Dynamics of Mechanical Systems 2009 Part 8 docx

Because gravity is the only force applied to P and because the path of movement of Pis parallel to the weight force through a distance h, the work done W is simply: 10.7.1 Because P is r

If we multiply the terms of Eq (10.6.12) by vG , the velocity of G in R, we have:

From Eqs (10.4.1), (10.4.2), and (10.4.3) we can recognize the left side of Eq (10.6.16) as

the derivative of the work W of the force system acting on B Also, from Eq (10.5.7) we recognize the right side of Eq (10.6.16) as the derivative of the kinetic energy K of B.

Hence, Eq (10.6.16) takes the form:

In the remaining sections of this chapter we will consider several examples illustratingapplication of this principle We will also consider combined application of this principlewith the impulse–momentum principles of Chapter 9

10.7 Elementary Example: A Falling Object

Consider first the simple case of a particle P with mass m released from rest at distance

h above a horizontal surface S as in Figure 10.7.1 The objective is to determine the speed

v of P when it reaches S.

F v⋅ G=MaG⋅vG= d vG

dt M

12

2

T⋅ = ⋅ωω ( )I αα ωω ωω⋅ +[ × ⋅( )I ωω ωω]⋅ = d  ωω⋅ ⋅I ωω

dt

12

2

12

2

dW dt

dK dt

=

W=K2−K1=∆K

Because gravity is the only force applied to P and because the path of movement of P

is parallel to the weight force through a distance h, the work done W is simply:

(10.7.1)

Because P is released from rest, its initial kinetic energy is zero When P reaches S, its

kinetic energy may be expressed as:

10.8 Elementary Example: The Simple Pendulum

Consider next the simple pendulum depicted in Figure 10.8.1 where the mass m of the pendulum is concentrated in the bob P which is supported by a pinned, massless rod of

length  as shown Let θ measure the inclination of the pendulum to the vertical Supposethe pendulum is held in a horizontal position and released from rest The objective is to

determine the speed v of P as it passes through the equilibrium position θ = 0 If we

consider a free-body diagram of P as in Figure 10.8.2, we see that of the two forces applied

to P the tension of the connecting rod does no work on P because its direction is dicular to the movement of P From Eq (10.2.18), we see that the work W done by the

perpen-weight force is:

Because the pendulum is released from rest its initial kinetic energy is zero Its kinetic

energy K as it passes through the equilibrium position may be expressed as:

Observe that this speed is the same as that of an object freely falling through a distance

 (see Eq (10.7.4)), even though the direction of the velocity is different

As a generalization of this example, consider a pendulum released from rest at an angle

θi with the objective of determining the speed of P when it falls to an angle θf, as in Figure10.8.3 The work iWf done by gravity as the pendulum falls from θi to θf is:

(10.8.5)

where ∆h is the change in elevation of P as the pendulum falls (see Eq (10.2.20)) The kinetic energies K i and K f of P when θ is θi and θf are:

(10.8.6)

where i and f are the values of when θ is θi and θf, respectively

The work–energy principle then leads to:

12

(10.8.8)Solving for f we have:

(10.8.9)

The result of Eq (10.8.9) could also have been obtained by integrating the governingdifferential equations of motion obtained in Chapter 8 Recall from Eq (8.4.4) that for asimple pendulum the governing equation is:

(10.8.10)Then, by multiplying both sides of this equation by , we have:

(10.8.11)Because may be recognized as being (1/2)d 2/dt, we can integrate the equation and

obtain:

(10.8.12)Because is zero when θ is θi, the constant is –(g/)cosθi Therefore, we have:

(10.8.13)

When θ is θf, Eqs (10.8.9) and (10.8.13) are seen to be equivalent

The work–energy principle may also be used to determine the pendulum rise anglewhen the speed at the equilibrium position (θ = 0) is known Specifically, suppose theangular speed of the pendulum when θ is zero is o Then, the work W done on the

pendulum as it rises to an angle θf is:

(10.8.14)

FIGURE 10.8.3

A simple pendulum released from

rest and falling through angle θ i – θ f

where the negative sign occurs because the upward movement of the pendulum isopposite to the direction of gravity, producing negative work The work–energy principle,then, is:

(10.8.15)or

(10.8.16)or

(10.8.17)

If the pendulum is to rise all the way to the vertical equilibrium position (θ = π), we have:

(10.8.18)

If is exactly 4g, the pendulum will rise to the vertical equilibrium position and come

to rest at that position If exceeds 4g, the pendulum will rise to the vertical positionand rotate through it with an angular speed given by Eq (10.8.18) (sometimes called the

rotating pendulum).

