Chapter 17. Plane Motion of Rigid Bodies: Energy and Momentum Methods Introduction Principle of Work and Energy for a Rigid Body Work of Forces Acting on a Rigid Body Kinetic Energy of a Rigid Body in Plane Motion Systems of Rigid Bodies Conservation of Energy Power Principle of Impulse and Momentum Systems of Rigid Bodies Conservation of Angular Momentum Impulsive Motion Eccentric Impact
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Introduction Principle of Work and Energy for a Rigid Body ... 17...Principle of Work and Energy • Work and kinetic energy are scalar quantities. • Assume that the rigid body is
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Chapter 17. Plane Motion of Rigid Bodies:
Energy and Momentum Methods
Introduction
Principle of Work and Energy for a Rigid Body
Work of Forces Acting on a Rigid Body
Kinetic Energy of a Rigid Body in Plane Motion
Systems of Rigid Bodies
Conservation of Energy
Power
Principle of Impulse and Momentum
Systems of Rigid Bodies
Conservation of Angular Momentum
Impulsive Motion
Eccentric Impact
Introduction
• Method of work and energy and the method of impulse and momentum will be used
to analyze the plane motion of rigid bodies and systems of rigid bodies.
• Principle of work and energy is well suited to the solution of problems involving
displacements and velocities.
• Principle of impulse and momentum is appropriate for problems involving velocities
and time.
• Problems involving eccentric impact are solved by supplementing the principle of
impulse and momentum with the application of the coefficient of restitution.
2211 TUT =+ →
( ) ( )21212
1
2
1
O
t
tOO
t
tHdtMHLdtFL
=+=+ ∑ ∫∑ ∫
Approaches to Rigid Body Kinetics Problems
Forces and Accelerations -> Newton’s Second Law (last chapter)
Velocities and Displacements -> Work-Energy
Velocities and Time -> Impulse-Momentum
1 1 2 2T U T→+ =
G
G G
F ma
M H
=
=
∑∑
2
11 2
t
tmv F dt mv+ =∫
2
11 2
t
G G GtI M dt Iω ω+ =∫
Principle of Work and Energy
• Work and kinetic energy are scalar quantities.
• Assume that the rigid body is made of a large number of particles.
initial and final total kinetic energy of particles forming body
total work of internal and external forces acting on particles of body.
• Internal forces between particles A and B are
equal and opposite.
• Therefore, the net work of internal forces is zero.
2211 TUT =+ →=21, TT
=→21U
Work of Forces Acting on a Rigid Body
• Work of a force during a displacement of its
point of application,
• Consider the net work of two forces
forming a couple of moment during a
displacement of their points of application.
if M is constant.
FF
− and M
( )12
212
1
θθ
θθ
θ
−=
= ∫→
M
dMU
( )2 2
1 1
1 2 cosA s
A s
U F dr F dsα→ = ⋅ =∫ ∫
1 1 2
2
dU F dr F dr F drF ds Fr dM d
θθ
= ⋅ − ⋅ + ⋅= ==
Ex:
Do the pin forces at point A do work?
Yes No
Does the force P do work?
Yes No
answer ; N/ Y
Does the normal force N do work on the disk?
Yes No
Does the weight W do work?
Yes No
If the disk rolls without slip, does the friction force F do work?
Yes No
answer ; N/N/N
( ) 0=== dtvFdsFdU cC
Kinetic Energy of a Rigid Body in Plane Motion
• Consider a rigid body of mass m in plane
motion consisting of individual particles i.
The kinetic energy of the body can then be
expressed as:
• Kinetic energy of a rigid body can be
separated into:
- the kinetic energy associated with the
motion of the mass center G and
- the kinetic energy associated with the rotation of the body about G.
Translation + Rotation
( )2 21 1
2 2
2 2 21 12 2
2 21 12 2
Δ
Δi i
i i
T mv m v
mv r m
mv I
ω
ω
′= +
′= +
= +
∑∑
2 21 12 2 T mv Iω= +
• Consider a rigid body rotating about a fixed axis through O.
• This is equivalent to using:
• Remember to only use
when O is a fixed axis of rotation
( ) ( )22 2 21 1 12 2 2
212
Δ Δ Δi i i i i i
O
T m v m r r m
I
ω ω
ω
= = =
=
∑ ∑ ∑
2 21 12 2 T mv Iω= +
212 OT I ω=
Systems of Rigid Bodies
• For problems involving systems consisting of several rigid bodies, the principle of
work and energy can be applied to each body.
