M1=100 Kg Adult, M2=10 Kg Baby. the Seesaw Starts from Rest. Which Direction Will It Rotates?

Total Page:16

File Type:pdf, Size:1020Kb

M1=100 Kg Adult, M2=10 Kg Baby. the Seesaw Starts from Rest. Which Direction Will It Rotates? m1 m2 m1=100 kg adult, m2=10 kg baby. The seesaw starts from rest. Which direction will it rotates? (a) Counter-Clockwise (b) Clockwise ()(c) NttiNo rotation (d) Not enough information Effect of a Constant Net Torque 2.3 A constant non-zero net torque is exerted on a wheel. Which of the following quantities must be changing? 1. angular position 2. angular velocity 3. angular acceleration 4. moment of inertia 5. kinetic energy 6. the mass center location A. 1, 2, 3 B. 4, 5, 6 C. 1,2, 5 D. 1, 2, 3, 4 E. 2, 3, 5 1 Example: second law for rotation PP10601-50: A torque of 32.0 N·m on a certain wheel causes an angular acceleration of 25.0 rad/s2. What is the wheel's rotational inertia? Second Law example: α for an unbalanced bar Bar is massless and originally horizontal Rotation axis at fulcrum point L1 N L2 Î N has zero torque +y Find angular acceleration of bar and the linear m1gmfulcrum 2g acceleration of m1 just after you let go τnet Constraints: Use: τnet = Itotα ⇒ α = Itot 2 2 Using specific numbers: where: Itot = I1 + I2 = m1L1 + m2L2 Let m1 = m2= m L =20 cm, L = 80 cm τnet = ∑ τo,i = + m1gL1 − m2gL2 1 2 θ gL1 − gL2 g(0.2 - 0.8) What happened to sin( ) in moment arm? α = 2 2 = 2 2 L1 + L2 0.2 + 0.8 2 net = − 8.65 rad/s Clockwise torque m gL − m gL a ==+ -α L 1.7 m/s2 α = 1 1 2 2 11 2 2 Accelerates UP m1L1 + m2L2 total I about pivot What if bar is not horizontal? 2 See Saw 3.1. Suppose everything is as it was in the preceding example, but the bar is NOT horizontal. Assume both masses are equal. Which of the following is the correct equation for the angular acceleration? L (L − L ) 1 N θ A) α = 1 2 g L2 + L2 1 2 m1g fulcrum L2 (L1 − L2 ) g m2g B) α = 2 2 L1 + L2 cos(θ) (L − L ) C) 1 2 gsin( ) α = 2 2 θ L1 + L2 (L − L ) D) 1 2 gcos( ) α = 2 2 θ L1 + L2 (L1 − L2 ) g E) α = 2 2 L1 + L2 sin(θ) τnet = Itotα See Saw - Solution 3.1. Suppose everything is as it was in the preceding example, but the bar is NOT horizontal. Assume both masses are equal. Which of the following is the correct equation for the angular acceleration? L1 N θ α = τ / I net tot m1g fulcrum L2 m2g τnet = ∑ τo,i = + m1gL1 cos(θ) − m2gL2 cos(θ) 2 2 Itot = I1 + I2 = m1L1 + m2L2 [m L − m L ] 1 1 2 2 g cos( ) α = 2 2 θ m1L1 + m2L2 For equal masses m1 –m2 = m (L − L ) If bar is balanced when it’s ∴ α = 1 2 gcos(θ) L2 + L2 horizontal, does that change 1 2 when it is not horizontal? 3 Method for solving (complex) problems using the Second Law Many components in the system means several (N) unknowns…. … need an equal number of independent equations Method: • Draw or sketch system. Adopt coordinates, name the variables, indicate rotation axes, list the known and unknown quantities, … • Draw free bbdody diagrams of key parts. Show forces at thiheir points of application. find torques about a (common) axis Note: can have • May need to apply Second Law twice to each part Fnet .eq. 0 G G but τ .ne. 0 ¾ Translation: net Fnet = ∑Fi = ma G G G ¾ Rotation: τnet = ∑ τi = Iα • Make sure there are enouggqh (N) equations; there may be constraint equations (extra conditions connecting unknowns) • Simplify and solve the set of (simultaneous) equations. • Interpret the final formulas. Do they make intuitive sense? Refer back to the sketches and original problem • Calculate numerical results, and sanity check anwers (e.g., right order of magnitude?) Example: Torque and Angular Acceleration of a Wheel Analysis approach: Break into two