10.9 Elementary Example — A Mass–Spring System

For a third fundamental example, consider the mass–spring system depicted in Figure

10.9.1 It consists of a block B with mass m and a linear spring, with modulus k, moving without friction or damping in a horizontal direction Let x measure the displacement of

B away from equilibrium.

Suppose B is displaced to the right (positive x displacement with the spring in tension)

a distance δ away from equilibrium Let B be released from rest in this position Questions

arising then are what is the speed v of B as it returns to the equilibrium position (x = 0), and how far to the left of the equilibrium position does B go?

m

x

B

k

To answer these questions using the work–energy principle, recall from Eq (10.2.22)that when a linear spring is stretched (or compressed) a distance δ, the corresponding

work W done by the stretching (or compressing) force is (1/2)kδ2 Because the force exerted

on the spring is equal, but oppositely directed, to the force exerted on B, the work done

on B as the spring is relaxed is also (1/2)kδ2 (That is, the work on B is positive because the force of the spring on B is in the same direction as the movement of B.)

Because B is released from rest, its initial kinetic energy is zero The kinetic energy at

the equilibrium position is:

(10.9.1)Then, from the work–energy principle, we have:

(10.9.2)or

(10.9.3)

where the minus sign is selected because B is moving to the left.

Next, as B continues to move to the left past the equilibrium position, the spring force will be directed opposite to the movement of B Therefore, the work W done on B as B moves to the left a distance d from the equilibrium position is:

(10.9.4)

When B moves to its leftmost position, its kinetic energy is zero From Eqs (10.9.1) and (10.9.2), the kinetic energy of B at the equilibrium position is:

(10.9.5)The work–energy principle then produces:

(10.9.6)or

(10.9.7)That is, the block moves to the left by precisely the same amount as it was originallydisplaced to the right

The usual explanation of this phenomenon is that when the spring is stretched (or

compressed) the work done by the stretching (or compressing) force stores energy (potential energy) in the spring This stored energy in the spring is derived from the kinetic energy

of the block Then, as the spring is relaxing, its potential energy is transferred back tokinetic energy of the block There is thus a periodic transfer of energy between the spring

and the block, with the sum of the potential energy of the spring and the kinetic energy

of the block being constant (We will discuss potential energy in the next chapter.)

As another example of work–energy transfer, consider the mass–spring system arranged

vertically as in Figure 10.9.2 Suppose B is held in a position where the spring is unstretched If B is then released from rest from this position, it will fall and stretch the

spring and eventually come to rest at an extreme downward position Questions arising

then include how far does B fall and what is the spring force when B reaches this maximum downward displacement? To answer these questions, consider that as B falls, the weight (or gravity) force on B is in the direction of the movement of B, whereas the spring force

on B is opposite to the movement of B Because B is at rest at both the beginning and the end of the movement, there is no change in the kinetic energy of B The net work on B is

then zero That is,

(10.9.8)

where d is the distance B moves downward By solving for d we obtain:

(10.9.9)The spring force in this extended position is, then,

(10.9.10)

The result of Eq (10.9.10) shows that a suddenly applied weight load on a spring creates

a force twice that of the weight This means that if a weight is suddenly placed on amachine or structure the force generated is twice that required to support the weight in

a static equilibrium configuration

10.10 Skidding Vehicle Speeds: Accident Reconstruction Analysis

The work–energy principle is especially useful in determining speeds of accident cles by using measurements of skid-mark data Indeed, the work–energy principletogether with the conservation of momentum principles are the primary methods used

vehi-by accident reconstructionists when attempting to determine vehicle speeds at variousstages of an accident

To illustrate the procedure, suppose an automobile leaves skid marks from all four

wheels in coming to an emergency stop Given the length d of the skid marks, the objective

is to determine the vehicle speed when the marks first began

Skid marks are created by abrading and degrading tires sliding on a roadway surface.The tire degradation is due to friction forces and heat abrading the rubber The frictionforces are proportional to the normal (perpendicular to the roadway surface) forces onthe tires and to the coefficient of friction µ The friction coefficient, ranging in value from

0 to 1.0, is a measure of the relative slipperiness between the tires and the roadway

pavement If F and N are equivalent friction and normal forces (see Section 6.5), they are

related by the expression:

(10.10.1)

If an automobile is sliding on a flat, level (horizontal) roadway, a free-body diagram of

the vehicle shows that the normal force N is equal to the vehicle weight w Then, as the vehicle slides a distance d, the work W done by the friction force (acting opposite to the

direction of the sliding) is:

(10.10.2)

where m is the mass of the automobile.

