Chapter 12: Kinematics of a Particle

Particle:

• A particle has a mass but negligible size and shape.

• A particle can undergo only translational motion.

• Objects like rockets, projectile, or vehicle may be considered as particles provided motion of the body is characterized by motion of its mass center (a point) and any rotation of the body is neglected.

• Chapter 12 – 15 are devoted to dynamics of particles.

Rigid Body:

A rigid body has mass, size and shape. A rigid body can undergo both translational and rotational motion.

Chapters 16 - 21 are devoted to rigid body dynamics.

12.1: Introduction - An Overview of Mechanics

Statics: The study of bodies in equilibrium.

Dynamics:1. Kinematics – concerned with the geometric aspects of motion2. Kinetics - concerned with the forces causing the motion

Mechanics: The study of how bodies react to forces acting on them.

RECTILINEAR KINEMATICS(Section 12.2)

Linear motion (also called rectilinear motion) is motion along a straight line, and can therefore be described mathematically using only one spatial dimension.

Position of a Particle

This is step – 1 in particle analysis.

The position of the particle at any instant, relative to the fixed origin, O, is defined by the position vector r, or the scalar s. Scalar s can be positive or negative. Typical units for r and s are meters (m) or feet (ft).

Displacement of a Particle

Displacement is a vector quantity.

Displacement = Final Position – Initial Position

The total distance traveled by the particle, sT, is a positive scalar that represents the total length of the path over which the particle travels.

The displacement of the particle is defined as the change in its position.

Vector form: r = r’ - r Scalar form: s = s’ - s

A particle is moving in a circular path of radius r. The magnitude of displacement and distance after half a circle would be:

(a) Zero

(b) π r

(c) 2 r

(d) 2π r

VELOCITY

Velocity is a measure of the rate of change in the position of a particle. It is a vector quantity (it has both magnitude and direction).

Mathematically, v = ds/dt

Direction: A positive value of velocity indicates that the particle is moving along the assumed positive direction of the coordinate axis. A negative value indicates that the particle moves in the opposite direction.

The magnitude of the velocity is called speed, with units of m/s or ft/s. Speed is a scalar quantity.

ACCELERATION

Acceleration is the rate of change in the velocity of a particle. It is a vector quantity. Typical units are m/s2 or ft/s2.

As the text indicates, the derivative equations for velocity and acceleration can be manipulated to get a ds = v dv

The instantaneous acceleration is the time derivative of velocity.

Vector form: a = dv / dt

Scalar form: a = dv / dt = d2s / dt2

Acceleration can be positive (speed increasing) or negative (speed decreasing, also called as decelerating).

SUMMARY OF KINEMATIC RELATIONS:RECTILINEAR MOTION

Differentiate position to get velocity and acceleration.

v = ds/dt a = dv/dt a = v dv/ds

Example: The acceleration of a particle is a function of its displacement and is given as: a = 600x(1 + kx2); where k is a constant. Given: v(x = 0) = 7.5 ft/s & v(x = 0.45ft) = 15ft/s. Find k.

Think: Which one of the following three equation will help you find k.

v = ds/dt a = dv/dt a ds = v dv

SPECIAL CASE: CONSTANT ACCELERATION MOTION

The three kinematic equations can be integrated for the special case

when acceleration is constant (a = ac) to obtain very useful equations.

A common example of constant acceleration is gravity; i.e., a body

freely falling toward earth. In this case, ac = g = 9.81 m/s2 = 32.2 ft/s2

downward. These equations are:

ta v v co

yieldst

oc

v

v

dtadvo

2coo

s

t(1/2) a t v s s yieldst

os

dtvdso

)s - (s2a )(v v oc

2o

2 yieldss

sc

v

v oo

dsadvv

The three kinematic equations are:

v = v0 + act

s = s0 +v0t + 0.5 act2

v2 = v20 + 2ac (s – s0)

Remember: To use the above three equations:

1) The particle must be moving with constant acceleration.

2) You must define a fixed coordinate axis and specify positive/ negative directions before you use these eqns.

3) The magnitudes and signs of s0, v0 and ac, used in the above three equations are determined w.r.t to the chosen coordinate axis.

Example: An airplane begins its take-off run at A with zero velocity and a constant acceleration a. Knowing that it becomes airborne 30 s later at B and that the distance AB is 2700 ft, determine (a) the acceleration a, (b) the take-off velocity @ B.

