Download KINETIC ENERGY, WORK, PRINCIPLE OF WORK AND ENERGY

READING QUIZ
1
CEE 271: Applied Mechanics II, Dynamics
– Lecture 27: Ch.18, Sec.1–5 –
Kinetic energy due to rotation (only) of the body is defined
as
(a) 12 m(vG )2
(b) 12 m(vG )2 + (1/2)IG ω 2
(c) 12 IG ω 2
(d) IG ω 2
ANS: (c)
Prof. Albert S. Kim
2
Civil and Environmental Engineering, University of Hawaii at Manoa
When calculating work done by forces, the work of an
internal force does not have to be considered because
(a)
(b)
(c)
(d)
Date: __________________
internal forces do not exist
the forces act in equal but opposite collinear pairs
the body is at rest initially
the body can deform
ANS: (b)
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KINETIC ENERGY, WORK, PRINCIPLE OF WORK
AND ENERGY
Today’s objectives: Students
will be able to
1
2
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APPLICATIONS
In-class activities:
• Reading Quiz
Define the various ways a
force and couple do work.
• Applications
Apply the principle of work
and energy to a rigid body.
• Work of a Force or Couple
• Kinetic Energy
• Principle of Work and
Energy
• The work of the torque (or moment) developed by the
• Concept Quiz
driving gears on the two motors on the concrete mixer is
transformed into the rotational kinetic energy of the mixing
drum.
• Group Problem Solving
• Attention Quiz
• If the motor gear characteristics are known, how would you
find the rotational velocity of the mixing drum?
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READING QUIZ
1
Elastic potential energy is defined as
(a)
(b)
(c)
(d)
2
APPLICATIONS(continued)
.
• Two torsional springs are used to assist
+ 12 k(s)2
− 12 k(s)2
+ 12 k(v)2
None of the above
ANS: (a)
in opening and closing the hood of the
truck.
• Assuming the springs are uncoiled
when the hood is opened, can we
determine the stiffness of each spring
so that the hood can easily be lifted,
i.e., practically no external force applied
to it, when a person is opening it?
The kinetic energy of a rigid body consists of the kinetic
energy due to
.
(a)
(b)
(c)
(d)
translational motion and rotational motion
only rotational motion
only translational motion
the deformation of the body
ANS: (a)
• Are the gravitational potential energy of
the hood and the torsional spring
stiffness related to each other? If so,
how?
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APPLICATIONS
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CONSERVATION OF ENERGY (Section 18.5)
• The torsion springs located at the
• The conservation of energy theorem is a ‘simpler’ energy
top of the garage door wind up as
the door is lowered.
method (recall that the principle of work and energy is also
an energy method) for solving problems.
• When the door is raised, the
potential energy stored in the spring
is transferred into the gravitational
potential energy of the door’s weight,
thereby making it easy to open.
• Are parameters such as the torsional
spring stiffness and initial rotation
angle of the spring important when
you install a new door?
• Once again, the problem parameter of distance is a key
indicator for when conservation of energy is a good
method to solve a problem.
• If it is appropriate for the problem, conservation of energy is
easier to use than the principle of work and energy.
• This is because the calculation of the work of a
conservative force is simpler. But, what makes a force
conservative?
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EXAMPLE II(continued)
GROUP PROBLEM SOLVING
• Given: The 30 kg pendulum has its
mass center at G and a radius of
gyration about point G of
kG = 0.3 meter. It is released from
rest when θ = 0◦ . The spring is
un-stretched when θ = 0◦ .
• Now all terms in the conservation of energy equation have
been formulated. Writing the general equation and then
substituting into it yields:
T 1 + V1 = T 2 + V2
0 + 0 = 11.25(ω2 )2 + (−110.4 + 11.09)
• Find: The angular velocity of the pendulum when θ = 90◦ .
• Solving for ω2 = 2.97 rad/s
• Plan: Conservative forces and distance (θ) leads to the
use of conservation of energy. First, determine the
potential energy and kinetic energy for both positions.
Then apply the conservation of energy equation.
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UNDERSTANDING QUIZ
1
GROUP PROBLEM SOLVING: Potential Energy
At the instant shown, the spring is undeformed. Determine
the change in potential energy if the 20 kg disk
(kG = 0.5 meter) rolls 2 revolutions without slipping.
(a)
+ 12 (200)(1.2π)2
• Let’s put the datum when θ = 0◦ .
Then, the gravitational potential
energy and the elastic potential
energy will be zero. So,
Vg1 = Ve1 = 0
Note that the un-stretched length
of the spring is 0.15 meter.
+ (20)9.81(1.2π sin 30 )
◦
(b) − 12 (200)(1.2π)2 − (20)9.81(1.2π sin 30◦ )
(c) + 12 (200)(1.2π)2 − (20)9.81(1.2π sin 30◦ )
(d) + 12 (200)(1.2π)2
ANS: (c)
2
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• Gravitational potential energy at θ = 90◦ :
Determine the kinetic energy of the disk at this instant.
Vg2 = −(30 kg)(9.81)(0.35) = −103.0 N · m
(a)
(b)
( 12 )(20)(3)2
1
2
2
2 (20)(0.5 )(10)
(c) Answer (a) + Answer (b)
(d) None of the above
ANS: (c)
• Elastic potential energy at θ = 90◦ is :
p
1
Ve2 = (300 N/m)( 0.62 + 0.452 − 0.15)2 = 54.0 N · m
2
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GROUP PROBLEM SOLVING: Kinetic Energy
• When θ = 0◦ , the pendulum is
released from rest. Thus, T1 = 0.
• When θ = 90◦ , the pendulum has
a rotational motion about point O.
Thus, T2 = 21 IO (ω2 )2
where
IO = IG + m(dOG )2 = (30)0.32 + 30(0.35)2 = 6.375kg · m2
2
1
2 6.375(ω2 )
T2 =
• Now, substitute into the conservation of energy equation.
T1 + V1 = T2 + V2
0+0 =
2
1
2 6.375(ω2 )
+ (−103 + 54.0)
• Solving for ω yields ω = 3.92 rad/s.
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ATTENTION QUIZ
1
Blocks A and B are released from rest and the disk turns 2
revolutions. The V2 of the system includes a term for ....?
1m
11
00
00
11
40kg
datum
(c) the disk and
both blocks
(a) only the 40 kg
block
(b) only the 80 kg
block
11
00
00
11
00
11
00
11
00
11
(d) only the two
blocks
ANS: (d)
80kg
2
A slender bar is released from rest while in the horizontal
position. The kinetic energy (T2 ) of the bar when it has
rotated through 90◦ is?
m
11111111111111
00000000000000
00000000000000
11111111111111
L
(a)
(b)
1
2
2 m(vG2 )
1
2
2 IG (ω2 )
(c)
(d)
1
2
2 k(s1 ) − W (L/2)
1
1
2
2
2 mvG2 + 2 IG ω2
ANS: (d)
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