sub-systems wheel is accelerated angularly by tension T block is accelerated linearly by weight mg, with tension opposing Free body diagrams shown r The wheel is rotating and so we apply Στ = Ια a The tension supplies the torque via tangential force The mass is moving in a straight line, so apply Newton’ s Second Law ΣFy = may = mg -T How to connect the two problems above? mg Need a constraint linking linear acceleration to α a = αr 4 Application of Method using the Second Law • Cord wrapped around disk, hanging weight • Cord does not slip or stretch Æ constraint • Disk’s rotational inertia slows accelerations • Let m = 1.2 kg, M = 2.5 kg, r =0.2 m •ForFind mass acceleration m: of mass m, find α for disk F = ma = mg − T r T ∑ y y mg T = m (g − a) Unknowns: T, a a support force at axis “O” has FBD for disk, with axis at “o”: zero torque 1 2 N τ = + Tr = Iα I = Mr ∑ 0 2 Tr m(g − a)r α = = 1 Unknowns: a, α mg T Mg I Mr2 2 from “no So far: 2 Equations, 3 unknowns ÆNeed a constraint: a = + αr slipping” Substitute and solve: assumption 2 mg 2mgr 2mαr 2 α = - α = (= 24 rad/s ) Mr2 Mr2 r(m + M/2) α proportional to g m 2mg α(1 + 2 ) = mg 2 a = (= 4.8 m/s ) M Mr (m + M/2) Example: Heavy pulley Atwood’s machine application of Newton’s 2nd Law Given the numerical values: • m1= 5.0 kg, • m2= 5.5 kg r • r = 0.2 m, • M=60kgM = 6.0 kg O • I = 0.20 kg-m2 M, I a) Find the accelerations of the hanging weights. Are they up or down? b) Find the angular acceleration of the pulley. Is it CW or CCW? m2 c) Find the tensions in each cord. d) How long does ittkit take the 55k5.5 kg wei ght htt to m fall 0.6 m from rest? 1 Strategy: Apply Newton’s Second Law (linear and rotational form) to all three bodies, connected by constraints. Make sure there are as many independent equations as there are unknowns 5 Heavy pulley Atwood’s machine: solution Apply Second Law: F = F = ma N (I) net ∑ i (each component) τ = τ = I α α (II) net ∑ i (pulley) r If I = 0 problem simplifies O M, I For m1 use I: T1 m a = T - m g y 1 1 1 1 y a1 Mg m g 1. T1 = m1 (a1 + g) 1 m2 • positive α Æ negative a1 (down) m1 For m2 use I again: T • choose y positive up for m 2a 2 = T2 - m 2g 2 both masses a 2. T2 = m 2 (a 2 + g) 2 • CCW rotation Æ positive α m2g • positive α Æ positive a2 (up) • Free fall if T1 or T2 = 0 Constraints: cord cannot stretch or slip Note: if I = 0, then T1 = T2. Result using 1 & 2 becomes: a 2 = −a1 only 2 of these are independent m1 - m 2 a 2 = g a 2 = + αr a1 = −αr m1 + m 2 Atwood’s solution, continued For the Pulley use II: N NOTE: τnet = + T1r − T2ri = I α α r T ≠ m g (T − T ) r 1 1 3. α = 1 2 O T ≠ m g I M, I 2 2 T T • positive α Æ T > T 1 2 1 2 Mg Check ability to solve: • 5 unknowns: a1, a2, T1, T2, α • Have 5 independent equations: 1,2,3 and 2 constraints OK! Solution: Begin eliminating unknowns (tension equations) I α = T1r − T2ri = (m1g + m1a1 - m 2a 2 - m2g)r = (m 1 - m 2 )gr + (m1 + m 2 )a 1r a 2 = −a1 2 I α - (m1 + m2 )a 1r = (m 1 - m2 )gr I α + (m1 + m2 )αr = (m 1 - m2 )gr a1 = −αr Result: (m1 − m 2 ) g r Net torque of hanging weights interpretation α = 2 Total moment of Inertia is simple I + (m1 + m 2 ) r Limiting case: (m1 − m 2 ) g m1 - m2 Let I Æ 0: α → a1 = −αr = − g massless (m1 + m2 ) r m1 + m 2 pulley Negative if m1 > m2: 6 Atwood’s solution, numerical evaluation (5.0 − 5.5) (9.8) (0.2) α = (b) α = - 1.58 rad/s 2 (CW ) 0.20 + (5.0 − 5.5) 0.2 2 2 (a) a1 = - αr = + 0.32 m/s (up) 2 a2 = - a1 = − 0320.32 m/s (down) T1 = 5.0 (9.8 + 0.32) (c) T1 = 506 N. (a1 increases tension) T2 = 5.5 (9.8 − 0.32) T2 = 521 N. (a 2 decreases tension) How long does m2 take to fall 0.6 m from rest? 1 2 2d d = a 2t (d) t = = 1.94 s 2 a 2 7.
