Kinematics of Particles

Lecture II

Subjects Covered in Kinematics of Particles

Rectilinear motion Curvilinear motion

Rectangular coords n-t coords Polar coords

Relative motion Constrained motion

Introduction Kinematics: Branch of dynamics that describes the

motion of bodies without reference to the forces which either cause the motion or are generated as a result of the motion.

Applications: The design of cams, gears, linkages, and other machines elements to control or produce certain desired motion, and the calculation of flight trajectories for aircraft, rockets, etc.

Particle: As mentioned, a particle is a body whose physical dimensions are so small compared with the radius of curvature of its path, e.g. an aircraft and its flight path.

Studying the Motion: Studying the motion of a body includes studying its displacement from one location to another, its velocity, and its acceleration.

Introduction (Cont.) Choice of Coordinates: The

position of a particle P at any time t can be described by specifying its rectangular coordinates (x, y, z), its cylindrical coordinates (r, , z), or its spherical coordinates (R, , ). The motion of P can also be described by measurements along the tangent (t) and normal (n) to a curved path. These two are path variables since they move with the particle on the path. The motion of a body described by fixed reference axes known as absolute motion, while the motion described by a moving reference axes known as relative motion.

Rectilinear Motion

Studying the motion of a particle moving in a straight line (1-D)

s s

O P P

t t+t

-S +S

Displacement

Rectilinear Motion - Velocity

Average velocity (vav):

Instantaneous velocity (v): as t approaches

zero in the limit, which is or

sdt

dsv

t

sv

t

lim0

t

sv

av

Note: the velocity is positive or negative depending on the displacement

(1)

Rectilinear Motion – Acceleration

Average acceleration (aav):

Instantaneous acceleration (a): as t

approaches zero in the limit, which is or v

dt

dva

t

va

t

lim0

t

va

av

Note: the acceleration is positive or negative depending whether the velocity is increasing or decreasing

sdt

sda

2

2or (2)

s s

O P P

t t+t

-S +Sv1 v2

v

Rectilinear Motion – Acceleration (Cont.)

Velocity & acceleration: are vector quantities, as we will see in the study of curvilinear motion; however, since in rectilinear motion, the particle is moving in straight line path, the sense of direction is described by plus or minus sign.

To obtain differential equation relating displacement, velocity, and acceleration: dt has to be eliminated from equation (2) (using Chain rule)

Equations (1), (2), and (3): known as the differential equations for the rectilinear motion.

dsadvv or

dsssds (3) dt

ds

ds

dva

v

Rectilinear Motion – Graphical Interpretation

The net displacement of a particle during interval t:

The net change in velocity of a particle during interval t:

When the acceleration is a function of the position coordinates S:

2

1

2

1

t

t

s

s

dtvds curve under area 12 tvss or

2

1

2

1

t

t

v

v

dtadv curve under area 12 tavv or

2

1

2

1

s

s

v

v

dsadvv curve under area 2

1 21

22 savv o

r

Rectilinear Motion – Problems Classifications

Rectilinear Motion(Problems Classifications)

Given s(t)Required v(t) and/or a(t) Given a Given v

a (t)Required v(t) and/or s(t)

a (v)Required v(t) or v(s) and/or s(t)

a (s)Required v(s) and/or s(t)

a = constant

v (t)Required s(t) and/or a(t) v = constant

Rectilinear Motion – Problems Classifications (Cont.)

1) Determining the velocity, v(t), and the acceleration , a(t), of a particle for a given position coordinate , s(t): the successive differentiation of s(t) will give the velocity and the acceleration of the particle, i.e.

sdt

dstv v

dt

dvta

Rectilinear Motion – Problems Classifications (Cont.)

2) Determining the velocity, v(t), and the position , s(t), of a particle for a given acceleration , a = f(t): The nature of the forces acting on the body specifies its acceleration. For simplicity, s = so, v = vo, and t = 0 at the beginning of the interval. From Eq. (2).

and hence,

If desired, the displacement, s, can be obtained by a direct solution of a second order differential equation

tv

v

dttfdvo 0

)( t

o dttfvv0

ts

s

vdtdso 0

t

o vdtss0

)(tfas 0)( tfs

v is a function of t

Rectilinear Motion – Problems Classifications (Cont.)

3) Determining the velocity, v(s), v(t), and the position , s(t), of a particle for a given acceleration , a = f(v): from Eq. (2).

and hence, we have to solve for v as a function of t to get the position, s, as a function of t from Eq. (1).

from Eq. (3).

tv

v

dtvf

dv

o 0)( v

vovf

dvt

)( t is a function of v

s

s

v

v oo

dsvf

dvv

)(

v

v

o

ovf

dvvss

)(

s is a function of v

Rectilinear Motion – Problems Classifications (Cont.)

4) Determining the velocity, v(s), and the position , s(t), of a particle for a given acceleration , a = f(s): from Eq. (3).

and hence, we have to solve for t from Eq. (1).

Rearrange to get s as function of t

s

s

v

v oo

dssfdvv )( v is a function of s

i.e. v = g(s)

ts

s

dtsg

ds

o 0)( s

sosg

dst

)( t is a function of s

s

s

o

o

dssfvv )(222

Rectilinear Motion – Special Cases

1) Constant acceleration, a = constant, (Uniformly Accelerated Rectilinear Motion):

, the position, s, then can be derived as follows:

tv

v

dtadvo 0

atvv o

s

s

v

v oo

dsadvv oo ssavv 222

v is a function of t

v is a function

of s

t

o

s

s

dtatvdso 0

2

2

1attvss oo

dtdsv

From Eq. (2)

From Eq. (3)

Rectilinear Motion – Special Cases

2) Constant velocity, v = constant, (Uniform Rectilinear Motion):

The acceleration (a) = zero

ts

s

dtvdso 0

vtss o s is a function of tFrom Eq. (1)

Exercises

Exercise # 1 The position of a particle which moves along a straight line is defined by the relation,

s = t3 - 6t2 - 15t + 40, where s is expressed in meters and t in seconds. Determine (a) the time at which the velocity will be zero, (b) the position and distance traveled by the particle at that time, (c) the acceleration of the particle at that time, (d) the distance traveled by the particle from t = 4 s to t = 6 s.

Exercise # 2 A ball is tossed with a velocity of 10 m/s directed vertically upward from a window located 20 m

above the ground. Knowing that the acceleration of the ball is constant and equal to 9.81 m/s2 downward, determine (a) the velocity v and elevation y of the ball above the ground at any time t, (b) the highest elevation reached by the ball and the corresponding value of t, (c) the time when the ball will hit the ground and the corresponding velocity. Draw the v-t and y-t curves.

Exercise # 3

o

+s

Exercise # 4

Exercise # 5