Kinematics of Particles

Kinematics of Particles

Today y

• Motion of several particles: Dependent motion

• Definition of curvilinear motion

• Representation of curvilinear motion – Using rectangular components

U i t ti l d l t

– Using tangential and normal components – Using radial and transverse components

Kinematics of Particles

Motion of Several Particles: Dependent Motion p

• Position of a particle may depend on position of one or more other particles.

or more other particles.

• Position of block B depends on position of block A.

• Why? What’s the relation? Rope length is constant!

• Why? What s the relation? Rope length is constant!

= + _B

A x

x 2 constant (one degree of freedom)

• Positions of three blocks are dependent.

= +

+ _{B C}

A x x

x 2

2 constant (two degrees of freedom)

• For linearly related positions, similar relations hold between velocities and accelerations.

0 2

2 or 0

2

2 + + = + + =

C B

A

C B

C A B

A

dv dv dv

v v

dt v dx dt

dx dt

dx

0 2

2 or 0

2

2 ^A + ^B + ^C = a_A + a_B + a_C =

dt dv dt

dv dt

dv

Kinematics of Particles

Sample Problem 11.5 p

SOLUTION:

• Define origin at upper horizontal surface g pp with positive displacement downward.

• Collar A has uniformly accelerated

rectilinear motion. Solve for acceleration and time t to reach L.

P ll D h if tili ti

Pulley D is attached to a collar which is pulled down at 3 cm/s. At t = 0,

ll A i d f K

• Pulley D has uniform rectilinear motion.

Calculate change of position at time t.

• Block B motion is dependent on motions collar A starts moving down from K

with constant acceleration and zero initial velocity. Knowing that velocity

• Block B motion is dependent on motions of collar A and pulley D. Write motion relationship and solve for change of block B iti t ti

of collar A is 12 cm/s as it passes L, determine the change in elevation, velocity, and acceleration of block B

B position at time t.

• Differentiate motion relation twice to develop equations for velocity and

y, develop equations for velocity and

Kinematics of Particles

Graphical Solution of Rectilinear-Motion Problems p

• Given the x-t curve, the v-t curve is equal to the x-t curve slope.

• Given the v-t curve, the a-t curve is equal to the v-t curve slope.

Kinematics of Particles

Graphical Solution of Rectilinear-Motion Problems p

Gi h h h i l i b d i

• Given the a-t curve, the change in velocity between t₁ and t₂ is equal to the area under the a-t curve between t₁ and t₂.

• Given the v-t curve, the change in position between t₁ and t₂ is equal to the area under the v-t curve between t₁ and t₂.

Kinematics of Particles

Curvilinear Motion: Position, Velocity & Acceleration , y

• Particle moving along a curve other than a straight line is in curvilinear motion.

• How is a Position vector of a particle at time t defined?

a vector between origin Og of a fixed reference frame and the position occupied by particle.

• Consider particle which occupies positionp p p P defined by at time trr and P’ defined by at t + rr′ Δt,

Δ =

= r dr

v lim

r r r

=

Δ

→

Δ t dt v

t 0

lim

instantaneous velocity (vector) Δ =

= Δ

→

Δ dt

ds t

v s

t 0

lim

= instantaneous speed (scalar)

Kinematics of Particles

Curvilinear Motion: Position, Velocity & Acceleration , y

• Consider velocity of particle at time t and velocity at t + Δt,

vr v^r′

Δ =

= Δ

→

Δ dt

v d t

a v

t

r r r

0

lim

= instantaneous acceleration (vector)

I l it t t t t th th?

• Is velocity vector tangent to the path?

• Is acceleration vector tangent to the path?