Recommended publications
  • Rotational Motion (The Dynamics of a Rigid Body)
    University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Robert Katz Publications Research Papers in Physics and Astronomy 1-1958 Physics, Chapter 11: Rotational Motion (The Dynamics of a Rigid Body) Henry Semat City College of New York Robert Katz University of Nebraska-Lincoln, [email protected] Follow this and additional works at: https://digitalcommons.unl.edu/physicskatz Part of the Physics Commons Semat, Henry and Katz, Robert, "Physics, Chapter 11: Rotational Motion (The Dynamics of a Rigid Body)" (1958). Robert Katz Publications. 141. https://digitalcommons.unl.edu/physicskatz/141 This Article is brought to you for free and open access by the Research Papers in Physics and Astronomy at DigitalCommons@University of Nebraska - Lincoln. It has been accepted for inclusion in Robert Katz Publications by an authorized administrator of DigitalCommons@University of Nebraska - Lincoln. 11 Rotational Motion (The Dynamics of a Rigid Body) 11-1 Motion about a Fixed Axis The motion of the flywheel of an engine and of a pulley on its axle are examples of an important type of motion of a rigid body, that of the motion of rotation about a fixed axis. Consider the motion of a uniform disk rotat­ ing about a fixed axis passing through its center of gravity C perpendicular to the face of the disk, as shown in Figure 11-1. The motion of this disk may be de­ scribed in terms of the motions of each of its individual particles, but a better way to describe the motion is in terms of the angle through which the disk rotates.
    [Show full text]
  • Physics B Topics Overview ∑
    Physics 106 Lecture 1 Introduction to Rotation SJ 7th Ed.: Chap. 10.1 to 3 • Course Introduction • Course Rules & Assignment • TiTopics Overv iew • Rotation (rigid body) versus translation (point particle) • Rotation concepts and variables • Rotational kinematic quantities Angular position and displacement Angular velocity & acceleration • Rotation kinematics formulas for constant angular acceleration • Analogy with linear kinematics 1 Physics B Topics Overview PHYSICS A motion of point bodies COVERED: kinematics - translation dynamics ∑Fext = ma conservation laws: energy & momentum motion of “Rigid Bodies” (extended, finite size) PHYSICS B rotation + translation, more complex motions possible COVERS: rigid bodies: fixed size & shape, orientation matters kinematics of rotation dynamics ∑Fext = macm and ∑ Τext = Iα rotational modifications to energy conservation conservation laws: energy & angular momentum TOPICS: 3 weeks: rotation: ▪ angular versions of kinematics & second law ▪ angular momentum ▪ equilibrium 2 weeks: gravitation, oscillations, fluids 1 Angular variables: language for describing rotation Measure angles in radians simple rotation formulas Definition: 360o 180o • 2π radians = full circle 1 radian = = = 57.3o • 1 radian = angle that cuts off arc length s = radius r 2π π s arc length ≡ s = r θ (in radians) θ ≡ rad r r s θ’ θ Example: r = 10 cm, θ = 100 radians Æ s = 1000 cm = 10 m. Rigid body rotation: angular displacement and arc length Angular displacement is the angle an object (rigid body) rotates through during some time interval..... ...also the angle that a reference line fixed in a body sweeps out A rigid body rotates about some rotation axis – a line located somewhere in or near it, pointing in some y direction in space • One polar coordinate θ specifies position of the whole body about this rotation axis.