Let v be the desired speed of the automobile when the skid marks first appear Then, the kinetic energy Ki of the vehicle at that point is:

(10.10.3)

Because the kinetic energy K f at the end of the skid marks is zero (the vehicle is thenstopped), the work energy principle produces:

(10.10.4)or

(10.10.5)

Observe that the calculated speed is independent of the automobile mass

To illustrate how the work–energy principle may be used in conjunction with the

momentum conservation principles, suppose an automobile with mass m1 slides a distance

d1 before colliding with a stopped automobile having mass m2 Suppose further that the

two vehicles then slide together (a plastic collision) for a distance d2 before coming to rest.The questions arising then are what were the speeds of the vehicles just before and justafter impact and what was the speed of the first vehicle when it first began to slide?

To answer these questions, consider first from Eq (10.10.5) that the speed v a of thevehicles just after impact is:

Next, during impact, the momentum is conserved That is,

(10.10.7)

where v b is the speed of the first vehicle just before impact (see Eq (9.7.1))

From Eq (10.10.2), the work W done by friction forces on the first vehicle as it slides a distance d1 before the collision is:

(10.10.8)

If v0 is the speed of the first vehicle when skidding begins, the change in kinetic energy

of the vehicle from the beginning of skidding until the collision is:

(10.10.9)The work–energy principle then gives:

(10.10.10)Finally, using Eqs (10.10.6) and (10.10.7), we can solve Eq (10.10.10) for :

(10.10.11)

Observe from Eq (10.10.7) that the speed va of the first vehicle just after the collision with the second vehicle is reduced by the factor [m1/(m1 + m2)] That is, the change inspeed ∆v is:

(10.10.12)

Observe further that because the velocity changes during the impact the kinetic energyalso changes That is, even though the momentum is conserved, the kinetic energy is notconserved Indeed, the change in kinetic energy ∆K just before and just after the impact is:

W=∆K or −µm gd1 1=( )m v1 b−( )m v

2

1 0 2

2 2

10.11 A Wheel Rolling Over a Step

For a second example illustrating the tandem use of the work–energy principle and theconservation of momentum principle, consider again the case of the wheel rolling over astep as in Figure 10.11.1 (recall that we considered this problem in Section 9.9) Let the

wheel W have a radius r, mass m, and axial moment of inertia I Suppose we are interested

in knowing the speed v of the wheel center required for the wheel to roll over the step.

Recall in Section 9.9 that when the wheel encounters the step its angular momentum

about the corner (or nose) O of the step is conserved By using the conservation of angular momentum principle we found that the angular speed of W just after impact is (see

Eq (9.9.8)):

(10.11.1)

where ω is the angular speed of W before impact and h is the height of the step.

After impact, W rotates about the nose O of the step For W to roll over the step it must

have enough kinetic energy after impact to overcome the negative work of gravity as it

rises up over the step The work W g of gravity as W rolls completely up the step is:

(10.11.4)Solving we have:

ω =[2mgh (I mr+ 2) ]1 2

Observe also that, as with the colliding vehicles, the energy is not conserved when the

wheel impacts the step From Eq (10.11.1), the energy loss L is seen to be:

(10.11.8)

For a uniform circular disk, this becomes:

(10.11.9)

10.12 The Spinning Diagonally Supported Square Plate

For a third example of combined application of the work–energy and momentum vation principles, consider again the spinning square plate supported by a thin wire along

conser-a diconser-agonconser-al conser-as in Figure 10.12.1 (recconser-all thconser-at we studied this problem in Section 9.12) As

before, let the plate have side length a, mass m, and initial angular speed Ω Let the plate

be arrested along an edge so that it rotates about that edge as in Figure 10.12.2

Ω

Recall that in Section 9.12 we discovered through the conservation of angular tum principle that the post-seizure rotation speed is related to the pre-seizure speed Ω

momen-by the expression:

(10.12.1)

In view of the result of Eq (10.12.1), a question that may be posed is what should thepre-seizure speed be so that after seizure the plate rotates through exactly 180°, coming

to rest in the upward configuration shown in Figure 10.12.3?

