DYNAMICS Ferdinand P. Beer Kinematics of Particles E. Russell - - PowerPoint PPT Presentation

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

Eighth Edition Ferdinand P. Beer

• E. Russell Johnston, Jr.

Lecture Notes:

• J. Walt Oler

Texas Tech University CHAPTER

• Kinematics of Particles

Vector Mechanics for Engineers: Dynamics

Contents

Introduction Rectilinear Motion: Position, Velocity & Acceleration Determination of the Motion of a Particle Sample Problem 11.2 Sample Problem 11.3 Uniform Rectilinear-Motion Uniformly Accelerated Rectilinear- Motion Motion of Several Particles: Relative Motion Sample Problem 11.4 Motion of Several Particles: Dependent Motion Sample Problem 11.5 Graphical Solution of Rectilinear-Motion Problems Other Graphical Methods Curvilinear Motion: Position, Velocity & Acceleration Derivatives of Vector Functions Rectangular Components of Velocity and Acceleration Motion Relative to a Frame in Translation Tangential and Normal Components Radial and Transverse Components Sample Problem 11.10 Sample Problem 11.12

Vector Mechanics for Engineers: Dynamics

Introduction

• Dynamics includes:
• Kinematics: study of the geometry of motion. Kinematics is used to relate

displacement, velocity, acceleration, and time without reference to the cause of motion.

• Kinetics: study of the relations existing between the forces acting on a body,

the mass of the body, and the motion of the body. Kinetics is used to predict the motion caused by given forces or to determine the forces required to produce a given motion.

• Rectilinear motion: position, velocity, and acceleration of a particle as it moves

along a straight line.

• Curvilinear motion: position, velocity, and acceleration of a particle as it moves

along a curved line in two or three dimensions.

Vector Mechanics for Engineers: Dynamics

Rectilinear Motion: Position, Velocity & Acceleration

• Particle moving along a straight line is said to

be in rectilinear motion.

• Position coordinate of a particle is defined by

positive or negative distance of particle from a fixed origin on the line.

• The motion of a particle is known if the position

coordinate for particle is known for every value

• f time t. Motion of the particle may be

expressed in the form of a function, e.g.,

3 2

6 t t x − =

• r in the form of a graph x vs. t.

Vector Mechanics for Engineers: Dynamics

Rectilinear Motion: Position, Velocity & Acceleration

• Instantaneous velocity may be positive or
• negative. Magnitude of velocity is referred to

as particle speed.

• Consider particle which occupies position P at

time t and P’ at t+∆t, t x v t x

t

∆ ∆ = = ∆ ∆ =

→ ∆

lim Average velocity Instantaneous velocity

• From the definition of a derivative,

dt dx t x v

t

= ∆ ∆ =

→ ∆

lim e.g.,

2 3 2

3 12 6 t t dt dx v t t x − = = − =

Vector Mechanics for Engineers: Dynamics

Rectilinear Motion: Position, Velocity & Acceleration

• Consider particle with velocity v at time t and v’

at t+∆t, Instantaneous acceleration t v a

t

∆ ∆ = =

→ ∆

lim t dt dv a t t v dt x d dt dv t v a

t

6 12 3 12 e.g. lim

2 2 2

− = = − = = = ∆ ∆ =

→ ∆

• From the definition of a derivative,
• Instantaneous acceleration may be:
• positive: increasing positive velocity
• r decreasing negative velocity
• negative: decreasing positive velocity
• r increasing negative velocity.

Vector Mechanics for Engineers: Dynamics

Rectilinear Motion: Position, Velocity & Acceleration

• Consider particle with motion given by

3 2

6 t t x − =

2

3 12 t t dt dx v − = = t dt x d dt dv a 6 12

2 2

− = = =

• at t = 0, x = 0, v = 0, a = 12 m/s2
• at t = 2 s, x = 16 m, v = vmax = 12 m/s, a = 0
• at t = 4 s, x = xmax = 32 m, v = 0, a = -12 m/s2
• at t = 6 s, x = 0, v = -36 m/s, a = 24 m/s2

Vector Mechanics for Engineers: Dynamics

Determination of the Motion of a Particle

• Recall, motion of a particle is known if position is known for all time t.
• Typically, conditions of motion are specified by the type of acceleration

experienced by the particle. Determination of velocity and position requires two successive integrations.