Example: Automobiles A and B are traveling in adjacent highway lanes and at t = 0 have the positions and speeds shown. Knowing that automobile A has a constant acceleration of 0.6 m/s2 and that B has a constant deceleration of 0.4 m/s2, determine (a) when and where A will overtake B, (b) the speed of each automobile at that time.

12.6 MOTION OF A PROJECTILE

MOTION OF A PROJECTILE (Section 12.6)

0

0

0

tan

x x

x

x

x

Horizontal Component

F Ma

a

dv

dtv Cons t

tan

y y

y

y

Vertical Component

F Ma Mg

a g

a Cons t

Constant acceleration motion along y axis

If we ignore the air drag/ air-friction on the particle, then the only force acting on a projectile is its weight

Projectile motion can be treated as two rectilinear motions, one in the horizontal direction experiencing zero acceleration and the other in the vertical direction experiencing constant acceleration (i.e., from gravity).

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

v = v0 + act

s = s0 +v0t + 0.5 act2

v2 = v20 + 2ac (s – s0)

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)

v = v0 + act

s = s0 +v0t + 0.5 act2

v2 = v20 + 2ac (s – s0)

IMPORTANT EQUATIONS FOR PROJECTILE MOTION

For Horizontal Motion: x = x0 + (v0x) t

For Vertical Motion:vy = v0y – g t

y = y0 + (v0y) t – ½ g t2

vy2 = v0y

2 – 2 g (y – y0)

Important: Time, “t” is the only common term that appears in the horizontal and vertical motion equations.

Example: A ski jumper starts with a horizontal take-off velocity of 25 m/s and lands on a straight landing hill inclined at 30º. Determine (a) the time between take-off and landing, (b) the length d of the jump.

Example: A golf ball is struck with a velocity of 80 ft/s as shown. Determine the distance, d, to where it will land.

Example: A pump is located near the edge of the horizontal platform shown. The nozzle at A discharges water with an initial velocity of 25 ft/s at an angle of 55° with the vertical. Determine the range of values of the height h for which the water enters the opening BC.

12.4: GENRAL CURVILINEAR MOTION - Introduction

The path of motion of a planeA roller coaster car

Curvilinear motion occurs when the particle moves along a curved path.

Curved road

CURVILINEAR MOTION

Since the path is often described in 3-D, vector analysis will be used to formulate the particle’s position, velocity and acceleration.

The position of the particle at any instant is designated by the vectorr = r(t). Both the magnitude and direction of r may vary with time.

A particle moves along a curve defined by the path function, s.

VELOCITY

Velocity represents the rate of change in the position of a particle.

The instantaneous velocity is the time-derivative of position

v = dr/dt .

The velocity vector, v, is always tangent to the path of motion.

ACCELERATION

Acceleration represents the rate of change in the velocity of a particle.

The instantaneous acceleration is the time-derivative of velocity:

a = dv/dt = d2r/dt2

ANALYSIS OF CURVILINEAR MOTION

1) Using Rectangular (x, y, z) Components

2) Using Normal and Tangential Components (n-t)

3) Cylindrical Coordinates (r, θ, z)

CURVILINEAR MOTION: RECTANGULAR COMPONENTS(Section 12.5)

It is often convenient to describe the motion of a particle in terms of its x, y, z or rectangular components, relative to a fixed frame of reference.

The magnitude of the position vector is: r = (x2 + y2 + z2)0.5

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) .

RECTANGULAR COMPONENTS: VELOCITY

The magnitude of the velocity vector is

v = [(vx)2 + (vy)2 + (vz)2]0.5

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

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 + vy j + vz k

where vx = = dx/dt, vy = = dy/dt, vz = = dz/dtx y z•••

RECTANGULAR COMPONENTS: ACCELERATION

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

The acceleration vector is the time derivative of the velocity vector (second derivative of the position vector):

a = dv/dt = d2r/dt2 = ax i + ay j + az k

where ax = = = dvx /dt, ay = = = dvy /dt,

az = = = dvz /dt

vx x vy y

vz z

• •• ••

•••

•

The magnitude of the acceleration vector is

a = [(ax)2 + (ay)2 + (az)2 ]0.5

Example: The box slides down the slope described by the equation y = (0.05x2) m, where x is in meters. vx = -3 m/s, ax = -1.5 m/s2 at x = 5 m. Find the y components of the velocity and the acceleration of the box at x = 5 m.

