READING QUIZ. CEE 271: Applied Mechanics II, Dynamics Lecture 27: Ch.18, Sec.1 5 APPLICATIONS KINETIC ENERGY, WORK, PRINCIPLE OF WORK AND ENERGY

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(d) I G ω 2 ANS: (c) 2 When calculating work done by forces, the work of an internal force does not have to be considered because (a) internal forces do not exist (b) the forces act in equal but

1 READING QUIZ CEE 27: Applied Mechanics II, Dynamics Lecture 27: Ch.8, Sec. 5 Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa Date: Kinetic energy due to rotation (only) of the body is defined as (a) 2 m(v G) 2 (b) 2 m(v G) 2 +(/2)I G ω 2 (c) 2 I Gω 2 (d) I G ω 2 ANS: (c) 2 When calculating work done by forces, the work of an internal force does not have to be considered because (a) internal forces do not exist (b) the forces act in equal but opposite collinear pairs (c) the body is at rest initially (d) the body can deform ANS: (b) / 42 3 / 42 KINETIC ENERGY, WORK, PRINCIPLE OF WORK AND ENERGY APPLICATIONS Today s objectives: Students will be able to Define the various ways a force and couple do work. 2 Apply the principle of work and energy to a rigid body. In-class activities: Reading Quiz Applications Kinetic Energy Work of a Force or Couple Principle of Work and Energy Concept Quiz Group Problem Solving Attention Quiz The work of the torque (or moment) developed by the driving gears on the two motors on the concrete mixer is transformed into the rotational kinetic energy of the mixing drum. If the motor gear characteristics are known, how would you find the rotational velocity of the mixing drum? 2 / 42 4 / 42

READING QUIZ APPLICATIONS(continued) Elastic potential energy is defined as.

(a) translational motion and rotational motion (b) only rotational motion (c) only translational motion (d) the deformation of the body ANS: (a) Two torsional springs are used to assist in opening

Are the gravitational potential energy of the hood and the torsional spring stiffness related to each other? If so, how? 2 / 42 23 / 42 APPLICATIONS CONSERVATION OF ENERGY (Section 8.

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.

6 READING QUIZ APPLICATIONS(continued) Elastic potential energy is defined as. (a) + 2 k(s)2 (b) 2 k(s)2 (c) + 2 k(v)2 (d) None of the above ANS: (a) 2 The kinetic energy of a rigid body consists of the kinetic energy due to. (a) translational motion and rotational motion (b) only rotational motion (c) only translational motion (d) the deformation of the body ANS: (a) Two torsional springs are used to assist 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? Are the gravitational potential energy of the hood and the torsional spring stiffness related to each other? If so, how? 2 / / 42 APPLICATIONS CONSERVATION OF ENERGY (Section 8.5) The torsion springs located at the top of the garage door wind up as the door is lowered. 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? The conservation of energy theorem is a simpler energy method (recall that the principle of work and energy is also an energy method) for solving problems. 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? 22 / / 42

10 EXAMPLE II(continued) GROUP PROBLEM SOLVING Now all terms in the conservation of energy equation have been formulated. Writing the general equation and then substituting into it yields: T +V = T 2 +V =.25(ω 2 ) 2 +( ) Solving for ω 2 = 2.97rad/s Given: The 30 kg pendulum has its mass center at G and a radius of gyration about point G of k G = 0.3meter. It is released from rest when θ = 0. The spring is un-stretched when θ = 0. Find: The angular velocity of the pendulum when θ = 90. 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. 37 / / 42 UNDERSTANDING QUIZ At the instant shown, the spring is undeformed. Determine the change in potential energy if the 20 kg disk (k G = 0.5meter) rolls 2 revolutions without slipping. (a) + 2 (200)(.2π)2 +(20)9.8(.2πsin30 ) (b) 2 (200)(.2π)2 (20)9.8(.2πsin30 ) (c) + 2 (200)(.2π)2 (20)9.8(.2πsin30 ) (d) + 2 (200)(.2π)2 ANS: (c) 2 Determine the kinetic energy of the disk at this instant. (a) ( 2 )(20)(3)2 (c) Answer (a) + Answer (b) (b) 2 (20)(0.52 )(0) 2 (d) None of the above ANS: (c) GROUP PROBLEM SOLVING: Potential Energy Let s put the datum when θ = 0. Then, the gravitational potential energy and the elastic potential energy will be zero. So, V g = V e = 0 Note that the un-stretched length of the spring is 0.5meter. Gravitational potential energy at θ = 90 : V g2 = (30kg)(9.8)(0.35) = 03.0 N m Elastic potential energy at θ = 90 is : V e2 = 2 (300N/m)( ) 2 = 54.0 N m 38 / / 42

GROUP PROBLEM SOLVING: Kinetic Energy When θ = 0, the pendulum is released from rest. Thus, T = 0. When θ = 90, the pendulum has a rotational motion about point O.

11 GROUP PROBLEM SOLVING: Kinetic Energy When θ = 0, the pendulum is released from rest. Thus, T = 0. When θ = 90, the pendulum has a rotational motion about point O. Thus, T 2 = 2 I O(ω 2 ) 2 where I O = I G +m(d OG ) 2 = (30) (0.35) 2 = 6.375kg m 2 T 2 = (ω 2) 2 Now, substitute into the conservation of energy equation. T +V = T 2 +V = (ω 2) 2 +( ) Solving for ω yields ω = 3.92rad/s. 4 / 42 ATTENTION QUIZ Blocks A and B are released from rest and the disk turns 2 revolutions. The V 2 of the system includes a term for...? kg m kg datum (a) only the 40 kg block (b) only the 80 kg block (c) the disk and both blocks (d) only the two blocks ANS: (d) 2 A slender bar is released from rest while in the horizontal position. The kinetic energy (T 2 ) of the bar when it has rotated through 90 is? m L (a) 2 m(v G2) 2 (b) 2 I G(ω 2 ) 2 (c) 2 k(s ) 2 W(L/2) (d) 2 mv2 G2 + 2 I Gω 2 2 ANS: (d) 42 / 42

CEE 271: Applied Mechanics II, Dynamics Lecture 27: Ch.18, Sec.1 5

1 / 42 CEE 271: Applied Mechanics II, Dynamics Lecture 27: Ch.18, Sec.1 5 Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa Tuesday, November 27, 2012 2 / 42 KINETIC