    [Show full text]
  • Rotation: Moment of Inertia and Torque
    Rotation: Moment of Inertia and Torque Every time we push a door open or tighten a bolt using a wrench, we apply a force that results in a rotational motion about a fixed axis. Through experience we learn that where the force is applied and how the force is applied is just as important as how much force is applied when we want to make something rotate. This tutorial discusses the dynamics of an object rotating about a fixed axis and introduces the concepts of torque and moment of inertia. These concepts allows us to get a better understanding of why pushing a door towards its hinges is not very a very effective way to make it open, why using a longer wrench makes it easier to loosen a tight bolt, etc. This module begins by looking at the kinetic energy of rotation and by defining a quantity known as the moment of inertia which is the rotational analog of mass. Then it proceeds to discuss the quantity called torque which is the rotational analog of force and is the physical quantity that is required to changed an object's state of rotational motion. Moment of Inertia Kinetic Energy of Rotation Consider a rigid object rotating about a fixed axis at a certain angular velocity. Since every particle in the object is moving, every particle has kinetic energy. To find the total kinetic energy related to the rotation of the body, the sum of the kinetic energy of every particle due to the rotational motion is taken. The total kinetic energy can be expressed as ..
    [Show full text]
  • Rotational Motion and Angular Momentum 317
    CHAPTER 10 | ROTATIONAL MOTION AND ANGULAR MOMENTUM 317 10 ROTATIONAL MOTION AND ANGULAR MOMENTUM Figure 10.1 The mention of a tornado conjures up images of raw destructive power. Tornadoes blow houses away as if they were made of paper and have been known to pierce tree trunks with pieces of straw. They descend from clouds in funnel-like shapes that spin violently, particularly at the bottom where they are most narrow, producing winds as high as 500 km/h. (credit: Daphne Zaras, U.S. National Oceanic and Atmospheric Administration) Learning Objectives 10.1. Angular Acceleration • Describe uniform circular motion. • Explain non-uniform circular motion. • Calculate angular acceleration of an object. • Observe the link between linear and angular acceleration. 10.2. Kinematics of Rotational Motion • Observe the kinematics of rotational motion. • Derive rotational kinematic equations. • Evaluate problem solving strategies for rotational kinematics. 10.3. Dynamics of Rotational Motion: Rotational Inertia • Understand the relationship between force, mass and acceleration. • Study the turning effect of force. • Study the analogy between force and torque, mass and moment of inertia, and linear acceleration and angular acceleration. 10.4. Rotational Kinetic Energy: Work and Energy Revisited • Derive the equation for rotational work. • Calculate rotational kinetic energy. • Demonstrate the Law of Conservation of Energy. 10.5. Angular Momentum and Its Conservation • Understand the analogy between angular momentum and linear momentum. • Observe the relationship between torque and angular momentum. • Apply the law of conservation of angular momentum. 10.6. Collisions of Extended Bodies in Two Dimensions • Observe collisions of extended bodies in two dimensions. • Examine collision at the point of percussion.