• Three classes of motion may be defined for:
• acceleration given as a function of time, a = f(t)
• acceleration given as a function of position, a = f(x)
• acceleration given as a function of velocity, a = f(v)

Vector Mechanics for Engineers: Dynamics

Determination of the Motion of a Particle

• Acceleration given as a function of time, a = f(t):

( ) ( )

( )

( ) ( ) ( ) ( ) ( )

( )

( ) ( ) ( )

• =

− = = = = − = = = =

t t t x x t t t v v

dt t v x t x dt t v dx dt t v dx t v dt dx dt t f v t v dt t f dv dt t f dv t f a dt dv

• Acceleration given as a function of position, a = f(x):

( ) ( )

( )

( ) ( ) ( )

• =

− = = = = = = =

x x x x x v v

dx x f v x v dx x f dv v dx x f dv v x f dx dv v a dt dv a v dx dt dt dx v

2 2 1 2 2 1

• r
• r

Vector Mechanics for Engineers: Dynamics

Determination of the Motion of a Particle

• Acceleration given as a function of velocity, a = f(v):

( ) ( ) ( )

( )

( )

( )

( ) ( )

( )

( )

( )

( ) ( )

( )

• =

− = = = = = = = = =

t v v t v v t x x t v v t t v v

v f dv v x t x v f dv v dx v f dv v dx v f a dx dv v t v f dv dt v f dv dt v f dv v f a dt dv

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.2

Determine:

• velocity and elevation above ground at time

t,

• highest elevation reached by ball and

corresponding time, and

• time when ball will hit the ground and

corresponding velocity. Ball tossed with 10 m/s vertical velocity from window 20 m above ground. SOLUTION:

• Integrate twice to find v(t) and y(t).
• Solve for t at which velocity equals zero

(time for maximum elevation) and evaluate corresponding altitude.

• Solve for t at which altitude equals zero

(time for ground impact) and evaluate corresponding velocity.

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.2

( )

( )

t v t v dt dv a dt dv

t t v v

81 . 9 81 . 9 s m 81 . 9

2

− = − − = − = =

• ( )

t t v

• −

=

2

s m 81 . 9 s m 10

( )

( ) ( )

2 2 1

81 . 9 10 81 . 9 10 81 . 9 10 t t y t y dt t dy t v dt dy

t t y y

− = − − = − = =

• ( )

2 2

s m 905 . 4 s m 10 m 20 t t t y

• −
• +

= SOLUTION:

• Integrate twice to find v(t) and y(t).

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.2

• Solve for t at which velocity equals zero and evaluate

corresponding altitude.

( )

s m 81 . 9 s m 10

2

=

• −

= t t v s 019 . 1 = t

• Solve for t at which altitude equals zero and evaluate

corresponding velocity.

( ) ( ) ( )2

2 2 2

s 019 . 1 s m 905 . 4 s 019 . 1 s m 10 m 20 s m 905 . 4 s m 10 m 20

• −
• +

=

• −
• +

= y t t t y m 1 . 25 = y

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.2

• Solve for t at which altitude equals zero and evaluate

corresponding velocity.

( )

s m 905 . 4 s m 10 m 20

2 2

=

• −
• +

= t t t y

( )

s 28 . 3 s meaningles s 243 . 1 = − = t t

( ) ( ) ( )

s 28 . 3 s m 81 . 9 s m 10 s 28 . 3 s m 81 . 9 s m 10

2 2

• −

=

• −

= v t t v s m 2 . 22 − = v

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.3

Brake mechanism used to reduce gun recoil consists of piston attached to barrel moving in fixed cylinder filled with oil. As barrel recoils with initial velocity v0, piston moves and oil is forced through orifices in piston, causing piston and cylinder to decelerate at rate proportional to their velocity. Determine v(t), x(t), and v(x). kv a − = SOLUTION:

• Integrate a = dv/dt = -kv to find v(t).
• Integrate v(t) = dx/dt to find x(t).
• Integrate a = v dv/dx = -kv to find v(x).

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.3

SOLUTION:

• Integrate a = dv/dt = -kv to find v(t).

