Ch11 Kinematics of Particles

Tenth Edition

CHAPTER

11

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS Ferdinand P. Beer E. Russell Johnston, Jr. Phillip J. Cornwell Lecture Notes: Brian P. Self

Kinematics of Particles

California Polytechnic State University

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Tenth Edition

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 RectilinearMotion Motion of Several Particles: Relative Motion Sample Problem 11.4 Motion of Several Particles: Dependent Motion

Sample Problem 11.5 Graphical Solution of RectilinearMotion 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

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11 - 2

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Vector Mechanics for Engineers: Dynamics Kinematic relationships are used to help us determine the trajectory of a golf ball, the orbital speed of a satellite, and the accelerations during acrobatic flying.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Introduction • Dynamics includes: Kinematics: study of the geometry of motion. Relates displacement, velocity, acceleration, and time without reference to the cause of motion. Fthrust

Fdrag Flift 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. © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics Introduction • Particle kinetics includes: • 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. © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics Rectilinear Motion: Position, Velocity & Acceleration • Rectilinear motion: particle moving along a straight line • Position coordinate: defined by positive or negative distance 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 of time t. • May be expressed in the form of a function, e.g., 2 3

x  6t  t

or in the form of a graph x vs. t. © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics Rectilinear Motion: Position, Velocity & Acceleration • Consider particle which occupies position P at time t and P’ at t+Dt, Dx  Average velocity Dt Dx  v  lim Instantaneous velocity Dt 0 Dt • Instantaneous velocity may be positive or negative. Magnitude of velocity is referred to as particle speed. • From the definition of a derivative, Dx dx v  lim  dt Dt 0 Dt e.g., x  6t 2  t 3 dx v  12t  3t 2 dt © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Rectilinear Motion: Position, Velocity & Acceleration • Consider particle with velocity v at time t and v’ at t+Dt, Dv Instantaneous acceleration  a  lim Dt 0 Dt • Instantaneous acceleration may be: - positive: increasing positive velocity or decreasing negative velocity - negative: decreasing positive velocity or increasing negative velocity.

• From the definition of a derivative, Dv dv d 2 x a  lim   2 dt dt Dt 0 Dt

e.g. v  12t  3t 2 dv a  12  6t dt © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Rectilinear Motion: Position, Velocity & Acceleration • From our example,

x  6t 2  t 3 v

dx  12t  3t 2 dt

dv d 2 x a   12  6t dt dt 2 • What are x, v, and a at t = 2 s ? - at t = 2 s, x = 16 m, v = vmax = 12 m/s, a = 0 • Note that vmax occurs when a=0, and that the slope of the velocity curve is zero at this point. • What are x, v, and a at t = 4 s ? - at t = 4 s, x = xmax = 32 m, v = 0, a = -12 m/s2 © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Concept Quiz

What is true about the kinematics of a particle? a) The velocity of a particle is always positive b) The velocity of a particle is equal to the slope of the position-time graph c) If the position of a particle is zero, then the velocity must zero d) If the velocity of a particle is zero, then its acceleration must be zero

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Determination of the Motion of a Particle • We often determine accelerations from the forces applied (kinetics will be covered later) • Generally have three classes of motion - 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) • Can you think of a physical example of when force is a When force is a function of velocity? function of position?

a spring © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

drag

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Acceleration as a function of time, position, or velocity If…. a  a t 

a  a  x

a  a v

Kinematic relationship

Integrate

dx dv and a  v dt

v dv  a  x  dx dv  a (v ) dt dv v  a v dx

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t

v0

0

 dv   a  t  dt

dv  a(t ) dt dt 

v

v

x

v0

x0

 v dv   a  x  dx v

t

dv v a  v   0 dt 0 x

v

v dv a v v0

 dx  

x0

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Vector Mechanics for Engineers: Dynamics Sample Problem 11.2 SOLUTION: • Integrate twice to find v(t) and y(t).

• Solve for t when velocity equals zero (time for maximum elevation) and evaluate corresponding altitude. Ball tossed with 10 m/s vertical velocity from window 20 m above ground.

• Solve for t when altitude equals zero (time for ground impact) and evaluate corresponding velocity.