    [Show full text]
  • 0230 Lecture Notes - a Tale of Three Accelerations Or the Differences Between Angular, Tangential, and Centripetal Accelerations.Docx Page 1 of 1
    Flipping Physics Lecture Notes: A Tale of Three Accelerations or The Differences between Angular, Tangential, and Centripetal Accelerations https://www.flippingphysics.com/3-accelerations.html An object moving in a circle can have three different types of accelerations: Δω rad • Angular Acceleration: in is an angular quantity. α = 2 Δt s m • Tangential Acceleration: in is a linear quantity. at = rα 2 s v 2 m • Centripetal Acceleration: t 2 in is a linear quantity. ac = = rω 2 r s Angular acceleration separates itself from the others: 1) Because it is an angular quantity, whereas the other two are linear quantities. 2) Because angular acceleration applies to the whole rigid object, however, tangential acceleration and centripetal acceleration are for a specific radius. A major difference between tangential acceleration and centripetal acceleration is their direction. • Centripetal means “center seeking”. Centripetal acceleration is always directed inward. • Tangential acceleration is always directed tangent to the circle. o By definition, tangential acceleration and centripetal acceleration are perpendicular to one another. Another major difference between tangential acceleration and centripetal acceleration is that circular motion cannot exist without centripetal acceleration. • No centripetal acceleration means the object is not moving in a circle. o Centripetal acceleration results from the change in direction of the tangential velocity. If the tangential velocity is not changing directions, then the object is not moving in a circle. • Tangential acceleration results from the change in magnitude of the tangential velocity of an object. An object can move in a circle and not have any tangential acceleration. No tangential acceleration simply means the angular acceleration of the object is zero and the object is moving with a constant angular velocity.
    [Show full text]
  • Rotational Motion Angular Position Simple Vs. Complex Objects Speed
    3/11/16 Rotational Motion Angular Position In physics we distinguish two types of motion for objects: • Translational Motion (change of location): • We will measure Whole object moves through space. angular position in radians: • Rotational Motion - object turns around an axis (axle); axis does not move. (Wheels) • Counterclockwise (CCW): positive We can also describe objects that do both at rotation the same time! • Clockwise (CW): negative rotation Linear Distance d vs. Angular Simple vs. Complex Objects distance Δθ Model motion with just Model motion with For a point at Position position and Rotation radius R on the wheel, R d = RΔθ ONLY works if Δθ is in radians! Speed in Rotational Motion Analogues Between Linear and Rotational Motion (see pg. 303) • Rotational Speed ω: rad per second • Tangential speed vt: distance per second • Two objects can have the same rotational speed, but different tangential speeds! 1 3/11/16 Relationship Between Angular and Linear Example: Gears Quantities • Displacements • Every point on the • Two wheels are rotating object has the connected by a chain s = θr same angular motion • Speeds that doesn’t slip. • Every point on the v r • Which wheel (if either) t = ω rotating object does • Accelerations not have the same has the higher linear motion rotational speed? a = αr t • Which wheel (if either) has the higher tangential speed for a point on its rim? Angular Acceleration Directions Corgi on a Carousel • If the angular acceleration and the angular • What is the carousel’s angular speed? velocity are in the same direction, the • What is the corgi’s angular speed? angular speed will increase with time.
    [Show full text]
  • Angular Acceleration Tangential Acceleration Additional Readings Radial Acceleration
    Acceleration - AccessScience from McGraw-Hill Education http://accessscience.com/content/acceleration/002500 (http://accessscience.com/) Article by: Rusk, Rogers D. Mount Holyoke College, South Hadley, Massachusetts. Howe, Carl E. Formerly, Physics and Astronomy Department, Oberlin College, Oberlin, Ohio. Stephenson, R. J. Department of Physics, Wooster College, Wooster, Ohio. Last updated: 2014 DOI: https://doi.org/10.1036/1097-8542.002500 (https://doi.org/10.1036/1097-8542.002500) Content Hide Angular acceleration Tangential acceleration Additional Readings Radial acceleration The time rate of change of velocity. Since velocity is a directed or vector quantity involving both magnitude and direction, a velocity may change by a change of magnitude (speed) or by a change of direction or both. It follows that acceleration is also a directed, or vector, quantity. If the magnitude of the velocity of a body changes from υ1 ft/s to υ2 ft/s in t seconds, then the average acceleration has a magnitude given by Eq. (1). (1) To designate it fully the direction should be given, as well as the magnitude. See also: Velocity (/content/velocity/729500) Instantaneous acceleration is defined as the limit of the ratio of the velocity change to the elapsed time as the time interval approaches zero. When the acceleration is constant, the average acceleration and the instantaneous acceleration are equal. If a body, moving along a straight line, is accelerated from a speed of 10 to 90 ft/s (3 to 27 m/s) in 4 s, then the average change in speed per second is = 20 ft/s or = 6 m/s in each second.