We can answer this question using the work–energy principle Because gravity is the

only force doing work on the plate, the work W is simply the plate weight multiplied by

the mass center elevation change (see Figure 10.12.4):

(10.12.2)where the negative sign occurs because the mass center elevation movement is opposite

to the direction of gravity

The final kinetic energy K f is zero because the plate comes to rest The initial kinetic

energy K i just after the plate edge is seized, is:

(10.12.3)The work–energy principle then produces:

(10.12.4)Solving for , we obtain:

(10.12.5)Then, from Eq (10.12.1), Ω, the pre-seizure angular speed, is:

Ω

Ω =[3 2g a]1 2

Ω=( )8 2 Ωˆ =[96 2g a]1 2/

10.13 Closure

The work–energy principle is probably the most widely used of all the principles ofdynamics The primary advantage of the work–energy principle is that it only requiresknowledge of velocities and not accelerations Also, calculation of the work done is oftenaccomplished by inspection of the system configuration

The major disadvantage of the work–energy principle is that only a single equation isobtained Hence, if there are several unknowns with a given mechanical system, at mostone of these can be obtained using the work–energy principle This in turn means thatthe principle is most advantageous for relatively simple mechanical systems However,the utility of the principle may often be enhanced by using it in tandem with otherdynamics principles — particularly impulse–momentum principles

In the next two chapters we will consider more general energy methods We will considerthe procedures of generalized dynamics, Lagrange’s equations, and Kane’s equations.These procedures, while not as simple as those of the work–energy principle, have theadvantage of still being computationally efficient and of producing the same number ofequations as there are degrees of freedom of a system

References (Accident Reconstruction)

10.1 Baker, J S., Traffic Accident Investigation Manual, The Traffic Institute, Northwestern University,

Evanston, IL, 1975.

10.2 Backaitis, S H., Ed., Reconstruction of Motor Vehicle Accidents: A Technical Compendium,

Publi-cation PT-34, Society of Automotive Engineers (SAE), Warrendale, PA, 1989.

10.3 Platt, F N., The Traffic Accident Handbook, Hanrow Press, Columbia, MD, 1983.

10.4 Moffatt, E A., and Moffatt, C A., Eds., Highway Collision Reconstruction, American Society of

Mechanical Engineers, New York, 1980.

10.5 Gardner, J D., and Moffatt, E A., Eds., Highway Truck Collision Analysis, American Society of

Mechanical Engineers, New York, 1982.

10.6 Adler, U., Ed., Automotive Handbook, Robert Bosch, Stuttgart, Germany, 1986.

10.7 Collins, J C., Accident Reconstruction, Charles C Thomas, Springfield, IL, 1979.

10.8 Limpert, R., Motor Vehicle Accident Reconstruction and Cause Analysis, The Michie Company,

Low Publishers, Charlottesville, VA, 1978.

10.9 Noon, R., Introduction to Forensic Engineering, CRC Press, Boca Raton, FL, 1992.

Problems

Sections 10.2 and 10.3 Work

P10.2.1: A particle P moves on a curve C defined by the parametric equations:

x=t, y=t2, z t= 3

where x, y, and z are coordinates, measured in meters, relative to an X, Y, Z Cartesian

system Acting on P is a force F given by:

where F is measured in Newtons, and nx, ny, and nz are unit vectors parallel to the X-, Y-,

and Z-axes Compute the work done by F in moving P from (0, 0, 0) to (2, 4, 8).

P10.2.2: The magnitude and direction of a force F acting on a particle P depend upon the coordinate position (x, y, z) of F (and P) in an X, Y, Z coordinate space as:

where nx, ny, and nz are unit vectors parallel to X, Y, and Z Suppose P moves from the

origin O to a point C (1, 2, 3) along two different paths as in Figure P10.2.2: (1) along the line segment OC, and (2) along the rectangular segments OA, AB, BC Calculate the work

done by F on P in each case (Assume that the coordinates are measured in meters.)

P10.2.3: See Problem P10.2.2 Suppose a force F acting on a particle P depends upon the position of F (and P) in an X, Y, Z space as:

Show that the work done by F on P as P moves from P1 (x1, y1, z1) to P2 (x2, y2, z2) is simply

φ (x2, y2, z2) – φ (x1, y1, z1) Comment: When a force F can be represented in the form F =

∇φ, F is said to be conservative.