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. © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics Sample Problem 11.2 SOLUTION: • Integrate twice to find v(t) and y(t).

dv  a  9.81m s 2 dt vt  t vt   v0  9.81t  dv    9.81dt v0

0

vt   10

dy  v  10  9.81t dt y t  t  dy   10  9.81t dt y0

0

m  m   9.81 2  t s  s 

yt   y0  10t  12 9.81t 2

m  m  yt   20 m  10 t   4.905 2 t 2  s  s  © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 11.2 • Solve for t when velocity equals zero and evaluate corresponding altitude.

vt   10

m  m   9.81 2  t  0 s  s 

t  1.019s

• Solve for t when altitude equals zero and evaluate corresponding velocity.

m  m  yt   20 m  10 t   4.905 2 t 2  s  s  m   m y  20 m  10 1.019 s    4.905 2 1.019 s 2  s  s 

y  25.1m © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 11.2 • Solve for t when altitude equals zero and evaluate corresponding velocity.

m  m  yt   20 m  10 t   4.905 2 t 2  0  s  s 

t  1.243s meaningless  t  3.28 s

vt   10

m  m   9.81 2  t s  s 

v3.28 s   10

m  m   9.81 2  3.28 s  s  s  v  22.2

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m s

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Vector Mechanics for Engineers: Dynamics Sample Problem 11.3 SOLUTION:

a  kv

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

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.

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

Determine v(t), x(t), and v(x).

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 11.3 SOLUTION: • Integrate a = dv/dt = -kv to find v(t). v

dv a   kv dt

t

dv v v  k 0 dt 0

ln

v t   kt v0

vt   v0e kt • Integrate v(t) = dx/dt to find x(t). v t  

dx  v0e kt dt

x

t

0

0

 kt dx  v e dt 0 

t

 1  x  t   v0  e kt   k 0

xt   © 2013 The McGraw-Hill Companies, Inc. All rights reserved.



v0 1  e kt k 11 - 18



Tenth Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 11.3 • Integrate a = v dv/dx = -kv to find v(x).

dv a  v  kv dx

dv  k dx

v

x

v0

0

 dv  k  dx

v  v0  kx

v  v0  kx

• Alternatively, with

and then

xt  



v0 1  e kt k



vt  vt   v0 e kt or e kt  v0 v  vt    xt   0 1  k  v0 

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v  v0  kx 11 - 19

Tenth Edition

Vector Mechanics for Engineers: Dynamics Group Problem Solving SOLUTION: • Determine the proper kinematic relationship to apply (is acceleration a function of time, velocity, or position?

The car starts from rest and accelerates according to the relationship

a  3  0.001v

2

• Determine the total distance the car travels in one-half lap

• Integrate to determine the velocity after one-half lap

It travels around a circular track that has a radius of 200 meters. Calculate the velocity of the car after it has travelled halfway around the track. What is the car’s maximum possible speed? © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics Group Problem Solving Given: a  3  0.001v 2 vo = 0, r = 200 m

Find:

v after ½ lap Maximum speed

Choose the proper kinematic relationship Acceleration is a function of velocity, and we also can determine distance. Time is not involved in the problem, so we choose:

dv v  a v dx

x

v

v dv a v v0

 dx  

x0

Determine total distance travelled x   r  3.14(200)  628.32 m

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Group Problem Solving Determine the full integral, including limits x

v

v dv dx  x v a  v  0 0

628.32

 0

v

v dv 2 0 3  0.001v

dx  

Evaluate the interval and solve for v 1 2 v 628.32   ln 3  0.001v  0 0.002

628.32(0.002)  ln 3  0.001v2   ln 3  0.001(0) ln 3  0.001v2   1.2566  1.0986=  0.15802

Take the exponential of each side © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

3  0.001v 2  e0.15802 2 - 22

Tenth Edition

Vector Mechanics for Engineers: Dynamics Group Problem Solving Solve for v

3  0.001v 2  e0.15802

3  e0.15802 v   2146.2 0.001 2

v  46.3268 m/s

How do you determine the maximum speed the car can reach?

a  3  0.001v 2

Velocity is a maximum when acceleration is zero

0.001v 2  3

This occurs when

vmax 

3

0.001

vmax  54.772 m/s

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Uniform Rectilinear Motion During free-fall, a parachutist reaches terminal velocity when her weight equals the drag force. If motion is in a straight line, this is uniform rectilinear motion.