• We may also apply the principle of work and energy to the entire system,
= arithmetic sum of the kinetic energies of all bodies
forming the system
= work of all forces acting on the various bodies, whether
these forces are internal or external to the system as a
whole.
2211 TUT =+ → 21,TT
21→U
T
T
Conservation of Energy
• Expressing the work of conservative forces as a
change in potential energy, the principle of work and
energy becomes
• Consider the slender rod of mass m.
• mass m
• released with zero velocity
• determine w at q
2211 VTVT +=+
0,0 11 == VT
( ) ( ) 22
22121
212
21
21
222
1222
12
321 ωωω
ω
mlmllm
IvmT
=+=
+=
θθ sinsin 21
21
2 mglWlV −=−=
1 1 2 22
21 10 sin2 3 23 sin
T V T Vml mgl
gl
ω θ
ω θ
+ = +
= −
=
Power
• Power = rate at which work is done
• For a body acted upon by force and moving with velocity ,
• For a rigid body rotating with an angular velocity and acted upon by a couple of
moment parallel to the axis of rotation,
F
v
vFdt
dU
⋅==Power
ω
M
ωθ MdtdM
dtdU
===Power
Sample Problem 17.1
For the drum and flywheel,
The bearing friction is equivalent to a couple of At the
instant shown, the block is moving downward at 2 m/s.
Determine the velocity of the block after it has moved 1.25 m
downward.
STRATEGY:
• Consider the system of the flywheel and block. The work
done by the internal forces exerted by the cable cancels.
• Note that the velocity of the block and the angular velocity
of the drum and flywheel are related by
• Apply the principle of work and kinetic energy to develop an expression for the final
velocity.
216kg m .I
ωrv =
MODELING and ANALYSIS:
• Consider the system of the flywheel and block. The work done
by the internal forces exerted by the cable cancels.
• Note that the velocity of the block and the angular velocity of
the drum and flywheel are related by
1 2 21 2
2m s 5rad s0.4m 0.4
v v vv rr r
• Apply the principle of work and kinetic energy to develop an
expression for the final velocity.
• Note that the block displacement and pulley rotation are
related by
Then,
2 21 11 1 12 2
2 221 1120 kg 2m s 16kg m 5rad s2 2440J
T mv I
2 21 11 1 12 2 440JT mv I
2 2 21 12 2 2 22 2 110T mv I v
22
1.25m 3.125rad0.4m
sr
( )
2 21 12 2 22 2
22 222 2
1 1120 kg 16 1102 2 0.4
T mv I
vv v
ω= +
= + =
• Principle of work and energy:
1 2 2 1 2 1
1177 2N 1.25m 90 N m 3.125rad1190J
U W s s M
1 1 2 222
2
440J 1190J 1103.85m s
T U T
vv
2 3.85m sv
REFLECT and THINK:
• The speed of the block increases as it falls, but much more slowly than if it were in
free fall. This seems like a reasonable result.
• Rather than calculating the work done by gravity, you could have also treated the
effect of the weight using gravitational potential energy, Vg.
Sample Problem 17.2
The system is at rest when a moment of is applied
to gear B.
Neglecting friction, a) determine the number of
revolutions of gear B before its angular velocity reaches
600 rpm, and b) tangential force exerted by gear B on
gear A.
STRATEGY:
• Consider a system consisting of the two gears. Noting that the gear rotational speeds are related,
evaluate the final kinetic energy of the system.
• Apply the principle of work and energy. Calculate the number of revolutions required for the work of
the applied moment to equal the final kinetic energy of the system.
• Apply the principle of work and energy to a system consisting of gear A. With the final kinetic energy
and number of revolutions known, calculate the moment and tangential force required for the indicated
work.
m
MODELING and ANALYSIS:
• Consider a system consisting of the two gears. Noting that
the gear rotational speeds are related, evaluate the final kinetic
energy of the system.