P10.2.4: See Problems P10.2.2 and P10.2.3 Show that the force F of Problem P10.2.2 is

conservative Determine the function φ Using the result of Problem P10.2.3, check theresult of P10.2.2

P10.2.5: A horizontal force F pushes a 50-lb cart C up a hill H which is modeled as a

sinusoidal curve with amplitude of 7 ft and half-period of 27 ft as shown in Figure P10.2.5

Assuming that there is no frictional resistance between C and H and that F remains directed

horizontally, find the work done by F.

FIGURE P10.2.2

A particle P moving from O to C along

two different paths.

P10.2.6: A force pushes a block B along a smooth horizontal slot from a position O to a position Q as in Figure P10.2.6 The movement of B is resisted by a linear spring with a

natural (unstretched) length of 6 in and modulus 12 lb/in Determine the work done by F.

P10.2.7: Repeat Problem P10.2.6 if the natural length of the spring is 4 in

P10.2.8: A motorist, in making a turn with a 15-in.-diameter steering wheel, exerts a force

of 8 lb with each hand tangent to the wheel as in Figure P10.2.8 If the wheel is turnedthrough an angle of 150°, determine the work done by the motorist

Section 10.4 Power

P10.4.1: An automobile with a 110-hp engine is traveling at 35 mph If there are no frictionallosses in the transmission or drive train, what is the tractive force exerted by the drivewheels?

P10.4.2: See Problem P10.4.1 Suppose the automobile of Problem P10.4.1 has diameter drive wheels, a transmission gear ratio of 8 to 1, and a drive axle gear ratio of4.5 to 1 What is the engine speed (in rpm)? What is the torque (in ft⋅lb) of the enginecrank shaft?

Section 10.5 Kinetic Energy

P10.5.1: An 2800-lb automobile starting from a stop accelerates at the rate of 3 mph persecond Find the kinetic energy of the vehicle after it has traveled 100 yards

P10.5.2: A double pendulum consists of two particles P1 and P2 having masses m1 and m2

supported by light cables with lengths 1 and 2, making angles θ1 and θ2 with the vertical

as in Figure P10.5.2 Find an expression for the kinetic energy of this system Express the

results in terms of m1, m2, 1, 2, θ1, θ2, 1, and 2

P10.5.3: Determine the kinetic energy of the rod pendulum of Figure P10.5.3 Let the rodhave length  and mass m Express the result in terms of m, , θ, and

P10.5.4: See Problem 10.5.3 Consider the double rod pendulum as in Figure 10.5.4 Leteach rod have length  and mass m Determine the kinetic energy of the system Express the results in terms of m, , θ1, θ2, 1, and 2

P10.5.5: See Problems P10.5.3 and P10.5.4 Extend the results of Problems P10.5.3 andP10.5.4 to the triple-rod pendulum of Figure P10.5.5

P10.5.6: An automobile with 25-in.-diameter wheels is traveling at 30 mph when theoperator suddenly swerves to the left, causing the vehicle to spin out and rotate at therate of 180°/sec If the wheels each weigh 62 lb and if their axial and diametral radii of

gyration are 10 and 7 in., respectively, determine the kinetic energy of one of the rearwheels for the spinning vehicle.

Sections 10.6 to 10.9: Work–Energy Principles and Applications

P10.6.1: A ball is thrown vertically upward with a speed of 12 m/sec Determine the maximum height h reached by the ball.

P10.6.2: An object is dropped from a window which is 45 ft above the ground What isthe speed of the object when it strikes the ground?

P10.6.3: See Problem P10.6.1 What is the speed of the ball when it is 3 m above the thrower

P10.6.4: A water faucet is dripping slowly When a drop has fallen 1 ft, a second drop

appears What are the speeds of the drops when the first drop has fallen 3 ft? What is theseparation between the drops at that time?

P10.6.5: A simple pendulum consists of a light string of length 3 ft and a concentrated mass P weighing 5 lb at the end as in Figure P10.6.5 Suppose the pendulum is displaced

through an angle θ of 60° and released from rest What is the speed of P when θ is (a) 45°,

(b) 30°, and (c) 0°?

P10.6.6: See Problem P10.6.5 Suppose P has a speed of 7.5 ft/sec when θ is 0 What is the

maximum angle reached by P?