For a particle in uniform rectilinear motion, the acceleration is zero and the velocity is constant.

dx  v  constant dt x

t

x0

0

 dx  v  dt x  x0  vt x  x0  vt Careful – these only apply to uniform rectilinear motion! © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Uniformly Accelerated Rectilinear Motion If forces applied to a body are constant (and in a constant direction), then you have uniformly accelerated rectilinear motion.

Another example is freefall when drag is negligible

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Uniformly Accelerated Rectilinear Motion For a particle in uniformly accelerated rectilinear motion, the acceleration of the particle is constant. You may recognize these constant acceleration equations from your physics courses.

dv  a  constant dt

dx  v0  at dt

v

t

v0

0

 dv  a dt

x

t

x0

0

 dx    v0  at  dt

dv v  a  constant dx

v

x

v0

x0

 v dv  a  dx

v  v0  at

x  x0  v0t  12 at 2

v2  v02  2a  x  x0 

Careful – these only apply to uniformly accelerated rectilinear motion! © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Motion of Several Particles We may be interested in the motion of several different particles, whose motion may be independent or linked together.

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Tenth Edition

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.

 x B  x A  relative position of B with respect to A xB  x A  xB A xB

A

 vB  v A  relative velocity of B with respect to A vB  v A  vB A vB

A

 a B  a A  relative acceleration of B with respect to A aB  a A  aB A aB

A

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Tenth Edition

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. • Substitute initial position and constant velocity of elevator into equation for uniform rectilinear motion. 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. © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

• 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 of ball with respect to elevator. 11 - 29

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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.

vB  v0  at  18

m  m   9.81 2 t s  s 

m  m  y B  y0  v0t  12 at 2  12 m  18 t   4.905 2 t 2  s  s  • Substitute initial position and constant velocity of elevator into equation for uniform rectilinear motion.

vE  2

m s

 m y E  y0  v E t  5 m   2 t  s © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

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.





y B E  12  18t  4.905t 2  5  2t   0 t  0.39 s meaningless  t  3.65 s • Substitute impact time into equations for position of elevator and relative velocity of ball with respect to elevator.

y E  5  23.65

y E  12.3 m

v B E  18  9.81t   2  16  9.813.65 vB © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

E

 19.81

m s 11 - 31

Tenth Edition

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.

x A  2 xB  constant (one degree of freedom) • Positions of three blocks are dependent.

2 x A  2 x B  xC  constant (two degrees of freedom) • For linearly related positions, similar relations hold between velocities and accelerations.

dx A dxB dxC 2 2   0 or 2v A  2vB  vC  0 dt dt dt dv dv dv 2 A  2 B  C  0 or 2a A  2a B  aC  0 dt dt dt © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Group Problem Solving Slider block A moves to the left with a constant velocity of 6 m/s. Determine the velocity of block B.

Solution steps • Sketch your system and choose coordinate system • Write out constraint equation • Differentiate the constraint equation to get velocity

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Group Problem Solving Given: vA= 6 m/s left Find: vB xA

This length is constant no matter how the blocks move

Sketch your system and choose coordinates yB Define your constraint equation(s)

xA  3 yB  constants  L Differentiate the constraint equation to get velocity

6 m/s + 3vB  0 v B  2 m/s  Note that as xA gets bigger, yB gets smaller.

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Graphical Solution of Rectilinear-Motion Problems Engineers often collect position, velocity, and acceleration data. Graphical solutions are often useful in analyzing these data. Data Fideltity / Highest Recorded Punch 180 160

Acceleration data from a head impact during a round of boxing.

140 Acceleration (g)

Tenth Edition

Vector Mechanics for Engineers: Dynamics

120 100 80 60 40 20 0 47.76

47.77

47.78

47.79

47.8

47.81

Time (s)

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Tenth Edition

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. © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

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.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Curvilinear Motion: Position, Velocity & Acceleration

The softball and the car both undergo curvilinear motion.