12.7 CURVILINEAR MOTIONNORMAL AND TANGENTIAL COMPONENTS

Objective: Determine the velocity and acceleration of a particle traveling along a curved path using normal and tangential components.

NORMAL AND TANGENTIAL COMPONENTS (Section 12.7)

When a particle moves along a curved path, it is sometimes convenient to describe its motion using coordinates other than Cartesian. When the path of motion is known, normal (n) and tangential (t) coordinates are often used.

In the n-t coordinate system, the origin, O, is located on the particle (the origin moves with the particle).

The t-axis is tangent to the path (curve) at the instant considered, positive in the direction of the particle’s motion.The n-axis is perpendicular to the t-axis with the positive direction toward the center of curvature of the curve.

NORMAL AND TANGENTIAL COMPONENTS (continued)

The positive n and t directions are defined by the unit vectors un and ut, respectively.

The center of curvature, O', always lies on the concave side of the curve.The radius of curvature, , is defined as the perpendicular distance from the curve to the center of curvature at that point.

VELOCITY IN THE n-t COORDINATE SYSTEM

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

v = v ut

ACCELERATION IN THE n-t COORDINATE SYSTEM

Acceleration is the time rate of change of velocity:a = dv/dt = d(vut)/dt = vut + vut

. .

. a = v ut + (v2/) un = at ut + an un.

After mathematical manipulation, the acceleration vector can be expressed as:

ACCELERATION IN THE n-t COORDINATE SYSTEM (continued)

So, there are two components to the acceleration vector:

a = at ut + an un

• The normal or centripetal component is always directed toward the center of curvature of the curve. an = v2/

• The tangential component is tangent to the curve and in the direction of increasing or decreasing velocity.

at = v .

• The magnitude of the acceleration vector is a = [(at)2 + (an)2]0.5

IMPORTANT EQUATIONS

The magnitude of the acceleration vector is a = [(at)2 + (an)2]0.5

Tangential Acceleration, at = dv/dt

Normal Acceleration, an = v2/ρ

SPECIAL CASES OF MOTIONThere are some special cases of motion to consider.

1) The particle moves along a straight line. => an = v2/ a = at = 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 = an = v2/.

SPECIAL CASES OF MOTION (continued)

3) Sometimes, the equation of the curve path followed by the particle may be given, y = f(x). The radius of curvature, ρ, at any point on the path can be calculated from the following equation:

32 2

2

2

1dydx

d ydx

Example: A boat travels around a circular path at a speed that increases with time, v = (0.0625 t2) m/s. Find the magnitudes of the boat’s velocity and acceleration at the instant t = 10 s.

Example: The motorcyclist travels along the curve at a constant speed of 30 ft/s. Determine his acceleration when he is located at point A. Hint: Treat the motorcycle and rider as a particle.

Example: The automobile has a speed of 80 ft/s at point A and an acceleration, a, having a magnitude of 10 ft/s2, acting in the direction shown. Determine the radius of curvature of the path at point A and the tangential component of acceleration.

RELATIVE-MOTION ANALYSIS OF TWO PARTICLES

All motion is relative to some frame of reference

2 trains approaching each other (along a line) at 95 km/h each. Observers on either train see the other coming at 95 + 95 = 190 km/h. Observer on ground sees 95 km/h.

Velocity depends on reference frame!!

RELATIVE POSITION (Section 12.10)

The absolute position of two particles A and B with respect to the fixed x, y, z reference frame are given by rA and rB. The position of B relative to A is represented by rB/A + rA = rB (vector addition)

rB/A = rB – rA

Therefore, if rB = (10 i + 2 j ) m

and rA = (4 i + 5 j ) m,

then rB/A = (6 i – 3 j ) m.

Note: rB/A means position of B as observed from A

rA/B means position of A as observed from B

RELATIVE VELOCITY

To determine the relative velocity of B with respect to A, the time derivative of the relative position equation is taken.

vB/A = vB – vA

orvB = vA + vB/A

In these equations, vB and vA are called absolute velocities and vB/A is the relative velocity of B with respect to A.

