EXAMPLE 18.2

The bar shown has a mass of 10-kg and is subjected to a couple moment of M = 50 N.m and a force of P = 80 N, which is always applied perpendicular to the end of the bar. Also, the spring has an unstretched length of 0.5 m and remains in the vertical position due to the roller guide at B. determine the total work done by all the forces acting on the bar when it has rotated downward from θ = 0 to θ° = 90°.

EXAMPLE 18.2

Solution

First the free-body diagram of the bar is drawn in order to account for all the forces that act on it.

View Free Body Diagram

EXAMPLE 18.2

Weight W.• Since the weight 10(9.81) = 98.1

N is displaced downward 1.5 m, the work is

Couple Moment M. • The couple moment rotates

through an angle of θ = π/2 rad. Hence,

JUW 2.147)5.1(1.98 ==

JU M 5.78)2/(50 == π

EXAMPLE 18.2

Spring Force Fs

• When θ = 0°the spring is stretched (0.75 – 0.5) = 0.25 m, and when θ = 90°, the stretched is (2 + 0.75) – 0.5 = 2.25 m. Thus,

• By inspection the spring does negative work on the bar since Fs acts in the opposite direction to displacement. This check with the result.

JUs 0.75)25.0)(30(21)25.2)(30(

21 22 −=⎥⎦

⎤⎢⎣⎡ −−=

EXAMPLE 18.2

Force P• As the bar moves downward, the

force is displaced through a distance of (π/2)(3) = 4.172 m.

• The work is positive.

JU P 0.377)172.4(80 ==

EXAMPLE 18.2

Pin Reactions• Forces Ax and Ay do no work since

they are not displaced.

Total Work• The work of all forces when the bar

is displaced is thus

JU 5280.3770.755.782.147 =+−+=

EXAMPLE 18.1

The system of three elements shown consists of a 6-kg block B, a 10-kg disk D and a 12-kg cylinder C. If no slipping occurs, determine the total kinetic energy of the system at the instant shown.

EXAMPLE 18.1

SolutionFirst determine ωD, ωC and vG. From the kinematics of the disk,

Since the cylinder rolls without slipping, the instantaneous center of zero velocity is at the point of contact with the ground

sradmsmrv DDDDB /8)1.0(/8.0; === ωωω

View Free Body Diagram

EXAMPLE 18.1

smvrvsradrv

GCICGG

CCCICEE

/4.0)4.0)(1.0(;/4)2.0(8.0;

/

/

======

ωωωω

JvmT BBB 92.121 2 ==

Block

Disk

JrmIT DDDDDD 60.121

21

21 222 =⎟

⎠⎞

⎜⎝⎛== ωω

EXAMPLE 18.1

Cylinder

JrmmvImvT CCCGCGGC 44.121

21

21

21

21 22222 =⎟

⎠⎞

⎜⎝⎛+=+= ωω

Therefore the total kinetic energy of the system is

J

TTTT CDB

96.444.160.192.1

=++=

++=

EXAMPLE 18.7

The 10-kg rod AB is confined so that its ends move in the horizontal and vertical slots. The spring has a stiffness of k = 800 N/m and is unstretched when θ = 0°. Determine the angular velocity of AB when θ = 0°, if the rod is released from rest when θ = 30°.

EXAMPLE 18.7

Potential Energy• The two diagrams of the rod, when it is

located at its initial and final positions as shown

• The datum, used to measure the gravitational potential energy, is placed in line with the rod when θ = 0°.

EXAMPLE 18.7

• When the rod is in position 1, the center of gravity G is located below the datum so that the gravitational potential energy is negative.