    [Show full text]
  • Development of Inference Methods for Rotational Motions on Ground Surface and Application to Microtremors Acquired with a Small-Size Dense Array
    DEVELOPMENT OF INFERENCE METHODS FOR ROTATIONAL MOTIONS ON GROUND SURFACE AND APPLICATION TO MICROTREMORS ACQUIRED WITH A SMALL-SIZE DENSE ARRAY Koji HADA1 and Masanori HORIKE2 1 Architectural Group, NEWJEC Inc., Osaka, Japan, [email protected] 2 Member, Advisor, Division of Engineering, Hanshin Consultants Co., Ltd, Osaka, Japan, [email protected] ABSTRACT: In this study, we develop two methods for the inference of rotation vector on ground surface, two rocking rotations and a single torsional rotation. The first, termed nth-order elastic method, is based on the elasticity of the ground surface. The rotation vector is constructed from the first derivative with respect to the space of the ground surface motions. The first derivative is calculated from simultaneous equations by n-th order Taylor expansion obtained by difference motion between multiple observation points. Meanwhile, the second, termed rigid method, is based on the rigidity of ground surface and the rotation. Keywords: Inference methods of rotational motions, Ground surface, Microtremors, Small-size dense array 1. INTRODUCTION Over the years, many reports and studies have noted the possibility that buildings were damaged by torsional vibration1)-3), including many reports concerning the 1995 Hyogoken-Nanbu Earthquake4). Damage due to torsional vibration was also reported for the 2016 Kumamoto Earthquakes in April 20165). In these reports and studies, the causes of the torsional vibration have generally been sought in the eccentricity of the buildings, and virtually no consideration has been given to the influence of rotational inputs. For the earthquake-resistant design, there are regulations covering the translational components of seismic inputs, but the rotational inputs are not regulated.
    [Show full text]
  • Chapter 24 Physical Pendulum
    Chapter 24 Physical Pendulum 24.1 Introduction ........................................................................................................... 1 24.1.1 Simple Pendulum: Torque Approach .......................................................... 1 24.2 Physical Pendulum ................................................................................................ 2 24.3 Worked Examples ................................................................................................. 4 Example 24.1 Oscillating Rod .................................................................................. 4 Example 24.3 Torsional Oscillator .......................................................................... 7 Example 24.4 Compound Physical Pendulum ........................................................ 9 Appendix 24A Higher-Order Corrections to the Period for Larger Amplitudes of a Simple Pendulum ..................................................................................................... 12 Chapter 24 Physical Pendulum …. I had along with me….the Descriptions, with some Drawings of the principal Parts of the Pendulum-Clock which I had made, and as also of them of my then intended Timekeeper for the Longitude at Sea.1 John Harrison 24.1 Introduction We have already used Newton’s Second Law or Conservation of Energy to analyze systems like the spring-object system that oscillate. We shall now use torque and the rotational equation of motion to study oscillating systems like pendulums and torsional springs. 24.1.1 Simple
    [Show full text]
  • Rotation E Work Done on a Rotating Object
    -28- PHY 131 Section 9: More Rotation e Work done on a rotating object. (This picture is supposed to stand for any rotating object.) Rotational Kinetic Energy. Ex. 9-1: The wheel starts from rest. Find ω after 10 revolutions. (No friction.) If an object is moving from place to place and rotating at the same time, like a rolling wheel: Add the translational KE of its center of mass and the rotational energy about the center of mass to get the total. Ex. 9-2: The solid sphere is released from rest at the top of the ramp. It rolls down without slipping. If no significant energy is lost to friction, what is its speed (in m/s) at the bottom? Ex. 9-3: A .80 kg disk rolls along a level floor, starting at 4.0 m/s. If the friction force on it is 1.5 N, how far will