Note that vB/A = - vA/B .

RELATIVE ACCELERATION

The time derivative of the relative velocity equation yields a similar vector relationship between the absolute and relative accelerations of particles A and B.

These derivatives yield: aB/A = aB – aA

or

aB = aA + aB/A

SOLVING PROBLEMS

Since the relative motion equations are vector equations, problems involving them requires resolving velocity (or acceleration) vector along x and y and then performing the vector operation:

vB = vA + vB/A

aB = aA + aB/A

EXAMPLE: Shown in figure are two airplanes flying at same altitude with their respective velocities and direction shown. Find vB/A.

Given: vA = 650 km/h vB = 800 km/h

Find: vB/A

EXAMPLE (continued)

Solution:

vA = (650 i ) km/h

vB = –800 cos 60 i – 800 sin 60 j

= ( –400 i – 692.8 j) km/h

vB/A = vB – vA = (–1050 i – 692.8 j) km/h

v AB /km/h

= tan-1( ) = 33.4

Example: At the instant shown, race car A is passing race car B with a relative velocity (vA/B) of 1 m/s. Knowing that the speeds of both cars are constant and that the relative acceleration of car A with respect to car B (aA/B) is 0.25 m/s2 directed toward the center of curvature, determine the speed of cars A and B at the instant shown.

Example: At a given instant in an airplane race, airplane A is flying horizontally in a straight line, and its speed is being increased at a rate of 6 m/s2. Airplane B is flying at the same altitude as airplane A and is following a circular path of 200-m radius. Knowing that at the given instant the speed of B is being decreased at the rate of 2 m/s2, determine, for the positions shown, the acceleration of B relative to A.

ABSOLUTE DEPENDENT MOTION ANALYSIS OF TWO PARTICLES (Section-12.9)

In many kinematics problems, the motion of one object will depend on the motion of another object.

• If block A moves downward along the inclined plane, block B will move up the other incline.

• This dependency commonly occurs if the particles are interconnected by inextensible/ inelastic cords which are wrapped around pulley(s).

PROCEDURE FOR ANALYSIS- DEPENDENT MOTION

The motion of each block can be related mathematically by defining position coordinates, sA and sB. Each coordinate axis is defined from a fixed point or datum line, measured positive along each plane in the direction of motion of each block.

Note: The length of the cord connecting the two particles will always remain same/ constant, irrespective of the position/ location of the two particles.

DEPENDENT MOTION (continued)

In this example, position coordinates sA and sB can be defined from fixed datum lines extending from the center of the pulley along each incline to blocks A and B.

Since, the cord has a fixed length, the position coordinates sA and sB are related mathematically by the equation

sA + lCD + sB = lT

Here lT is the total cord length and lCD is the length of cord passing over the arc CD on the pulley.

DEPENDENT MOTION (continued)

The negative sign indicates that as A moves down the incline (positive sA direction), B moves up the incline (negative sB direction).

Accelerations can be found by differentiating the velocity expression. aB = -aA .

dsA/dt + dsB/dt = 0 => vB = -vA

The velocities of blocks A and B can be related by differentiating the position equation. Note that lCD and lT remain constant, so dlCD/dt = dlT/dt = 0

DEPENDENT MOTION: PROCEDURES

These procedures can be used to relate the dependent motion of particles moving along rectilinear paths (only the magnitudes of velocity and acceleration change, not their line of direction).

4. Differentiate the position coordinate equation(s) to relate velocities and accelerations. Keep track of signs!

3. If a system contains more than one cord, relate the position of a point on one cord to a point on another cord. Separate equations are written for each cord.

2. Relate the position coordinates to the cord length. Segments of cord that do not change in length during the motion may be left out.

1. Define position coordinates from fixed datum lines, along the path of each particle. Different datum lines can be used for each particle.

EXAMPLE-1

Given: In the figure on the left, the cord at A is pulled down with a speed of 2 m/s.

Find: The speed of block B.

Example - 2: Slider block B moves to the right with a constant velocity of 300 mm/s. Determine (a) the velocity of slider block A, (b) the velocity of point C on the cable, (c) the relative velocity of C w.r.t. A (vC/A).

Example-3: In the figure, particle A is pulled to the right with a speed of 2 m/s.

Find: The speed of block B.