• A particle moving along a curve other than a straight line is in curvilinear motion. © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Curvilinear Motion: Position, Velocity & Acceleration • The 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 a particle which occupies position P defined by r at time t  and P’ defined by r  at t + Dt,

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Curvilinear Motion: Position, Velocity & Acceleration Instantaneous velocity (vector) Dr dr v  lim  Dt 0 Dt dt

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

Instantaneous speed (scalar) Ds ds  Dt 0 Dt dt

v  lim

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Vector Mechanics for Engineers: Dynamics Curvilinear Motion: Position, Velocity & Acceleration   • Consider velocity v of a particle at time t and velocity v at t + Dt,

Dv dv   instantaneous acceleration (vector) Dt 0 Dt dt

a  lim

• In general, the acceleration vector is not tangent to the particle path and velocity vector. © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics Rectangular Components of Velocity & Acceleration • When position vector of particle P is given by its rectangular components,     r  xi  y j  zk • Velocity vector,     dx  dy  dz  v  i  j  k  xi  y j  zk dt dt dt     vx i  v y j  vz k • Acceleration vector,     d 2 x d 2 y  d 2 z  a  2 i  2 j  2 k  xi  y j  zk dt dt dt     ax i  a y j  az k © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

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, a x  x  0 a y  y   g a z  z  0

with initial conditions, v x 0 , v y 0 , v z 0  0 x0  y0  z0  0

 

Integrating twice yields

v x  v x 0 v y  v y   gt vz  0 0 x  v x 0 t y  v y  y  12 gt 2 z  0 0

• Motion in horizontal direction is uniform. • Motion in vertical direction is uniformly accelerated. • Motion of projectile could be replaced by two independent rectilinear motions. © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 11.7 SOLUTION: • Consider the vertical and horizontal motion separately (they are independent) • Apply equations of motion in y-direction

• Apply equations of motion in x-direction • A projectile is fired from the edge of a 150-m cliff with an initial velocity of 180 m/s at an angle of 30°with the horizontal. Neglecting • air resistance, find (a) the horizontal distance from the gun to the point where the projectile strikes the ground, (b) the greatest elevation above the ground reached by the projectile.

Determine time t for projectile to hit the ground, use this to find the horizontal distance Maximum elevation occurs when vy=0

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 11.7 SOLUTION:

Given: (v)o =180 m/s (a)y = - 9.81 m/s2

(y)o =150 m (a)x = 0 m/s2

Vertical motion – uniformly accelerated:

Horizontal motion – uniformly accelerated:

Choose positive x to the right as shown

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 11.7 SOLUTION:

Horizontal distance Projectile strikes the ground at: Substitute into equation (1) above Solving for t, we take the positive root

Substitute t into equation (4)

Maximum elevation occurs when vy=0

Maximum elevation above the ground = © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Group Problem Solving

SOLUTION: A baseball pitching machine “throws” baseballs with a horizontal velocity v0. If you want the height h to be 42 in., determine the value of v0.

• Consider the vertical and horizontal motion separately (they are independent) • Apply equations of motion in y-direction • Apply equations of motion in x-direction • Determine time t for projectile to fall to 42 inches • Calculate v0=0

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Group Problem Solving Given: x= 40 ft, yo = 5 ft, yf= 42 in. Find: vo Analyze the motion in the y-direction 1 y f  y0  (0)t  gt 2 2 1 2 3.5  5  gt 2 1 1.5 ft   (32.2 ft/s2 )t 2 2

Analyze the motion in the x-direction x  0  (vx )0 t  v0t 40 ft  (v0 )(0.305234 s)

v0  131.047 ft/s  89.4 mi/h

t  0.305234 s © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Motion Relative to a Frame in Translation A soccer player must consider the relative motion of the ball and her teammates when making a pass.

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

It is critical for a pilot to know the relative motion of his aircraft with respect to the aircraft carrier to make a safe landing.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics 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 are rA and rB .  r • Vector B A joining A and B defines the position of B with respect to the moving frame Ax’y’z’ and    rB  rA  rB A • Differentiating twice,     vB  v A  vB A v B A  velocity of B relative to A.     a B  a A  a B A a B A  acceleration of B relative to 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. © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Tangential and Normal Components If we have an idea of the path of a vehicle, it is often convenient to analyze the motion using tangential and normal components (sometimes called path coordinates).

