity is defined as:

v = dr/dt = d(rur)/dt

v = r ur + r

dur

dt

.

Using the chain rule:

dur/dt = (du

r/d)(d/dt)

We can prove that dur/d = u

θ so du

r/dt = u

θ

Therefore: v = rur + ru

θ

Thus, the velocity vector has two components: r, called the radial

component, and r called the transverse component. The speed

of the particle at any given instant is the sum of the squares of

both components or

v = [(r )2 + ( r )

2 ]1/2

The instantaneous acceleration is defined as:

a = dv/dt = (d/dt)(rur + ru

θ)

After manipulation, the acceleration can be

expressed as

a = (r – r 2)u

r + (r + 2r)u

θ

ACCELERATION in POLAR COORDINATES

.. ..

The term (r – r 2)

is the radial acceleration

or ar

The term (r + 2r) is the transverse

acceleration or a

..

..

The magnitude of acceleration is: a = [(r – r 2)

2 + (r + 2r)

2 ]1/2

.. ..

Cylindrical Coordinates:

r = r ur + z u

z

Sections 12.9 and 12.10 Skipped

Chapter 13: Equations of Motion [FBD and Kinetic diagram]

[13.1 – 13.3]: Newton’s 2nd Law of Motion, Equation of Motion, Equation of Motion for

System of Particles:

F = m a

F = ma

F = maG

[13.4]: Equation of Motion: Rectangular Coordinates:

F = ma or Fx i + F

y j + F

z k = m(a

x i + a

y j + a

z k)

Scalar equations: Fx =

ma

x,

Fy =

ma

y, and

F

z =

ma

z

Curvilinear motion equations (n-t coordinates)

The tangential acceleration, at = dv/dt, represents the time rate of change in the

magnitude of the velocity. Depending on the direction of Ft, the particle’s speed will either be

increasing or decreasing.

Since the equation of motion is a vector equation,

F = ma,

it may be written in terms of the n & t coordinates as:

Ftu

t + F

nu

n+ F

bu

b = ma

t+ma

n

This vector equation will be satisfied provided the individual components on each side of the

equation are equal, resulting in the two scalar equations:

Ft = ma

t and F

n = ma

n

The normal acceleration, an = v

2/, represents the time rate of

change in the direction of the velocity vector. Remember, an always acts toward the path’s center of

curvature. Thus, Fn will always be directed toward the center of the path.

SOLVING PROBLEMS WITH n-t COORDINATES

Recall, if the path of motion is defined as y = f(x), the radius of

curvature at any point can be obtained from

• Use n-t coordinates when a particle is moving along a known, curved path.

• Establish the n-t coordinate system on the particle.

[13.6]: Curvilinear motion equations (Cylindrical coordinates)

• Apply the equations of motion in scalar form and solve.

• It may be necessary to employ the kinematic relations:

at = dv/dt = v dv/ds a

n = v

2/

• Draw free-body and kinetic diagrams of the particle. The normal acceleration (an)

always acts “inward” (the positive n-direction). The tangential acceleration (at)

may act in either the positive or negative t direction.

Equilibrium equations or “Equations of Motion” in cylindrical

coordinates (using r, , and z coordinates) may be expressed in

scalar form as:

Fr = ma

r = m (r – r

2 )

F = ma

= m (r + 2 r )

Fz = ma

z = m z

[No HW assigned for section 13.6] and Section 13.7 Skipped

Chapter 14: Work of a Force, the Principal of Work and Energy and Conservative

Forces, Potential Energy and Conservation of Energy

Work & Kinetic Energy: Principal of Work & Energy, Work of a Force, and Power & Systems of

Particles

[14.1]: Work Done by Particular Forces:

2

11 2  

s

sU F ds→ =

Work of a weight, work of spring force, work of friction caused by sliding forces

WORK OF A WEIGHT: U1-2 = - W (y2 − y1) = - W Δy

WORK OF A FORCE: U1-2 = Fc cos Ɵ (s2 - s1)

WORK OF A SPRING FORCE: U1-2 = – [ 0.5 k (s2)2 – 0.5 k (s1)2 ]

[14.2]: Principals of Work & Energy:

T1 + U1-2 = T2 or 0.5 m (v1)2 + U1-2 = 0.5 m (v2)2

• Note that the principle of work and energy (T1 + U1-2 = T2) is not a vector equation! Each

term results in a scalar value.

