MASS MOMENT OF INERTIA (Section 10.9) Today’s Objectives: Students will be able to : a) Explain the concept of the mass moment of inertia (MMI). b) Determine the MMI of a composite body. In-Class Activities: Check homework, if any Reading quiz Applications MMI: concept and definition Determining the MMI Concept quiz Group problem solving Attention quiz Statics:The Next Generation Mehta & Danielson Lecture Notes for Section 10.9

2. The parallel-axis theorem can be applied to determine ________ . READING QUIZ 1. The formula definition of the mass moment of inertia about an axis is ___________ . A)  r dm B)  r2 dm C)  m dr D)  m2 dr 2. The parallel-axis theorem can be applied to determine ________ . A) only the MoI B) only the MMI C) both the MoI and MMI D) None of the above. Note: MoI is the moment of inertia of an area and MMI is the mass moment inertia of a body Answers : 1.B 2.C Statics:The Next Generation Mehta & Danielson Lecture Notes for Section 10.9

APPLICATIONS The large flywheel in the picture is connected to a large metal cutter. The flywheel is used to provide a uniform motion to the cutting blade. What property of the flywheel is most important for this use? How can we determine a value for this property? Why is most of the mass of the flywheel located near the flywheel’s circumference? Statics:The Next Generation Mehta & Danielson Lecture Notes for Section 10.9

APPLICATIONS (continued) If a torque M is applied to a fan blade which is initially at rest, its angular speed begins to increase. On which property (P) of the fan blade does the angular acceleration () depend? How can we determine a value for P? What is the relationship between M, P, and ? Statics:The Next Generation Mehta & Danielson Lecture Notes for Section 10.9

· CONCEPT OF THE MMI G T Consider a rigid body with a center of mass at G. It is free to rotate about the z axis, which passes through G. Now, if we apply a torque T about the z axis to the body, the body begins to rotate with an angular acceleration . T and  are related by the equation T = I  . In this equation, I is the mass moment of inertia (MMI) about the z axis. The MMI of a body is a property that measures the resistance of the body to angular acceleration. This is similar to the role of mass in the equation F = m a. The MMI is often used when analyzing rotational motion (done in dynamics). Statics:The Next Generation Mehta & Danielson Lecture Notes for Section 10.9

DEFINITION OF THE MMI p Consider a rigid body and the arbitrary axis p shown in the figure. The MMI about the p axis is defined as I = m r2 dm, where r, the “moment arm,” is the perpendicular distance from the axis to the arbitrary element dm. The MMI is always a positive quantity and has a unit of kg ·m2 or slug · ft2. Statics:The Next Generation Mehta & Danielson Lecture Notes for Section 10.9

p’ · RELATED CONCEPTS Parallel-Axis Theorem: Just as with the MoI for an area, the parallel-axis theorem can be used to find the MMI about a parallel axis p’ that is a distance d from the axis through the body’s center of mass G. The formula is Ip’ = IG + (m) (d)2 (where m is the mass of the body). d p’ G · m The radius of gyration is similarly defined as k = (I / m) Finally, the MMI can be obtained by integration or by the method for composite bodies. The latter method is easier for many practical shapes. Statics:The Next Generation Mehta & Danielson Lecture Notes for Section 10.9

Plan: Follow steps similar to finding the MoI for a composite area. q EXAMPLE Given: The wheel consists of a thin ring with a mass 10 kg and four spokes (slender rods) with a mass 2 kg each. Find: The wheel’s MMI about an axis perpendicular to the screen and passing through point A. Plan: Follow steps similar to finding the MoI for a composite area. q r p Solution: 1. The wheel can be divided into a thin ring (p) and two slender rods (q and r). Will both rods be treated the same? Statics:The Next Generation Mehta & Danielson Lecture Notes for Section 10.9

IAq = IAr = (1/12) (4) (1)2 + 4 (0.5)2 = 1.333 kg·m2 EXAMPLE (continued) p q r 2. The center of mass for each of the three pieces is at point O, 0.5 m from Point A. 3. The MMI data for a thin ring and slender rod are given on the inside back cover of the textbook. Using those data and the parallel-axis theorem, calculate the following. O IA = IO + (m) (d) 2 IAp = 10 (0.5)2 + 10 (0.5)2 = 5.0 kg·m2 IAq = IAr = (1/12) (4) (1)2 + 4 (0.5)2 = 1.333 kg·m2 4. Now add the three MMIs about point A. IA = IAp + IAq + IAr = 7.67 kg·m2 Statics:The Next Generation Mehta & Danielson Lecture Notes for Section 10.9

1. Consider a particle of mass 1 kg CONCEPT QUIZ 1. Consider a particle of mass 1 kg located at point P, whose coordinates are given in meters. Determine the MMI of that particle about the z axis. A) 9 kg·m2 B) 16 kg·m2 C) 25 kg·m2 D) 36 kg·m2 z x y ·P(3,4,6) 2. Consider a rectangular frame made of four slender bars with four axes (zP, zQ, zR and zS) perpendicular to the screen and passing through the points P, Q, R, and S respectively. About which of the four axes will the MMI of the frame be the largest? A) zP B) zQ C) zR D) zS E) Not possible to determine. P • S • Q • R Answers: 1.C 2.B Statics:The Next Generation Mehta & Danielson Lecture Notes for Section 10.9

Find: The radius of gyration of the pendulum about an axis GROUP PROBLEM SOLVING R Given: The pendulum consists of a 24 lb plate and a slender rod weighing 8 lb. Find: The radius of gyration of the pendulum about an axis perpendicular to the screen and passing through point O. P Plan: Determine the MMI of the pendulum using the method for composite bodies. Then determine the radius of gyration using the MMI and mass values (check units!!). Solution 1. Separate the pendulum into a square plate (P) and a slender rod (R). Statics:The Next Generation Mehta & Danielson Lecture Notes for Section 10.9

2. The center of mass of the plate and rod are 3.5 ft and 0.5 ft from GROUP PROBLEM SOLVING P R 2. The center of mass of the plate and rod are 3.5 ft and 0.5 ft from point O, respectively. 3. The MMI data on plates and slender rods are given on the inside cover of the textbook. Using those data and the parallel-axis theorem, IP = (1/12) (24/32.2) (12 + 12) + (24/32.2) (3.5)2 = 9.254 slug·ft2 IR = (1/12) (8/32.2) (5)2 + (8/32.2) (0.5)2 = 0.5797 slug·ft2 4. IO = IP + IR = 9.254 + 0.5797 = 9.834 slug·ft2 5. Total mass (m) equals (24+8)/32.2 = 0.9938 slug Radius of gyration k = IO / m = 3.15 ft Statics:The Next Generation Mehta & Danielson Lecture Notes for Section 10.9

2. Consider a rectangular frame made of four ATTENTION QUIZ 1. A particle of mass 2 kg is located 1 m down the y-axis. What are the MMI of the particle about the x, y, and z axes, respectively? A) (2, 0, 2) B) (0, 2, 2) C) (0, 2, 2) D) (2, 2, 0) • 1 m x y z 2. Consider a rectangular frame made of four slender bars and four axes (zP, zQ, zR and zS) perpendicular to the screen and passing through points P, Q, R, and S, respectively. About which of the four axes will the MMI of the frame be the lowest? A) zP B) zQ C) zR D) zS E) Not possible to determine. P • S • Q • R Answers : 1.A 2.D Statics:The Next Generation Mehta & Danielson Lecture Notes for Section 10.9

End of the Lecture Let Learning Continue Statics:The Next Generation Mehta & Danielson Lecture Notes for Section 10.9