• (positive) elastic potential energy is stored in the spring, since it is stretched a distance of s1 = (0.4 sin 30°) m, thus

J

ksWyV

19.6

)30sin4.0)(800(21)30sin2.0(1.98

21

2

2111

=

+−=

−−=

oo

EXAMPLE 18.7

• When the rod is in position 2, the potential energy of the rod is zero, since the spring is unstretched, s2 = 0, and the center of gravity G is located at the datum. Thus,

Kinetic Energy• The rod is released from rest from

position 1, thus (vG)1 = 0 and ω1 = 0, and

02 =V

01 =T

EXAMPLE 18.7

• In position 2, the angular velocity is ω2 and the rod’s mass center has a velocity of (vG)2. Thus,

• Using kinematics, (vG)2 can be related to ω2 as shown

22

222

22

222

])4.0)(10(121[

21))(10(

21

21)(

21

ω

ω

+=

+=

G

GG

v

IvmT

EXAMPLE 18.7

• At the instant considered, the instantaneous center of zero velocity (IC) for the rod is at point A; hence (vG)2 =(rG/IC)ω2 = (0.2)ω2

• Substituting into the previous expression and simplfying, we get

222 267.0 ω=T

EXAMPLE 18.7

Conservation of Energy

{ } { } { } { }{ } { } { } { }

srad

VTVT

/82.40267.019.60

2

22

2211

=+=+

+=+

ωω

EXAMPLE 18.8

The disk has a mass of 15 kg and a radius of gyration of kG = 0.18 m, and it is attached to a spring which has a stiffness k = 30 N/m and an unstretched length of 0.3 m. If the disk is released from rest in the position shown and rolls without slipping, determine its angular velocity at the instant G moves 0.9 m to the left.

EXAMPLE 18.8

Potential Energy• Two diagrams of the disk, when it is

located in its initial and final positions, are shown

• A gravitational datum is not needed here since the weight is not displaced vertically.

EXAMPLE 18.8

• From the problem geometry the spring is stretched s1 = [(0.92 + 1.22)½ - 0.3] = 1.2 m and s2 = (1.2 –0.3) = 0.9 m in the initial and final positions respectively.

• Hence,

JksV

JksV

15.12)9.0)(30(21

21

6.21)2.1)(30(21

21

2222

2211

===

===

EXAMPLE 18.8

Kinetic Energy• The disk is released from rest so that

(vG)1 = 0, ω1 = 0, and

• In the final position,

01=T

22

222

22

222

])18.0)(15[(21))(15(

21

21)(

21

ω

ω

+=

+=

G

GG

v

IvmT

EXAMPLE 18.8

• Since the disk rolls without slipping, (vG)2 can be related to ω2 from the instantaneous center of zero velocity, (vG)2 = 0.225ω2

• Substituting and simplifying yields222 6227.0 ω=T

EXAMPLE 18.8

Conservation of Energy

{ } { } { } { }{ } { } { } { }

srad

VTVT

/90.315.126227.06.210

2

22

2211

=+=+

+=+

ωω

EXAMPLE 18.9

The 10-kg homogeneous disk is attached to a uniform 5-kg rod AB. If the assembly is released from rest when θ = 60°, determine the angular velocity of the rod when θ = 0°. Assume that the disk rolls without slipping. Neglect friction along the guide and the mass of the collar at B.

EXAMPLE 18.9

Potential Energy• When the system is in position 1, the

rod’s weight has a positive potential energy. Thus,

• When the system is in position 2, both the weight of the rod and the weight of the disk have zero potential energy. Thus,

JyWV R 74.12)60sin3.0(05.4911 === o

02 =V

View Free Body Diagram

EXAMPLE 18.9

Kinetic Energy• Since the entire system is at rest at

the initial position,

• In the final position the rod has an angular velocity (ωR)2 and its mass center has a velocity (vG)2.

01 =T

EXAMPLE 18.9

• In the final position the rod has an angular velocity (ωR)2 and its mass center has a velocity (vG)2.

• Since the rod is fully extended in this position, the disk is momentarily at rest, so (ωD)2 = 0 and (vA)2 = 0.