• T1 and T2 are the kinetic energies of the particle at the initial and final position, respectively.

Thus, T1 = 0.5 m (v1)2 and T2 = 0.5 m (v2)2. The kinetic energy is always a positive scalar

(velocity is squared!).

• U1-2 is the work done by all the forces acting on the particle as it moves from point 1 to

point 2. Work can be either a positive or negative scalar.

• Both kinetic energy and work have the same units, that of energy! In the SI system, the unit

for energy is called a joule (J), where 1 J = 1 N·m. In the FPS system, units are ft·lb.

• The principle of work and energy cannot be used, in general, to determine forces directed

normal to the path, since these forces do no work

• The principle of work and energy can also be applied to a system of particles by summing the

kinetic energies of all particles in the system and the work due to all forces acting on the

system.

[14.3]: Principle of Work and Energy for a System of Particles:

[14.4]: Power and Efficiency:

P = dU/dt = (F • dr)/dt = F • (dr/dt) = F • v

e = (power output) / (power input)

e = (energy output) / (energy input)

Machines will always have frictional forces. Since frictional forces dissipate energy,

additional power will be required to overcome these forces. Consequently, the efficiency of

a machine is always less than 1.

Work of Friction caused by Sliding

0.5m (v)2 + P s – (

k N) s = 0.5m (v)

2

[14.5]: Potential Energy & Conservation of Energy/Conservative Forces

The “conservative” potential energy of a particle/system is typically written using the potential

function V. There are two major components to V commonly encountered in mechanical systems,

the potential energy from gravity and the potential energy from springs or other elastic elements.

Vtotal

= Vgravity

+ Vsprings

Select a Datum line of Reference

Potential energy is a measure of the amount of work a conservative force will do when a

body changes position

POTENTIAL ENERGY DUE TO GRAVITY

Vg = ± W y

ELASTIC POTENTIAL ENERGY

0.5k s2

V e =

Ve (where ‘e’ denotes an elastic spring) has the distance “s”

raised to a power (the result of an integration) or

[14.6]: Principle of Conservation of Energy and is expressed as:

Chapter 15 Kinetics of a Particle Impulse and Momentum

[Section 15.1]: Principal of Linear Impulse & Momentum

Principal of Impulse & Momentum, Impulse and momentum – scalar equations

m(vx)1 + ∫ F𝑋𝑡2𝑡1

dt = m(vx)2

m(vy)1 + ∫ F𝑌𝑡2𝑡1

dt = m(vy)2

m(vz)1 + ∫ F𝑍𝑡2𝑡1

dt = m(vz)2

• Establish the x, y, z coordinate system.

2 2 1 1 V T V T + = +

2

11 2 

t

tmV F dt mV+ =

• Draw the particle’s free body diagram and establish the direction of the particle’s initial and

final velocities, drawing the impulse and momentum diagrams for the particle. Show the

linear momenta and force impulse vectors.

• Resolve the force and velocity (or impulse and momentum) vectors into their x, y, z

components, and apply the principle of linear impulse and momentum using its scalar form.

• Forces as functions of time must be integrated to obtain impulses. If a force is constant, its

impulse is the product of the force’s magnitude and time interval over which it acts.

[Section 15.2]: Principle of Linear Impulse & Momentum for System of Particles

This equation states that the initial linear momentum of the system plus

the impulses of all External Forces acting on the system from t1 to t

2 is

equal to the systems’ final linear momentum

mi(v

i)

2 dt

Fi m

i(v

i)

1

t2

t1

= +

[Section 15.3]: Conservation of Linear Momentum for System of Particles

When the sum of external impulses acting on a system of objects

is zero, the linear impulse-momentum equation simplifies to:

mi(v

i)

1 = m

i(v

i)

2

[Section 15.4]: Impact:

Impacts, central impact, coefficient of restitution, impact energy losses, oblique impact,

Central Impact: Conservation of momentum and Coefficient of Restitution

(mA vA)1 + (mB vB)1 = (mA vA)2 + (mB vB)2

This provides one equation, but there are usually two unknowns, (vA)2 and (vB)2. So

another equation is needed. The principle of impulse and momentum is used to

develop this equation, which involves the coefficient of restitution, or e

The equation defining the coefficient of restitution, e, is

(vA)

