Rigid Body F=ma: Fixed Axis Rotation

Similar documents
Midterm Solutions. mvr = ω f (I wheel + I bullet ) = ω f 2 MR2 + mr 2 ) ω f = v R. 1 + M 2m

Lab 7: Rotational Motion

PHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013

Equivalent Spring Stiffness

Physics 41 HW Set 1 Chapter 15

Lecture L22-2D Rigid Body Dynamics: Work and Energy

PHYSICS 111 HOMEWORK SOLUTION #9. April 5, 2013

Torque Analyses of a Sliding Ladder

Solution Derivations for Capa #11

Rotational Motion: Moment of Inertia

Linear Motion vs. Rotational Motion

Mechanical Principles

Rotation: Moment of Inertia and Torque

Hand Held Centripetal Force Kit

Figure 1.1 Vector A and Vector F

Physics 1A Lecture 10C

Center of Gravity. We touched on this briefly in chapter 7! x 2

Practice Exam Three Solutions

11. Rotation Translational Motion: Rotational Motion:

Physics 201 Homework 8

Angular acceleration α

3600 s 1 h. 24 h 1 day. 1 day

Lecture 17. Last time we saw that the rotational analog of Newton s 2nd Law is

Problem Set 1. Ans: a = 1.74 m/s 2, t = 4.80 s

Lab 8: Ballistic Pendulum

Rotational Inertia Demonstrator

3 Work, Power and Energy

Let s first see how precession works in quantitative detail. The system is illustrated below: ...

Problem 6.40 and 6.41 Kleppner and Kolenkow Notes by: Rishikesh Vaidya, Physics Group, BITS-Pilani

Lecture 16. Newton s Second Law for Rotation. Moment of Inertia. Angular momentum. Cutnell+Johnson: 9.4, 9.6

If you put the same book on a tilted surface the normal force will be less. The magnitude of the normal force will equal: N = W cos θ

Chapter 10 Rotational Motion. Copyright 2009 Pearson Education, Inc.

Mechanics lecture 7 Moment of a force, torque, equilibrium of a body

4.2 Free Body Diagrams

HW Set VI page 1 of 9 PHYSICS 1401 (1) homework solutions

Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx Acceleration Velocity (v) Displacement x

Chapter 11. h = 5m. = mgh mv Iω 2. E f. = E i. v = 4 3 g(h h) = m / s2 (8m 5m) = 6.26m / s. ω = v r = 6.

D Alembert s principle and applications

EXPERIMENT: MOMENT OF INERTIA

Vector Algebra II: Scalar and Vector Products

Unit 4 Practice Test: Rotational Motion

PHY231 Section 2, Form A March 22, Which one of the following statements concerning kinetic energy is true?

TOP VIEW. FBD s TOP VIEW. Examination No. 2 PROBLEM NO. 1. Given:

PHY231 Section 1, Form B March 22, 2012

MECHANICAL PRINCIPLES OUTCOME 4 MECHANICAL POWER TRANSMISSION TUTORIAL 1 SIMPLE MACHINES

Copyright 2011 Casa Software Ltd. Centre of Mass

Physics 1120: Simple Harmonic Motion Solutions

TORQUE AND FIRST-CLASS LEVERS

Chapter 6. Work and Energy

PHYS 101-4M, Fall 2005 Exam #3. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

State Newton's second law of motion for a particle, defining carefully each term used.

C B A T 3 T 2 T What is the magnitude of the force T 1? A) 37.5 N B) 75.0 N C) 113 N D) 157 N E) 192 N