EXAMPLE 18.9

• For the rod (vG)2 can be related to (ωR)2 from the instantaneous center of zero velocity, which is located at point A

• Hence, (vG)2 =(rG/IC)(ωR)2 or (vG)2 = (0.3)(ωR)2

22

22

222

22

22

22

222

)(3.0

00)]()6.0)(5(121[

21])(3.0)[5(

21

)(21)(

21)(

21)(

21

R

RR

DAADRGGR IvmIvmT

ω

ωω

ωω

=

+++=

+++=

EXAMPLE 18.9

Conservation of Energy

{ } { } { } { }{ } { } { } { }

srad

VTVT

R

R

/52.6)(0)(3.074.120

2

22

2211

=+=+

+=+

ωω

EXAMPLE 18.3

The 30-kg disk shown is pin supported at its center. Determine the number of revolutions it must take to attain an angular velocity of 20 rad/s starting from rest. It is acted upon by a constant force F = 10 N, which is applied to a cord wrapped around its periphery, and a constant couple moment M = 5 N.m

EXAMPLE 18.3

SolutionKinetic Energy• Since the disk rotates about a fixed

axis, the kinetic energy can be computed using T = ½ IOω2, where the moment of inertia is IO = ½ mr2

• Initially the disk is at rest, so that

( ) JIT

T

O 12020)2.0)(30(21

21

210

22222

1

=⎥⎦⎤

⎢⎣⎡==

=

ω

EXAMPLE 18.3

Work (Free-Body Diagram)• As shown in figure, the pin Ox and Oy

and the weight (294.3 N) do no work, since they are not displaced.

• The couple moment, having a constant magnitude, does positive work UM = Mθas the disk rotates through a clockwise angle of θ rad, and the constant force Fdoes positive work UFC

= Fs as the cord moves downward s = θr = θ(0.2 m)

EXAMPLE 18.3

Principle of Work and Energy{ } { } { }

{ } { } { }{ } { } { }

revrad

revradrad

TFsMTTUT

73.2211.171.17

120)2.0()10()5(021

2211

=⎟⎟⎠

⎞⎜⎜⎝

⎛==

=++=++

=+ ∑ −

πθ

θθθ

EXAMPLE 18.4

The 700-kg pipe is equally suspended from the two tines of the fork lift. It is undergoing a swinging motion such that when θ = 30° it is momentarily at rest. Determine the normal and frictional forces acting on each tine which are needed to support the pipe at the instant θ = 0°. Measurement of the pipe and the suspender are shown in.

EXAMPLE 18.4

Solution• Using the equation of motion to find

the forces on the tines since these forces do no work.

• Before doing this, however, we will apply the principle of work and energy to determine the angular velocity of the pipe when θ = 0°.

Kinetic Energy (Kinematic Diagram)• Since the pipe is originally at rest,

then 01 =T

EXAMPLE 18.4

• The final kinetic energy may be computed with reference to either the fixed point O or the center of mass G.

• For the calculation, consider the pipe to be a thin ring so that IG = mr2. If point Gis considered, we have

[ ] [ ]22

22

222

22

222

875.63

)15.0)(700(21)4.0()700(

21

21)(

21

ω

ωω

ω

=

+=

+= GG IvmT

EXAMPLE 18.4

• If point O is considered then the parallel-axis theorem must be used to determine IO. Hence,

[ ]22

22

22222

875.63

)4.0(700)15.0(70021

21

ω

ωω

=

+== OIT

Work (Free-Body Diagram)• The normal and frictional forces on the tines do no work since they do not move as the pipe swings.

EXAMPLE 18.4

• The weight, centered at G, does positive work since the weight moves downward through a vertical distance Δy = 0.4 m – 0.4 cos 30° m = 0.05359 m.

Principle of Work and Energy

{ } { } { }

{ } { } { }srad

TUT

/40.2875.63)05359.0)(81.9(7000

2

22

2211

==+

=+ ∑ −

ωω

EXAMPLE 18.4

Equations of Motion• Referring to the free-body and

kinetic diagrams as shown, we have

αα ])4.0(700)15.0(700[0;

)4.0()40.2(700)81.9(700;)(

)(700;)(

22

2

+==+

=−=↑+

==←

∑∑∑

+

OO

TnGn

tGTtGt

IM

NamF

aFamF

EXAMPLE 18.4

• Since (aG)t = 0.4α, then

• There are two tines used to support the load, therefore

kNNF

a

T

T

tG

48.80

0)(,0

==

==α

kNkNN

F

t

t

24.42

48.80

==′

=′

EXAMPLE 18.5

The wheel weighs 20 kg and has a radius of gyration kG = 0.18 m about its mass center G. if it is subjected to a clockwise couple moment of 22 N.mand rolls from rest with slipping, determine its angular velocity after its center G moves 0.15 m. The spring has a stiffness k = 160 N/m and is initially unstretched when the couple moment is applied.