P10.6.7: See Problems 10.6.5 and P10.6.6 What is the minimum speed v of P when θ is 0

so that the pendulum will make a complete loop without the string becoming slack even

in the topmost position (θ = 180°)?

P10.6.8: Repeat Problems P10.6.5 to 10.6.7 if the mass of P is 10 lb instead of 5 lb.

P10.6.9: Repeat Problems P10.6.5 to 10.6.7 if the length  of the pendulum is 4 ft

P10.6.10: Repeat Problems P10.6.5 and P10.6.6 if the simple pendulum of Figure P10.6.5

is replaced by a rod pendulum of length 3 ft and weight 5 lb as in Figure P10.6.10, with

P being the end point of the rod.

P10.6.11: A block B with mass m is attached to a vertical linear spring with modulus k as

in Figure P10.6.11 Suppose B is held in a position where the spring is neither stretched

nor compressed and is then released from rest Find:

a The maximum downward displacement of B

b The maximum force in the spring

c The maximum speed of B

d The position where the maximum speed of B occurs

P10.6.12: Solve Problem P10.6.11 for k = 7 lb/in and m = 0.25 slug.

P10.6.13: Solve Problem P10.6.11 for k = 12 N/cm and m = 2 kg.

P10.6.14: A 5-lb block B sliding in a smooth vertical slot is attached to a linear spring with modulus k of 4 lb/in as in Figure P10.6.14 Let the natural length  of the spring be 8 in

Let the displacement y of B be measured downward from O, opposite the spring anchor

Q as shown Find the speed of B when (a) y = 0, and (b) y = 5 in.

P10.6.15: Solve Problem P10.6.14 if the natural length of the spring is 6 in

P10.6.16: A 10-kg block B is placed at the top of an incline which has a smooth surface as represented in Figure P10.6.16 If B is released from rest and slides down the incline, what

will its speed be when it reaches the bottom of the incline?

FIGURE P10.6.10

A rod pendulum.

FIGURE P10.6.11

A vertical mass–spring system.

P10.6.17: See Problem P10.6.16 A 10-kg circular disk D with 0.25-m diameter is placed at

the top of an incline that has a perfectly rough surface as represented in Figure P10.6.17

If D is released from rest and rolls down the incline, what will be the speed of the center

O of D when it reaches the bottom of the incline? Compare the result with that of Problem

P10.6.16

P10.6.18: A solid half-cylinder C, with radius r and mass m, is placed on a horizontal surface S and held with its flat side vertical as represented in Figure P10.6.18 Let C be released from rest and let S be perfectly rough so that C rolls on S Determine the angular speed of C when its mass center G is in the lowest most position.

P10.6.19: Repeat Problem P10.6.18 if S is smooth instead of rough.

P10.6.20: A 7-kg, 1.5-m-long rod AB has its end pinned to light blocks that are free to move

in frictionless horizontal and vertical slots as represented in Figure P10.6.20 If the pins

are also frictionless, and if AB is released from rest in the position shown, determine the speed of the mass center G and the angular speed of AB when AB falls to a horizontal position and when AB has fallen to a vertical position Assume that the guide blocks at

A and B remain in their vertical and horizontal slots, respectively, throughout the motion.

P10.6.21: Repeat Problem P10.6.20 if the guide blocks at A and B are no longer light but

instead have masses of 1 kg each

A rod with ends moving in

frictionless guide slots.

P10.6.22: A 300-lb flywheel with radius of gyration of 15 in is rotating at 3000 rpm A

bearing failure causes a small friction moment of 2 ft⋅lb which in turn causes the flywheel

to slow and eventually stop How many turns does the flywheel make before coming to

a stop?

Section 10.10 Motor Vehicle Accident Reconstruction

P10.10.1: All four wheels of a car leave 75-ft-long skid marks in coming to a stop on a

level roadway If the coefficient of friction between the tires and the road surface is 0.75,determine the speed of the car at the beginning of the skid marks

P10.10.2: The front wheels of a car leave 70 ft of skid marks and its rear wheels leave 50

ft of skid marks in coming to a stop on a level roadway If the coefficient of friction betweenthe tires and the roadway is 0.80, and if 60% of the vehicle weight is on the front wheels,determine the speed of the car at the beginning of the skid marks