Kinematics of Particles

Derivatives of Vector Functions

( )

^u

Pr

• Let be a vector function of scalar variable u,

(

^{u u}

) ( )

^{P u}

P P

P

dr = Δ r = r + Δ − r lim

lim u u

du u = u Δ

= Δ

→ Δ

→

Δ 0 0

lim lim

• Derivative of vector sum,

( )

( )

du Q d du

P d du

Q P

d r r r r

+ + =

( )

du P f d du P

df du

P f

d r r r

+

=

• Derivative of product of scalar and vector functions, du

du du

• Derivative of scalar product and vector product,

(

^{P Q}

)

^{dP dQ}

d

(

r • r

)

r r

(

^{P Q}

)

^{dP dQ}

d

du Q P d

du Q P d du

Q P d

r r

r r

r r

×

• +

•

• =

( )

du Q P d

du Q P d du

Q P

d × = × r + r×

Kinematics of Particles

Rectangular Components of Velocity & Acceleration g p y

• When position vector of particle P is given by its rectangular components,

k z j y i x

rr = r + r + r

• Velocity vector,

k z j y i x dtk

j dz dt i dy dt

v dx r

&

&r

&r r r

r = r + + = + +

k v j v i v

j dt y

dt j dt

z y

x

r r

r + +

=

• Acceleration vector,

2 2

2

k z j y i x k dt

z j d

dt y i d

dt x a d

r

&&r

&&r

&&r r r

r = r + + = + +

2 2 2

2 2

2

r r

r + +

=

Kinematics of Particles

Rectangular Components of Velocity & Acceleration g p y

• Rectangular components particularly effective when component accelerations can be integrated independently, e.g., motion of a projectile,

0

0 = = − = =

=

= x a y g a z

a_x && _y && _z &&

i h i i i l di i with initial conditions,

( )

^,

( )

^,

( )

⁰

0 ₀

0 0 0

0

0 = y = z = v_x v_y v_z =

x

Integrating twice yields

( ) ( )

( )

^{0 2 0}

0 = − =

= v v v gt v

v_{x x y y z}

( )

₀ ⁼

( )

₀ ⁻ ₂^{1 2 = 0}

= v t y v y gt z

x _{x y}

• Motion in horizontal direction is uniform.

• Motion in vertical direction is uniformly accelerated.

• Motion of projectile could be replaced by two

• Motion of projectile could be replaced by two independent rectilinear motions.

Kinematics of Particles

Motion Relative to a Frame in Translation

• Designate one frame as the fixed frame of reference.

All other frames not rigidly attached to the fixed reference frame are moving frames of reference.

• Position vectors for particles A and B with respect to the fixed frame of reference Oxyz arerr_A and rr_B

the fixed frame of reference Oxyz are r_A and r_B.

• Vector joining A and B defines the position of B with respect to the moving frame Ax’y’z’ and

A

rrB

A B A

B r r

rr = r + r

• Differentiating twice,

A =

vrB velocity of B relative to A.

A B A

B v v

vr = r + r

A =

arB acceleration of B relative

A B A

B a a

ar = r + r

to A.

• Absolute motion of B can be obtained by combining motion of A with relative motion of B with respect to motion of A with relative motion of B with respect to

Kinematics of Particles

Tangential and Normal Components g p

• Velocity vector of particle is tangent to path of particle. In general, acceleration vector is not.

particle. In general, acceleration vector is not.

Wish to express acceleration vector in terms of tangential and normal components. Why?

• are tangential unit vectors for the particle path at P and P’. When drawn with

t t th i i d

t

t e

er and r′

r r

r _′

Δ

respect to the same origin, and is the angle between them. Why ?

t t

t e e

er ₌ r_{′ −} r θ Δ

Δ

(

Δθ

)

Δ 2 i 2

θ Δ

( )

( )

θ θ θ

θ

e e e

e

n t n

t

r r r

Δ =

= Δ Δ Δ

Δ

= Δ

2 2 lim sin

lim

2 sin

2

θ

θ

θ θ

θ

d e e_n d ^t

n n

r = r

Δ Δ Δ →

→

Δ 0 0 2

Direction of change of tangential term of the θ

n d tangential term of the acceleration

Kinematics of Particles

Tangential and Normal Components

• With the velocity vector expressed as

the particle acceleration may be written as

g p

et

v vr = r

dt ds ds d d

e v d dt e

dv dt

e v d dt e

dv dt

v

a d _t ^t _t ^t θ

θ r r

r r

r = r = + = +

but but

dt v ds ds

e d d

e d

n

t = = =

ρ θ

θ

r 1 r

Af b i i

After substituting,

ρ ρ

2

2 v

dt a a dv

v e dt e

ar = dv r_t + r_{n t} = _n = ρ ρ

• Tangential component of acceleration reflects change of speed and normal component reflects

h f di i

change of direction.