( )

( )

kt v t v dt k v dv kv dt dv a

t t v v

− = − = − = =

• ln

( )

kt

e v t v

−

=

• Integrate v(t) = dx/dt to find x(t).

( )

( )

( )

t kt t kt t x kt

e k v t x dt e v dx e v dt dx t v 1

• −

= = = =

− − −

• ( )

( )

kt

e k v t x

−

− = 1

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.3

• Integrate a = v dv/dx = -kv to find v(x).

kx v v dx k dv dx k dv kv dx dv v a

x v v

− = − − = − = − = =

• kx

v v − =

• Alternatively,

( ) ( )

• −

=

0 1

v t v k v t x kx v v − =

( ) ( )

• r

v t v e e v t v

kt kt

= =

− −

( )

( )

kt

e k v t x

−

− = 1 with and then

Vector Mechanics for Engineers: Dynamics

Uniform Rectilinear Motion

For particle in uniform rectilinear motion, the acceleration is zero and the velocity is constant. vt x x vt x x dt v dx v dt dx

t x x

+ = = − = = =

• constant

Vector Mechanics for Engineers: Dynamics

Uniformly Accelerated Rectilinear Motion

For particle in uniformly accelerated rectilinear motion, the acceleration of the particle is constant. at v v at v v dt a dv a dt dv

t v v

+ = = − = = =

• constant

( )

2 2 1 2 2 1

at t v x x at t v x x dt at v dx at v dt dx

t x x

+ + = + = − + = + =

• (

)

( ) ( )

2 2 2 2 2 1

2 constant x x a v v x x a v v dx a dv v a dx dv v

x x v v

− + = − = − = = =

Vector Mechanics for Engineers: Dynamics

Motion of Several Particles: Relative Motion

• For particles moving along the same line, time

should be recorded from the same starting instant and displacements should be measured from the same origin in the same direction. = − =

A B A B

x x x relative position of B with respect to A

A B A B

x x x + = = − =

A B A B

v v v relative velocity of B with respect to A

A B A B

v v v + = = − =

A B A B

a a a relative acceleration of B with respect to A

A B A B

a a a + =

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.4

Ball thrown vertically from 12 m level in elevator shaft with initial velocity of 18 m/s. At same instant, open-platform elevator passes 5 m level moving upward at 2 m/s. Determine (a) when and where ball hits elevator and (b) relative velocity of ball and elevator at contact. SOLUTION:

• Substitute initial position and velocity

and constant acceleration of ball into general equations for uniformly accelerated rectilinear motion.

• Substitute initial position and constant

velocity of elevator into equation for uniform rectilinear motion.

• Write equation for relative position of

ball with respect to elevator and solve for zero relative position, i.e., impact.

• Substitute impact time into equation for

position of elevator and relative velocity

• f ball with respect to elevator.

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.4

SOLUTION:

• Substitute initial position and velocity and constant

acceleration of ball into general equations for uniformly accelerated rectilinear motion.

2 2 2 2 1 2

s m 905 . 4 s m 18 m 12 s m 81 . 9 s m 18 t t at t v y y t at v v

B B

• −
• +

= + + =

• −

= + =

• Substitute initial position and constant velocity of

elevator into equation for uniform rectilinear motion. t t v y y v

E E E

• +

= + = = s m 2 m 5 s m 2

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.4

• Write equation for relative position of ball with respect to

elevator and solve for zero relative position, i.e., impact.

( ) (

)

2 5 905 . 4 18 12

2

= + − − + = t t t y

E B

( )

s 65 . 3 s meaningles s 39 . = − = t t

• Substitute impact time into equations for position of elevator and

relative velocity of ball with respect to elevator.

( )

65 . 3 2 5 + =

E

y m 3 . 12 =

E

y

( ) ( )

65 . 3 81 . 9 16 2 81 . 9 18 − = − − = t v

E B

s m 81 . 19 − =

E B

v

Vector Mechanics for Engineers: Dynamics

Motion of Several Particles: Dependent Motion

• Position of a particle may depend on position of one or

more other particles.

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

rope is of constant length, it follows that sum of lengths of segments must be constant. = +

B A

x x 2 constant (one degree of freedom)

• Positions of three blocks are dependent.