( )( )600 rpm 2 rad rev62.8rad s
60s min0.10062.8 25.1rad s0.250
B
BA B
A
rr
πω
ω ω
= =
= = =
( )( )( )( )
22 2
22 2
10kg 0.200m 0.400kg m
3kg 0.080m 0.0192kg mA A A
B B B
I m k
I m k
= = = ⋅
= = = ⋅
( )( ) ( )( )
2 21 12 2 2
2 21 12 20.400 25.1 0.0192 62.8163.9 J
A A B BT I Iω ω= +
= +
=
• Apply the principle of work and energy. Calculate
the number of revolutions required for the work.
• Apply the principle of work and energy to a system
consisting of gear A. Calculate the moment and
tangential force required for the indicated work.
( )1 1 2 2
0 6 J 163.9J27.32 rad
B
B
T U Tθ
θ
→+ =
+ =
=27.32 4.35rev
2Bθ π= =
( )( )221 12 2 2 0.400 25.1 126.0 JA AT I ω= = =
( )1 1 2 2
0 10.93rad 126.0J11.52 N m
A
A A
T U TM
M r F
→+ =
+ =
= = ⋅
0.10027.32 10.93rad0.250
BA B
A
rr
θ θ= = =
11.52 46.2 N0.250
F = =
REFLECT and THINK:
• When the system was both gears, the tangential force between the gears did not appear
in the work–energy equation, since it was internal to the system and therefore did no
work. If you want to determine an internal force, you need to define a system where the
force of interest is an external force. This problem, like most problems, also could have
been solved using Newton’s second law and kinematic relationships.
10kg 200mm3kg 80mm
A A
B B
m km k
= == =
Sample Problem 17.3
A sphere, cylinder, and hoop, each having the same
mass and radius, are released from rest on an incline.
Determine the velocity of each body after it has
rolled through a distance corresponding to a change
of elevation h.
STRATEGY:
• The work done by the weight of the bodies is
the same. From the principle of work and energy,
it follows that each body will have the same kinetic energy after the change of
elevation.
• Because each of the bodies has a different centroidal moment of inertia, the
distribution of the total kinetic energy between the linear and rotational components
will be different as well.
MODELING and ANALYSIS:
• The work done by the weight of the bodies is the same.
From the principle of work and energy, it follows that each
body will have the same kinetic energy after the change of
elevation.
With vr
ω =2
2 2 21 1 1 12 2 2 2 2
212 2
vT mv I mv Ir
Im vr
ω = + = +
= +
1 1 2 2
212 2
22 2
0
2 21
T U T
IWh m vr
Wh ghvm I r I mr
→+ =
+ = +
= =+ +
• Because each of the bodies has a different centroidal moment of inertia, the
distribution of the total kinetic energy between the linear and rotational components
will be different as well.
NOTE:
• For a frictionless block sliding through the same distance,
• The velocity of the body is independent of its mass and radius.
• The velocity of the body does depend on
22
12
mrIghv
+=
ghvmrIHoopghvmrICylinder
ghvmrISphere
2707.0:
2816.0:
2845.0:
2
221
252
==
==
==
ghv 2,0 ==ω
22
2 rk
mrI =
Sample Problem
A 15-kg slender rod pivots about the point O. The
other end is pressed against a spring (k = 300 kN/m)
until the spring is compressed 40 mm and the rod
is in a horizontal position.
If the rod is released from this position, determine
its angular velocity and the reaction at the pivot as
the rod passes through a vertical position.
STRATEGY:
• The weight and spring forces are conservative. The principle of work and energy
can be expressed as
• Evaluate the initial and final potential energy.
• Express the final kinetic energy in terms of the final angular velocity of the rod.
• Based on the free-body-diagram equation, solve for the reactions at the pivot.
2211 VTVT +=+
MODELING and ANALYSIS:
• The weight and spring forces are conservative. The
principle of work and energy can be expressed as
• Evaluate the initial and final potential energy.
• Express the final kinetic energy in
terms of the angular velocity of the rod.
2211 VTVT +=+
221 11 12 20 300,000 N m 0.040m
240Jg eV V V kx
2112
2
2
1 15kg 2.5m127.81kg m
I ml
( )( )2 0 147.15N 0.75m
110.4Jg eV V V Wh= + = + =
=
22 2 21 1 1 12 2 2 2 22 2 2 2
2 2 212 2 22
1 15 0.75 7.81 8.122
T mv I m r I
From the principle of work and energy,
• Based on the free-body-diagram equation, solve for