it go before stopping? Angular Momentum. Definition, for a pointlike particle. Angular momentum of a rigid body. Ex. 9-4: Find the magnitude of Earth's angular momentum, due to its spin. Conservation of Angular Momentum: If there are no external torques, a system's total 퐿⃑ is constant. Ex. 9-5: A cork initially moves as shown. You then pull the string, shortening the radius to 25 cm. Find the cork's new speed. -29- Ex. 9-6: If the tar sticks to the rim of the wheel, what is its final angular velocity? Summary of basic concepts: d = rate of change with respect to time dt d = rate of change with respect to position dx Position: Angle: d⃑r d⃑θ⃑ Velocity: v⃑ = Angular velocity: : ω⃑⃑ = dt dt dv⃑⃑ dω⃑⃑⃑ Acceleration: : a⃑ = Angular acceleration: : ⃑α⃑ = dt dt Momentum: p⃑ = m v⃑ Angular momentum: : L⃑ = r x p⃑ ( where m = mass.) ( = I ω⃑⃑ where I = ∫ r2dm) dp⃑⃑ d⃑L⃑ Force: : F⃑ = Torque: : τ⃑ = dt dt d dv⃑⃑ d dω⃑⃑⃑ ( = (mv⃑ ) = m = ma⃑ ) ( = (I ω⃑⃑ ) = I = I⃑α⃑ ) dt dt dt dt d dp⃑⃑ Work: dW = F⃑ ∙ dr (= (r x p⃑ ) = r x = r x F⃑ ) dt dt = d(½mv2) dW Work: dW = τ⃑ ∙ dθ Power: P = 2 dt = d(½Iω ) Notice similarities, for example: p⃑ = m v⃑ L⃑ = I ω⃑⃑ -30- Section 10: Springs and Vibration Hooke's law.
    [Show full text]
  • 15. Parallel Axis Theorem and Torque A) Overview B) Parallel Axis Theorem
    15. Parallel Axis Theorem and Torque A) Overview In this unit we will continue our study of rotational motion. In particular, we will first prove a very useful theorem that relates moments of inertia about parallel axes. We will then move on to develop the equation that determines the dynamics for rotational motion. In so doing, we will introduce a new quantity, the torque, that plays the role for rotational motion that force does for translational motion. We will find, once again, that the rotational analog of mass will be the moment of inertia. B) Parallel Axis Theorem We have shown earlier that the total kinetic energy of a system of particles in any reference frame is equal to the kinetic energy of the center of mass of the system, defined to be one half times the total mass times the square of the center of mass velocity, plus the kinetic energy of the motion of all of the parts relative to the center of mass. = + K system ,lab K REL KCM For a solid object the only possible motion relative to the center of mass is rotation. Therefore, the kinetic energy relative to the center of mass is just equal to one half times the moment of inertia about a rotation axis through the center of mass times the square of the angular velocity about the center of mass. = 1 ω 2 + 1 2 K system ,lab 2 ICM CM 2 Mv CM We can use this result to calculate the moment of inertia about a chosen axis if the moment of inertia about a parallel axis that passes through the center of mass is known.
    [Show full text]
  • Rotation: Kinematics
    Rotation: Kinematics Rotation refers to the turning of an object about a fixed axis and is very commonly encountered in day to day life. The motion of a fan, the turning of a door knob and the opening of a bottle cap are a few examples of rotation. Rotation is also commonly observed as a component of more complex motions that result as a combination of both rotation and translation. The motion of a wheel of a moving bicycle, the motion of a blade of a moving helicopter and the motion of a curveball are a few examples of combined rotation and translation. This module focuses on the kinematics of pure rotation. This module begins by defining the angular variables and then proceeds to describe the relation between these variables and the variables of linear motion. Angular Variables Angular Position Consider an object rotating about a fixed axis. Let us define a coordinate system such that the axis of rotation passes through the origin and is perpendicular to the x- and y- axes. Fig. 1 shows a view of the x-y plane. If a reference line is drawn through the origin and a fixed point in the object, the angle between this line and a fixed direction is used to define the angular position of the object. In Fig. 1, the angular position is measured from the positive x-direction. Fig. 1: Rotating object - Angular position. Also, from geometry, the angular position can be written as the ratio of the length of the path traveled by any point on the reference line measured from the fixed direction to the radius of the path, ...Eq.
    [Show full text]