= + +

C B A

x x x 2 2 constant (two degrees of freedom)

• For linearly related positions, similar relations hold

between velocities and accelerations. 2 2

• r

2 2 2 2

• r

2 2 = + + = + + = + + = + +

C B A C B A C B A C B A

a a a dt dv dt dv dt dv v v v dt dx dt dx dt dx

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.5

Pulley D is attached to a collar which is pulled down at 3 in./s. At t = 0, collar A starts moving down from K with constant acceleration and zero initial

• velocity. Knowing that velocity of

collar A is 12 in./s as it passes L, determine the change in elevation, velocity, and acceleration of block B when block A is at L. SOLUTION:

• Define origin at upper horizontal surface with

positive displacement downward.

• Collar A has uniformly accelerated rectilinear
• motion. Solve for acceleration and time t to

reach L.

• Pulley D has uniform rectilinear motion.

Calculate change of position at time t.

• Block B motion is dependent on motions of

collar A and pulley D. Write motion relationship and solve for change of block B position at time t.

• Differentiate motion relation twice to develop

equations for velocity and acceleration of block B.

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.5

SOLUTION:

• Define origin at upper horizontal surface with positive

displacement downward.

• Collar A has uniformly accelerated rectilinear motion.

Solve for acceleration and time t to reach L.

( ) ( )

[ ]

( )

2 2 2 2

s in. 9 in. 8 2 s in. 12 2 = =

• −

+ =

A A A A A A A

a a x x a v v

( )

s 333 . 1 s in. 9 s in. 12

2

= = + = t t t a v v

A A A

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.5

• Pulley D has uniform rectilinear motion. Calculate

change of position at time t.

( ) ( ) ( )

in. 4 s 333 . 1 s in. 3 =

• =

− + =

D D D D D

x x t v x x

• Block B motion is dependent on motions of collar A

and pulley D. Write motion relationship and solve for change of block B position at time t. Total length of cable remains constant,

( ) ( ) ( ) ( )

[ ]

( )

[ ]

( )

[ ]

( ) ( ) ( )

[ ]

in. 4 2 in. 8 2 2 2 = − + + = − + − + − + + = + +

B B B B D D A A B D A B D A

x x x x x x x x x x x x x x

( )

in. 16 − = −

B B

x x

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.5

• Differentiate motion relation twice to develop equations

for velocity and acceleration of block B. s in. 3 2 s in. 12 2 constant 2 = +

• +
• =

+ + = + +

B B D A B D A

v v v v x x x s in. 18 =

B

v s in. 9 2

2

= +

• =

+ +

B B D A

v a a a

2

s in. 9 − =

B

a

Vector Mechanics for Engineers: Dynamics

Graphical Solution of Rectilinear-Motion Problems

• 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.

Vector Mechanics for Engineers: Dynamics

Graphical Solution of Rectilinear-Motion Problems

• 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.

Vector Mechanics for Engineers: Dynamics

Other Graphical Methods

• Moment-area method to determine particle position at

time t directly from the a-t curve:

( )

• −

+ = − = −

1 1 1

curve under area

v v

dv t t t v t v x x using dv = a dt ,

( )

• −

+ = −

1 1 1 v v

dt a t t t v x x

( )

= −

• 1

1 v v

dt a t t first moment of area under a-t curve with respect to t = t1 line.

( )( )

C t t t a-t t v x x centroid

• f

abscissa curve under area

1 1 1

= − + + =

Vector Mechanics for Engineers: Dynamics

Other Graphical Methods

• Method to determine particle acceleration from

v-x curve: = = = = BC AB dx dv v a θ tan subnormal to v-x curve

Vector Mechanics for Engineers: Dynamics

Curvilinear Motion: Position, Velocity & Acceleration

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

curvilinear motion.

• Position vector of a particle at time t is defined by a vector

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

• Consider particle which occupies position P defined by

at time t and P’ defined by at t + ∆t, r

• r ′
• =

= ∆ ∆ = = = ∆ ∆ =

→ ∆ → ∆

dt ds t s v dt r d t r v

t t

lim lim

• instantaneous velocity (vector)

instantaneous speed (scalar)

Vector Mechanics for Engineers: Dynamics

Curvilinear Motion: Position, Velocity & Acceleration

= = ∆ ∆ =

→ ∆

dt v d t v a

t

• lim

instantaneous acceleration (vector)

• Consider velocity of particle at time t and velocity

at t + ∆t, v

• v

′

• In general, acceleration vector is not tangent to

particle path and velocity vector.