• Tangential component may be positive or negative Normal component always points negative. Normal component always points

Kinematics of Particles

Radial and Transverse Components p

• When particle position is given in polar coordinates, it is convenient to express velocity and acceleration with components parallel and perpendicular to OP.

• The particle velocity vector is

( )

^θ e_θ

dt r d dt e

dr dt

e r d dt e

e dr dt r

v d _{r r} ^r _r

&

r r r

r r

r = = + = +

p y

d

dr r

θ eθ

r e

r& r_r + & r

=

• Similarly, the particle acceleration vector is er

r rr = r

r er

d e e d

d e

d r r

r r

−

=

= θ

θ ^{θ θ}

d d

θ θ

dt e r d dt e

dr dt

ar d r_r r

2 2

⎟⎠

⎜ ⎞

⎝

⎛ +

=

dt e d dt

d d

e d dt

e

d _{r r} θ θ

θ ^rθ r

r = =

d d

e d e

d^r_θ ^r_θ θ θ

( ) ( )

θ θ θ

θ θ

θ

θ θ

θ

dt e d dt r d e

dt r d dt e

d dt dr dt

e d dt e dr

dt r

d _r

r

& r

&

&&

& r

&&

r r r r

r

2 2

2 2 2

2 + + + +

=

dt e d dt

d d

e d dt

e d

r

θ θ

θ^θ

θ = = −r =

(

r&&− rθ ²

)

er_r +

(

rθ + 2r&θ

)

erθ

Kinematics of Particles

Radial and Transverse Components p

• When particle position is given in cylindrical coordinates, it is convenient to express the

velocity and acceleration vectors using the unit vectors .er_R, er ,and kr

θ

• Position vector, k z e

R

rr = r_R+ r

• Velocity vector,

k R

r R

dr & r & r r

r Re Rθ e zk

dt r

vr = d = r_R + θ r_θ + &

• Acceleration vector

• Acceleration vector,

(

^{R R}

)

^e

(

^{R R}

)

^{e zk}

dt v

a d _R r

&&

& r

&

&&

& r

&&

r = r = − θ ² + θ + 2 θ _θ +

Kinematics of Particles

Sample Problem 11.10 p

• Calculate tangential and normal

• Calculate tangential and normal components of acceleration.

( )

m

75 s 2

m 88

66− =

Δ =

= v a

( )

2 2 2

2

10 m . s 3

m 88

s 75 . s 2

8

=

=

=

−

= Δ =

=

a v a t

n t

A motorist is traveling on curved section of highway at 60 km/h. The motorist applies brakes causing a

s2

10 . m 3

ρ 2500 a_n

• Determine acceleration magnitude and

di ti ith t t t t t

motorist applies brakes causing a constant deceleration rate.

Knowing that after 8 s the speed has

direction with respect to tangent to curve.

( )

^{2 2}

2

2 + = − 2 75 + 3 10

= a_t a

g p a

been reduced to 66 m/s, determine the acceleration of the automobile

immediately after the brakes are 2 75

10 . tan 3

tan⁻¹ = ⁻¹

= aⁿ α

(

^2.75

)

+ ^3.10

= +

= a_t a_n a

immediately after the brakes are

applied. a_t 2.75

Kinematics of Particles

Sample Problem 11.12 p

Approach:

Wh t ki d f ti i thi ? What kind of motion is this?

• Dependent motion

• Curvilinear motion

Rotation of the arm about O is defined

• Rotation + rectilinear on the rotation

Rotation of the arm about O is defined by θ = 0.15t² where θ is in radians and t in seconds. Collar B slides along the

arm such that r = 0 9 - 0 12t² where r is ^• Motion of the arm

M ti f th ll t t th

What are the knowns?

arm such that r = 0.9 - 0.12t where r is in meters.

After the arm has rotated through 30^o,

• Motion of the collar w.r.t. to the arm What do we want to know?

g ,

determine (a) the total velocity of the collar, (b) the total acceleration of the collar, and (c) the relative acceleration

• Absolute motion of the collar

• Relative acceleration of the collar w.r.t collar, and (c) the relative acceleration h

the arm