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Tangential and Normal Components y

r= the instantaneous radius of curvature

v  v et

en

v= vt et

dv v2 a  et  en dt r

et x

• The tangential direction (et) is tangent to the path of the particle. This velocity vector of a particle is in this direction • The normal direction (en) is perpendicular to et and points towards the inside of the curve. • The acceleration can have components in both the en and et directions © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 11.8 SOLUTION: • Define your coordinate system • Calculate the tangential velocity and tangential acceleration

• Calculate the normal acceleration A motorist is traveling on a curved section of highway of radius 2500 ft at the speed of 60 mi/h. The motorist suddenly applies the brakes, causing the automobile to slow down at a constant rate. Knowing that after 8 s the speed has been reduced to 45 mi/h, determine the acceleration of the automobile immediately after the brakes have been applied.

• Determine overall acceleration magnitude after the brakes have been applied

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 11.8 SOLUTION:

• Define your coordinate system

• Determine velocity and acceleration in the tangential direction

et

en • The deceleration constant, therefore

• Immediately after the brakes are applied, the speed is still 88 ft/s

a  an2  at2  2.752  3.102

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Group Problem Solving SOLUTION: • Define your coordinate system • Calculate the tangential velocity and tangential acceleration

• Calculate the normal acceleration The tangential acceleration of the centrifuge cab is given by

• Determine overall acceleration magnitude

at  0.5 t (m/s2 ) where t is in seconds and at is in m/s2. If the centrifuge starts from fest, determine the total acceleration magnitude of the cab after 10 seconds. © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Group Problem Solving Define your coordinate system In the side view, the tangential direction points into the “page”

en

Determine the tangential velocity at  0.5 t vt   0.5 t dt  0.25t t

2 t

0

0

 0.25t 2

Top View

vt  0.25 10  25 m/s 2

et en

Determine the normal acceleration vt   252 an    78.125 m/s2 r 8 2

Determine the total acceleration magnitude amag  an2  at2  78.1252 + (0.5)(10)

2

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

amag  78.285 m/s2 11 - 56

Tenth Edition

Vector Mechanics for Engineers: Dynamics Radial and Transverse Components By knowing the distance to the aircraft and the angle of the radar, air traffic controllers can track aircraft. Fire truck ladders can rotate as well as extend; the motion of the end of the ladder can be analyzed using radial and transverse components.

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Radial and Transverse Components • The position of a particle P is expressed as a distance r from the origin O to P – this defines the radial direction er. The transverse direction eq is perpendicular to er

  r  rer • The particle velocity vector is

v  r er  rq eq • The particle acceleration vector is









a  r  rq 2 er  rq  2rq eq © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 11.12 SOLUTION: • Evaluate time t for q = 30o.

• Evaluate radial and angular positions, and first and second derivatives at time t. Rotation of the arm about O is defined by q = 0.15t2 where q 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.

• Calculate velocity and acceleration in cylindrical coordinates. • Evaluate acceleration with respect to arm.

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. © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 11.12 SOLUTION: • Evaluate time t for q = 30o.

q  0.15t 2  30  0.524 rad

t  1.869 s

• Evaluate radial and angular positions, and first and second derivatives at time t.

r  0.9  0.12 t 2  0.481 m r  0.24 t  0.449 m s r  0.24 m s 2

q  0.15 t 2  0.524 rad q  0.30 t  0.561rad s q  0.30 rad s 2 © 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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Tenth Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 11.12 • Calculate velocity and acceleration. vr  r  0.449 m s vq  rq  0.481m 0.561rad s   0.270 m s v v  vr2  vq2   tan 1 q vr

v  0.524 m s

  31.0

ar  r  rq2  0.240 m s 2  0.481m0.561rad s 2  0.391m s 2 aq  rq  2rq





 0.481m 0.3 rad s2  2 0.449 m s 0.561rad s   0.359 m s 2 a  ar2  aq2

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

a   tan 1 q

ar a  0.531m s

  42.6 11 - 61

Tenth Edition

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.

aB OA  r  0.240 m s 2

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

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