Vector Mechanics for Engineers: Dynamics

Derivatives of Vector Functions

( )

u P

• Let

be a vector function of scalar variable u,

( ) ( )

u u P u u P u P du P d

u u

∆ − ∆ + = ∆ ∆ =

→ ∆ → ∆

• lim

lim

• Derivative of vector sum,

( )

du Q d du P d du Q P d

• +

= +

( )

du P d f P du df du P f d

• +

=

• Derivative of product of scalar and vector functions,
• Derivative of scalar product and vector product,

( ) ( )

du Q d P Q du P d du Q P d du Q d P Q du P d du Q P d

• ×

+ × = ×

• +
• =

Vector Mechanics for Engineers: Dynamics

Rectangular Components of Velocity & Acceleration

• When position vector of particle P is given by its

rectangular components, k z j y i x r

• +

+ =

• Velocity vector,

k v j v i v k z j y i x k dt dz j dt dy i dt dx v

z y x

• +

+ = + + = + + =

• Acceleration vector,

k a j a i a k z j y i x k dt z d j dt y d i dt x d a

z y x

• +

+ = + + = + + =

2 2 2 2 2 2

Vector Mechanics for Engineers: Dynamics

Rectangular Components of Velocity & Acceleration

• Rectangular components particularly effective when

component accelerations can be integrated independently, e.g., motion of a projectile, = = − = = = = z a g y a x a

z y x

• with initial conditions,

( ) (

) (

)

, , = = = =

z y x

v v v z y x Integrating twice yields

( )

( )

( )

( )

2 2 1

= − = = = − = = z gt y v y t v x v gt v v v v

y x z y y x x

• Motion in horizontal direction is uniform.
• Motion in vertical direction is uniformly accelerated.
• Motion of projectile could be replaced by two

independent rectilinear motions.

Vector Mechanics for Engineers: Dynamics

Motion Relative to a Frame in Translation

• Designate one frame as the fixed frame of reference. All
• ther 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 are . and

B A

r r

• Vector

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

A B

r

• A

B A B

r r r

• +

=

• Differentiating twice,

=

A B

v

• velocity of B relative to A.

A B A B

v v v

• +

= =

A B

a

• acceleration of B relative to

A.

A B A B

a a a

• +

=

• Absolute motion of B can be obtained by combining

motion of A with relative motion of B with respect to moving reference frame attached to A.

Vector Mechanics for Engineers: Dynamics

Tangential and Normal Components

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

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

• are tangential unit vectors for the particle

path at P and P’. When drawn with respect to the same origin, and is the angle between them.

t t

e e

• ′

and

t t t

e e e

• −

′ = ∆ θ ∆

( ) ( )

θ θ θ θ θ

θ θ

d e d e e e e e

t n n n t t

• =

= ∆ ∆ = ∆ ∆ ∆ = ∆

→ ∆ → ∆

2 2 sin lim lim 2 sin 2

Vector Mechanics for Engineers: Dynamics

Tangential and Normal Components

t

e v v

• =
• With the velocity vector expressed as

the particle acceleration may be written as dt ds ds d d e d v e dt dv dt e d v e dt dv dt v d a

t t

θ θ

• +

= + = = but v dt ds ds d e d e d

n t

= = = θ ρ θ

• After substituting,

ρ ρ

2 2

v a dt dv a e v e dt dv a

n t n t

= = + =

• Tangential component of acceleration reflects

change of speed and normal component reflects change of direction.

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

toward center of path curvature.

Vector Mechanics for Engineers: Dynamics

Tangential and Normal Components

ρ ρ

2 2

v a dt dv a e v e dt dv a

n t n t

= = + =

• Relations for tangential and normal acceleration also

apply for particle moving along space curve.

• Plane containing tangential and normal unit vectors

is called the osculating plane.

n t b

e e e

• ×

=

• Normal to the osculating plane is found from

binormal e normal principal e

b n

= =

• Acceleration has no component along binormal.