1 - (v

B)

1

(vB)

2 – (v

A)

2

e =

Oblique Impact:

Sections 15.5-15.9 Skipped

Momentum of each particle is conserved in the direction perpendicular to the

line of impact (y-axis):

mA(v

Ay)

1 = m

A(v

Ay)

2 and m

B(v

By)

1 = m

B(v

By)

2

Conservation of momentum and the coefficient of restitution

equation are applied along the line of impact (x-axis):

mA(v

Ax)

1 + m

B(v

Bx)

1 = m

A(v

Ax)

2 + m

B(v

Bx)

2

e = [(vBx

)2 – (v

Ax)

2]/[(v

Ax)

1 – (v

Bx)

1]

In an oblique impact, one or both of the particles’ motion is at an

angle to the line of impact. Typically, there will be four

unknowns: the magnitudes and directions of the final velocities.

Chapter 16 Planar Rigid Body Motion: Translation and rotation

[16.1-16.2]: Planar Ridig Bodies and types of motion

Translation:

Note, all points in a rigid body subjected to translation move with the same velocity and

acceleration

The velocity at B is vB

= vA+ dr

B/A/dt

.

Now drB/A

/dt = 0 since r

B/A is constant. So, v

B = v

A, and by following

similar logic, aB

= aA.

The positions of two points A and B on a translating body can be related by

rB

= rA + r

B/A

where rA

& rB are the absolute position vectors defined from the fixed x-y coordinate

system, and rB/A

is the relative-position vector between B and A.

[16.3]: RIGID-BODY MOTION: ROTATION ABOUT A FIXED AXIS:

Similarly, angular acceleration is

= d2/dt

2 = d/dt or = (d/d) rad/s

2 +

When a body rotates about a fixed axis, any point P in the body travels along a

circular path. The angular position of P is defined by

The change in angular position, d, is called the angular

displacement, with units of either radians or revolutions. They are

related by

1 revolution = (2) radians

Angular velocity, , is obtained by taking the time derivative of angular

displacement: = d/dt (rad/s) +

Rotation about fixed axis

RIGID-BODY ROTATION: VELOCITY OF POINT P

If the angular acceleration of the body is constant, = C,

the equations

for angular velocity and acceleration can be integrated to yield the set of

algebraic equations below.

= 0 +

C t

= 0 +

0 t + 0.5

C t2

2 = (

0)

2 + 2

C ( –

0)

0 and

0 are the initial values of the body’s angular position and angular

velocity. Note these equations are very similar to the constant acceleration

relations developed for the rectilinear motion of a particle.

The magnitude of the velocity of P is equal to r (the text provides the

derivation). The velocity’s direction is tangent to the circular path of P.

In the vector formulation, the magnitude and direction of v can be

determined from the cross product of and rp . Here r

p is a vector

from any point on the axis of rotation to P:

v = × rp = × r

RIGID-BODY ROTATION: ACCELERATION OF POINT P

The acceleration of P is expressed in terms of its normal (an) and tangential

(at) components:

In scalar form, these are at = r and a

n =

2 r.

The tangential component, at, represents the time rate of change in the

velocity's magnitude. It is directed tangent to the path of motion.

The normal component, an, represents the time rate of change in the

velocity’s direction. It is directed toward the center of the circular path.

Using the vector formulation, the acceleration of P can also be defined by

differentiating the velocity.

a = dv/dt = d/dt × rP + × dr

P/dt

= × rP + × ( × r

P)

It can be shown that this equation reduces to:

a = × r – 2r = a

t + a

n

ROTATION ABOUT A FIXED AXIS: PROCEDURE

The magnitude of the acceleration vector is a = (at)

2 + (a

n)

2

• Establish a sign convention along the axis of rotation.

• If a relationship is known between any two of the variables (, , , or t),

the other variables can be determined from the equations: = d/dt

= d/dt d = d

• If is constant, use the equations for constant angular acceleration.

• To determine the motion of a point, the scalar equations v = r, at = r,

an =

2r , and a = (a

t)

2 + (a

n)

2 can be used

• Alternatively, the vector form of the equations can be used (with i, j, k

components).

v = × rP = × r

a = at + a

n = × r

P + × ( × r

P) = × r –

2r