AP Physics C. Oscillations/SHM Review Packet

KINEMATICS OF PARTICLES RELATIVE MOTION WITH RESPECT TO TRANSLATING AXES

Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam

Chapter 11 Equilibrium

PHY121 #8 Midterm I

CHAPTER 15 FORCE, MASS AND ACCELERATION

Force on Moving Charges in a Magnetic Field

Chapter H - Problems

Chapter 4. Forces and Newton s Laws of Motion. continued

APPLIED MATHEMATICS ADVANCED LEVEL

Simple Harmonic Motion

CHAPTER 6 WORK AND ENERGY

Lecture L3 - Vectors, Matrices and Coordinate Transformations

1 One Dimensional Horizontal Motion Position vs. time Velocity vs. time

Experiment 9. The Pendulum

Lecture L30-3D Rigid Body Dynamics: Tops and Gyroscopes

Acceleration due to Gravity

Lecture Presentation Chapter 7 Rotational Motion


Lab #4 - Linear Impulse and Momentum

Chapter. 4 Mechanism Design and Analysis

Chapter 21 Rigid Body Dynamics: Rotation and Translation about a Fixed Axis

F N A) 330 N 0.31 B) 310 N 0.33 C) 250 N 0.27 D) 290 N 0.30 E) 370 N 0.26

Prelab Exercises: Hooke's Law and the Behavior of Springs

226 Chapter 15: OSCILLATIONS

1. Units of a magnetic field might be: A. C m/s B. C s/m C. C/kg D. kg/c s E. N/C m ans: D

So if ω 0 increases 3-fold, the stopping angle increases 3 2 = 9-fold.

Chapter 10: Linear Kinematics of Human Movement

Chapter 4. Moment - the tendency of a force to rotate an object

2 Session Two - Complex Numbers and Vectors

BUILD A TABLETOP LOOM

VELOCITY, ACCELERATION, FORCE

SOLID MECHANICS TUTORIAL MECHANISMS KINEMATICS - VELOCITY AND ACCELERATION DIAGRAMS

Unified Lecture # 4 Vectors

Sample Questions for the AP Physics 1 Exam

THEORETICAL MECHANICS

Weight The weight of an object is defined as the gravitational force acting on the object. Unit: Newton (N)

SOLID MECHANICS BALANCING TUTORIAL BALANCING OF ROTATING BODIES

8.012 Physics I: Classical Mechanics Fall 2008

Physics 9e/Cutnell. correlated to the. College Board AP Physics 1 Course Objectives

FXA UNIT G484 Module Simple Harmonic Oscillations 11. frequency of the applied = natural frequency of the

State Newton's second law of motion for a particle, defining carefully each term used.

Lecture L5 - Other Coordinate Systems

Wind Turbines. Wind Turbines 2. Wind Turbines 4. Wind Turbines 3. Wind Turbines 5. Wind Turbines 6

Awell-known lecture demonstration1

Chapter 18 Static Equilibrium

v v ax v a x a v a v = = = Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and ( )

ANALYTICAL METHODS FOR ENGINEERS

Transcription:

Rigid Body F=ma: Fixed xis Rotation Rigid Body F=ma Equations: In the one or two classes prior to this one, we gave the following scalar equations of motion for planar rigid body F=ma problems: F = a x F = a y M = I x y kinetic moment equation, useful for some problems, is: M = P ( ) M K P For fixed axis rotation problems, one additional moment equation is quite useful: (only for fixed axis rotation!) M = PIN IPIN This is not a new equation; it is basically the kinetic moment equation applied to fixed axis problems.

Two types of Problems: (1) at pin, or (2) not at pin. In the previous section we discussed translation problems, which had some a of the mass center, but their angular acceleration α was zero because they did not rotate. This section discusses fixed axis rotation problems which all have an angular acceleration, α, but may or may not have an a, depending on the location of their mass center. There are two classes of fixed axis problems, depending on the location of the mass center, : (1) If is at the pin, doesn t move, and a is zero. (2) If is not at the pin, moves in a circle about the pin, and is thus resolved into a t and a n components.

Two types of Problems: (a) at pin, or (b) not at pin. There are two classes of fixed axis problems, depending on the location of the mass center, : (1) If is at the pin, doesn t move, and a is zero. If is at the (fixed) pin at, then a = 0. r a n = 2r a t = r If is not at the pin, then have a t and a n., (2) If is not at the pin, moves in a circle about the pin, and is thus resolved into a t and a n components.