Vector Mechanics for Engineers: Dynamics

Radial and Transverse Components

• When particle position is given in polar coordinates, it

is convenient to express velocity and acceleration with components parallel and perpendicular to OP.

r r

e d e d e d e d

• −

= = θ θ

θ θ

dt d e dt d d e d dt e d

r r

θ θ θ

θ

• =

= dt d e dt d d e d dt e d

r

θ θ θ

θ θ

• −

= =

( )

θ θ

θ θ e r e r e dt d r e dt dr dt e d r e dt dr e r dt d v

r r r r r

• +

= + = + = =

• The particle velocity vector is
• Similarly, the particle acceleration vector is

( )

( ) θ

θ θ θ θ

θ θ θ θ θ θ θ e r r e r r dt e d dt d r e dt d r e dt d dt dr dt e d dt dr e dt r d e dt d r e dt dr dt d a

r r r r

• 2

2 2 2 2 2

+ + − = + + + + =

• +

=

r

e r r

• =

Vector Mechanics for Engineers: Dynamics

Radial and Transverse Components

• When particle position is given in cylindrical

coordinates, it is convenient to express the velocity and acceleration vectors using the unit vectors . and , , k e eR

• θ
• Position vector,

k z e R r

R

• +

=

• Velocity vector,

k z e R e R dt r d v

R

• +

+ = =

θ

θ

• Acceleration vector,

( )

( )

k z e R R e R R dt v d a

R

• +

+ + − = =

θ

θ θ θ 2

2

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.10

A motorist is traveling on curved section of highway at 60 mph. The motorist applies brakes causing a constant deceleration rate. Knowing that after 8 s the speed has been reduced to 45 mph, determine the acceleration of the automobile immediately after the brakes are applied. SOLUTION:

• Calculate tangential and normal

components of acceleration.

• Determine acceleration magnitude and

direction with respect to tangent to curve.

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.10

ft/s 66 mph 45 ft/s 88 mph 60 = = SOLUTION:

• Calculate tangential and normal components of

acceleration.

( ) ( )

2 2 2 2

s ft 10 . 3 ft 2500 s ft 88 s ft 75 . 2 s 8 s ft 88 66 = = = − = − = ∆ ∆ = ρ v a t v a

n t

• Determine acceleration magnitude and direction with

respect to tangent to curve.

( )

2 2 2 2

10 . 3 75 . 2 + − = + =

n t

a a a

2

s ft 14 . 4 = a 75 . 2 10 . 3 tan tan

1 1 − −

= =

t n

a a α ° = 4 . 48 α

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.12

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 in meters. After the arm has rotated through 30o, determine (a) the total velocity of the collar, (b) the total acceleration of the collar, and (c) the relative acceleration of the collar with respect to the arm. SOLUTION:

• Evaluate time t for θ = 30o.
• Evaluate radial and angular positions,

and first and second derivatives at time t.

• Calculate velocity and acceleration in

cylindrical coordinates.

• Evaluate acceleration with respect to

arm.

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.12

SOLUTION:

• Evaluate time t for θ = 30o.

s 869 . 1 rad 524 . 30 0.15 2 = = ° = = t t θ

• Evaluate radial and angular positions, and first and

second derivatives at time t.

2 2

s m 24 . s m 449 . 24 . m 481 . 12 . 9 . − = − = − = = − = r t r t r

• 2

2

s rad 30 . s rad 561 . 30 . rad 524 . 15 . = = = = = θ θ θ

• t

t

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.12

• Calculate velocity and acceleration.

( )( )

r r r

v v v v v r v s r v

θ θ θ

β θ

1 2 2

tan s m 270 . s rad 561 . m 481 . m 449 .

−

= + = = = = − = =

• °

= = . 31 s m 524 . β v

( )( ) ( )(

)

( )( )

r r r

a a a a a r r a r r a

θ θ θ

γ θ θ θ

1 2 2 2 2 2 2 2 2

tan s m 359 . s rad 561 . s m 449 . 2 s rad 3 . m 481 . 2 s m 391 . s rad 561 . m 481 . s m 240 .

−

= + = − = − + = + = − = − − = − =

• °

= = 6 . 42 s m 531 . γ a

Vector Mechanics for Engineers: Dynamics

Sample Problem 11.12

• Evaluate acceleration with respect to arm.

Motion of collar with respect to arm is rectilinear and defined by coordinate r.

2

s m 240 . − = = r a

OA B