EXAMPLE 18.5SolutionKinetic Energy (Kinematic Diagram)• Since the wheel is initially at rest, • The kinematic diagram of the wheel

when it is in the final position is shown, hence the final kinetic energy is

22

222

22

222

])18.0)(20[(21))(20(

21

21)(

21

ω

ω

+=

+=

G

GG

v

IvmT

01 =T

EXAMPLE 18.5

• The velocity of the mass center can be related to the angular velocity from the instantaneous center of zero velocity (IC), (vG)2 = 0.24ω2.

• Substituting into the equation and simplifying, we have

222 9.0 ω=T

EXAMPLE 18.5

Work (Free-body Diagram)• As shown, only the spring force Fs

and the couple moment do work• The normal force does not move

along its line of action and the frictional force does no work, since the wheel does not slip as it rolls.

• The work of Fs may be computed using Us = -½ ks2.

EXAMPLE 18.5

• Here the work is negative since Fs is in the opposite direction to displacement.

• Since the wheel does not slip when the center G moves 0.15 m, then the wheel rotates θ = sG/rG/IC= 0.15/0.24 = 0.625 rad

• Hence the spring stretches sA = θrA/IC = 0.625 (0.48) = 0.3 m

EXAMPLE 18.5

Principle of Work and Energy

{ } { } { }

{ } { }

{ }

srad

TksMT

TUT

/70.2

))(9.0()3.0)(160(21)625.0(220

21

2

22

2

22

1

2211

=

=−+

=⎭⎬⎫

⎩⎨⎧ −+

=+ ∑ −

ω

ω

θ

EXAMPLE 18.6

The 10-kg rod is constrained so that its ends move along the grooved slots. The rod is initially at rest when θ= 0°. If the slider block at B is acted upon by a horizontal force P = 50 N, determine the angular velocity of the rod at the instant θ = 45°.

EXAMPLE 18.6

• Two kinematic diagrams of the rod, when it is, in the initial position 1 and final position 2, are shown

• When the rod is in position 1, T1 = 0 since (vG)1= ω1 = 0. In position 2 the angular velocity is ω2 and the velocity of the mass center is (vG)2

Kinetic Energy (Kinematic Diagrams)

View Free Body Diagram

EXAMPLE 18.6

• Hence, the kinetic energy is

• The 2 unknowns may be related from the instantaneous center of zero velocity for the rod.

22

22

22

222

22

222

)(267.0)(5

])8.0)(10(121[

21))(10(

21

21)(

21

ω

ω

ω

+=

+=

+=

G

G

GG

v

v

IvmT

EXAMPLE 18.6

• It is seen that as A moves downward with a velocity (vA)2, B moves horizontally to the left with a velocity (vB)2

• Knowing these directions, the IC is determined as shown in the figure

• Therefore,22

22

222

2

22/2

067.1267.08.0

4.0)45tan4.0()(

ωωω

ωωω

=+=

===

T

rv ICGGo

EXAMPLE 18.6

Work (Free-body Diagram)• The normal forces NA and NB do no

work as the rod is displaced• The 98.1 N weight is displaced a

vertical distance of Δy = (0.4 – 0.4 cos45°) m; whereas the 50-N force moves a horizontal distance of s = (0.8 sin 45°) m

• Both of these forces do positive work

EXAMPLE 18.6

Principle of Work and Energy{ } { } { }

{ } { } { }{ } { } { }

srad

TPsyWTTUT

/11.6067.1)45sin8.0(50)45cos4.04.0(1.980

2

22

21

2211

==+−+

=+Δ+

=+ ∑ −

ωωoo