Where does the ΣM pin = I pin α equation come from? Consider a basic fixed axis problem, a slender bar, with not at the pin. Find α and a. y FBD KD x r r mg (given) ma n = m 2r ma t = m r I Write a Kinetic Moment equation about the pin at : (You could sum about if you wish...) M = ( M K ) ; mg (r cos ) = (m r)(r) + I

Write a Kinetic Moment equation about the pin at : (You could sum about if you wish...) a t = r M = ( M K ) ; mg (r cos ) = (m r)(r) + I for F rotation Do you recognize this? This is the Parallel xis Theorem! I PIN = I 2 + mr = I 2 + mr = I PIN This is the general form of the equation: M = PIN IPIN Only use this for Fixed xis Rotation, because we used a = r kinematics to obtain it. t

Summary of the Equations of Motion for this Problem y FBD KD x r r mg (given) ma n = m 2r ma t = m r I To solve for x, y and, write these equations: (Sum in either n-t or x-y) F = a ; = (m r) cos + (m r) sin x x x 2 F = a ; - mg = +(m r) sin - (m r) cos y y y M PIN = I PIN ; mg (r cos ) = I PIN 2 For a slender rod: I PIN = 1 3 ml2

M = PIN IPIN To review, this is not a radically new equation. It comes from combining: (a) Kinetic moment equation, (b) a t = αr, and (c) Parallel axis theorem. Finally, we applied this to a slender bar, but it applies to any shaped body rotating about a pin with its center of mass not at the pin. If you have a composite body made up of several mass shapes rotating about the pin, you may use this equation as well, where I pin is the sum of the I pin s for all of the mass centers, using the parallel axis theorem: For a composite body: I pin = Σ(I + md 2 )

Examples where is not at the pin. Unbalanced rotating equipment is a common, simple example of fixed axis rotation with the mass center not exactly at the pin. Here are some examples: 1. Washing machine: When washing clothes, have your wet clothes shifted to one side of the drum? On the spin cycle, what happens? The entire machine shakes dramatically, doesn t it? Why? The composite mass center of the drum plus the clothes has shifted well away from the axis of rotation. 2. utomobile tires: re you aware that tires even new ones are not perfectly balanced? When you buy new tires, after they are mounted on your rims, the mechanic balances the rim-tire assembly on a special balancing machine. The machine spins the tires (measuring dynamic loads) and computes the specific locations on the rim to clip on lead weights of various sizes to balance the tire.

Examples where may or may not be at the pin. 3. utomobile crankshafts: Special care is taken to use counter weights to balance these. 4. Fishing reels: Next time you are in the fishing section at a sporting good store, try out a spinning reel (the kind with an open view of the fishing line). The spool, bale and handle rotate on different axes, but together, they are balanced amazingly well. 5. In-class example: Create a 6-10 inch diameter wooden disk (or an 8 inch long stick) with a centered ¼ inch hole and two offcenter holes. Place a ¼ inch bolt, tightened with a nut and washers, in the centered hole. Chuck this in a cordless drill and spin the assembly. The drill should not vibrate very much. Place a bolt and nut in one of the other holes, tighten it, and spin the drill again. Try various speeds. The drill vibrates in your hand. What you are feeling is the rotating force at the drill s chuck needed to keep the mass center moving in the circle.

Disk in a Horizontal Plane: not at the pin. On rotating equipment, an off-centered creates a rotating (normal) force component at the pin, with n = m(ω 2 r). Example: motor drives an unbalanced disk in a HORIZ plane at a constant angular velocity, ω. Find the pin reaction at. Horizontal Plane: FBD KD = 0 = const n n n ma n = m( 2r) If the disk lies in a horizontal plane, the weight mg is perpendicular to the disk. The only pin reaction in the plane at is simply the rotating 2 normal reaction n equal to m( r).

Disk in a Vertical Plane: not at the pin. On rotating equipment, an off-centered creates a rotating (normal) force component at the pin, with n = m(ω 2 r). Example: motor drives an unbalanced disk in a VERTICL plane at a constant angular velocity, ω. Find the pin reaction at. Vertical Plane: FBD mg KD = 0 = const n y n ma n = m( 2r) If the disk lies in a vertical plane, the pin reaction at consists of a vertical component y = mg plus a rotating normal reaction equal to m( 2 n r).

machine designed to bounce. This machine is designed to bounce on the spring at C due to the rotation of the unbalanced wheel. Vibration like this is accompanied by pulsing force reactions everywhere--at B, C,, etc. In time, a machine like this will suffer breakdowns due to fatigue. B r C ssembly pivots at a pin at B. End C of the machine is supported by a spring.