P10.10.3: A car slides to a stop leaving skid marks of length s (measured in feet) on a level

roadway If the coefficient of friction between the tires and the roadway is µ, show that

the speed v (in miles per hour) of the car at the beginning of the skid marks is given by

the simple expression:

P10.10.4: See Problem P10.10.3 Suppose that the roadway, instead of being level, has aslight down slope in the direction of travel as represented in Figure P10.10.4 If the downslope angle is θ (measured in radians) as shown, show that the speed formula of ProblemP10.10.3 should be modified to:

P10.10.5: An automobile leaves 50 ft of skid marks before striking a pole at 20 mph Findthe speed of the vehicle at the beginning of the skid marks if the roadway is level and ifthe coefficient of friction between the tires and the roadway is 0.65 Assume the vehiclestops upon hitting the pole

P10.10.6: A pickup truck leaves 30 ft of skid marks before colliding with the rear of a stopped automobile Following the collision, the two vehicles slide together (a plastic

collision) for 25 ft Let the coefficient of friction between the pickup truck tires and theroadway be 0.7; after the collision, for the sliding vehicles together, let the coefficient offriction with the roadway be 0.5 Let the weights of the pickup truck and automobile be

3500 lb and 2800 lb, respectively Find the speed of the pickup truck at the beginning ofthe skid marks

P10.10.7: Repeat Problem P10.10.6 if just before collision, instead of being stopped, theautomobile is moving at 10 mph in the same direction as the pickup truck

P10.10.8: Repeat Problem P10.10.6 if just before collision, instead of being stopped, theautomobile is moving toward the pickup truck at 10 mph (that is, a head-on collision).

Sections 10.11 and 10.12 Work, Energy, and Impact

P10.11.1: A 60-gauge bullet is fired into a 20-kg block suspended by a 7-m cable as depicted

in Figure P10.11.1 The impact causes the block pendulum to swing through an angle of

15° Determine the speed v of the bullet.

P10.11.2: A 14-in diameter wheel W1 is rotating at 350 rpm when it is brought into contact

with a 10-in.-diameter wheel W2, which is initially at rest, as represented in Figure P10.11.2

After the wheels come into contact, they roll together without slipping Although W1 is free

to rotate, with negligible friction, W2 is subjected to a frictional moment of 2 ft⋅lb in its

bearings Let the weights of W1 and W2 be 28 lb and 20 lb, respectively Let the radii of

gyration of W1 and W2 be 5 in and 3.5 in., respectively Determine the number of revolutions

N1 and N2 turned by each wheel after the meshing contact until they come to rest

P10.11.3: A 1-m rod B with a mass of 1 kg is hanging

vertically and supported by a frictionless pin at O as

in Figure P10.11.3 A particle P with a mass of 0.25

kg moving horizontally with speed v collides with

the rod as also indicated in Figure P10.11.3 If the

collision is perfectly plastic (with coefficient of

resti-tution e = 0), determine v so that B completes exactly

one half of a revolution and then comes to rest in a

vertically up position Let the point of impact x be

(a) 0.5 m; (b) 0.667 m; and (c) 1.0 m

P10.11.4: Repeat Problem P10.11.3 for a perfectly

elastic collision (coefficient of restitution e = 1).

A particle P colliding with a vertical pin

supported rod initially at rest.

In Chapter 10, we found that the work–energy principle, like clever statics solutionprocedures, can often produce simple and direct solutions to dynamics problems We alsofound, however, that while the work–energy principle is simple and direct, it is also quiterestricted in its range of application The work–energy principle leads to a single scalarequation, thus enabling the determination of a single unknown Hence, if two or moreunknowns are to be found, the work–energy principle is inadequate and is restricted torelatively simple problems.

The objective of generalized dynamics is to extend the relatively simple analysis of thework–energy principle to complex dynamics problems having a number of unknowns.The intention is to equip the analyst with the means of determining unknowns with aminimal effort — as with insightful solutions of statics problems

In this chapter, we will introduce and discuss the elementary procedures of generalizeddynamics These include the concepts of generalized coordinates, partial velocities andpartial angular velocities, generalized forces, and potential energy In Chapter 12, we willuse these concepts to obtain equations of motion using Kane’s equations and Lagrange’sequations

11.2 Coordinates, Constraints, and Degrees of Freedom

In the context of generalized dynamics, a coordinate (or generalized coordinate) is a parameterused to define the configuration of a mechanical system Consider, for example, a particle

P moving on a straight line L as in Figure 11.2.1 Let x locate P relative to a fixed point O

on L Specifically, let x be the distance between O and P Then, x is said to be a coordinate of P.Next, consider the simple pendulum of Figure 11.2.2 In this case, the configuration ofthe system and, as a consequence, the location of the bob P are determined by the angle θ.Thus, θ is a coordinate of the system

354 Dynamics of Mechanical Systems

System coordinates are not unique For the systems of Figures 11.2.1 and 11.2.2 we couldalso define the configurations by the coordinates y and φ as in Figures 11.2.3 and 11.2.4.(In Figure 11.2.3, Q, like O, is fixed on L.)

As a mechanical system moves and its configuration changes, the values of the nates change This means that the coordinates are functions of time t In a dynamicalanalysis of the system, the coordinates become the dependent variables in the governingdifferential equations of the system From this perspective, constant geometrical parame-ters, such as the pendulum length  in Figure 11.2.2, are not coordinates

coordi-The minimum number of coordinates needed to define a system’s configuration is thenumber of degrees of freedom of the system Suppose, for example, that a particle P moves

in the X–Y plane as in Figure 11.2.5 Then, (x, y) or, alternatively, (r, θ) are coordinates of

P Because P has two coordinates defining its position, P is said to have two degrees offreedom

A restriction on the movement of a mechanical system is said to be a constraint Forexample, in Figure 11.2.5, suppose P is restricted to move only in the X–Y plane Thisrestriction is then a constraint that can be expressed in the three-dimensional X, Y, Z spaceas:

(11.2.1)Expressions describing movement restrictions, such as Eq (11.2.1) are called constraint equations A mechanical system may have any number of constraint equations, often morethan the number of degrees of freedom For example, the particle of Figure 11.2.1, restricted

to move on the straight line, has two constraint equations in the three-dimensional X, Y,

Z space That is,

Generalized Dynamics: Kinematics and Kinetics 355

The number of degrees of freedom of a mechanical system is the number of coordinates

of the system if it were unrestricted minus the number of constraint equations Forexample, if a particle P moves relative to a Cartesian reference frame R as in Figure 11.2.6,then it has, if unrestricted, three degrees of freedom If, however, P is restricted to move

in a plane (say, a plane parallel to the X–Z plane), then P is constrained, and its constraintmay be described by a single constraint equation of the form:

(11.2.3)

Hence, in this case there are three minus one, or two, degrees of freedom

To further illustrate these concepts, consider the system of two particles P1 and P2 atopposite ends of a light rod (a “dumbbell”) as in Figure 11.2.7 In a three-dimensionalspace, the two particles with unrestricted motion require six coordinates to specify theirpositions Let these coordinates be (x1, y1, z1) and (x2, y2, z2) as shown in Figure 11.2.7.Now, for the particles to remain at opposite ends of the rod, the distance between themmust be maintained at the constant value , the rod length That is,

(11.2.4)This is a single constraint equation; thus, the system has a net of five degrees of freedom

356 Dynamics of Mechanical Systems

If the movement of the dumbbell system is further restricted to the X–Y plane, additional

constraints occur, as represented by the equations:

(11.2.5)These expressions together with Eq (11.2.4) then form three constraint equations, leaving

the system with six minus three, or three, degrees of freedom These degrees of freedom

might be represented by either the parameters (x1, y1, θ) or (x2, y2, θ) as shown in Figure

11.2.8

As a final illustration of these ideas consider a rigid body B composed of N particles P i

(i = 1,…, N) moving in a reference frame R as in Figure 11.2.9 The rigidity of B requires

that the distances between the respective particles are maintained at constant values

Suppose, for example, that P1, P2, and P3 are noncollinear points Let p1, p2, and p3 locate

P1, P2, and P3 relative to O in R Then, the respective distances between these particles are

maintained by the equations:

(11.2.6)

where the distances d1, d2, and d3 are constants

P1, P2, and P3 thus form a rigid triangle The other particles of B are then maintained in

fixed positions relative to the triangle of P1, P2, and P3 by the expressions:

(11.2.7)

Equations (11.2.6) and (11.2.7) form 3 + 3(N – 3) or 3N – 6 constraint equations If the

particles of B are unrestricted in their movement, 3N coordinates would be required to

specify their position in R Hence, the number of degrees of freedom n of the rigid body is:

P

2 3 2