Chapter 10 Review - Chapter 10 Lecture Essential University...

Unformatted text preview: Chapter 10 Lecture Essential University Physics Richard Wolfson 2nd Edition Rotational Motion Angular Velocity - Angular velocity (a is the rate of change of angular position. . — _ A a The arm l'fll-illlfh- Iltmuglt AVEI'HgE . (D — — the angle AH in time .‘l.r. AI. so its average angular velocity in? T .l'ttlr'J-t. d6 Instantaneous: a) : ? r‘ - Angular and linear velocity Din-anon i5. — The linear speed of a point on a rotating ...................mas-""— body is proportional to its distance from _ _ _ _ _ LlIltftll' speed Ir. pruparttunal the rotatlfln aXIS: [u :liatance From the rotation axis. The full circumference iw-r ,-27rr. so i revolution i5. 21.- 5" radians. That makes] radian d9 1 d3 v Stifl‘r'l'rr or abuut 513*. a) _ = I A. 3. "1|”: 1"“ The point on the rim hate the same angular fipfllflj a; but a higher linear speed 1' than the inner paint. Angle in radians is the ratio t11' .'L|.'L' 3r It! radius r'. t'l‘ = .Itrrr: Hunt: ii iii a Iitlli: tom. than i radian. lamll-u I'FH-N'n-H-F‘ r: Angular Acceleration - Angular acceleration a is the rate of change of angular velocity. _ Am dw a. is the tangential Average: 0: : — Instantaneous: a : — “WWW ”i. _ A d acceleration a and is I I parallel to the linear velocity 13. - Angular and tangential acceleration — The linear acceleration of a point on a rotating body is proportional to its distance from the rotation axis: £11:F(I — A point on a rotating object also has radial acceleration: a: is the radial eurr'iplonent+ 2 perpendicular to 13. v 2 Emlmlm In: {I Z—Zfl)? I I" Constant Angular Acceleration - Problems with constant angular acceleration are exactly analogous to similar problems involving linear motion in one dimension. — The same equations apply, with the substitutions x—Hi', “if—>60, o—Hx Tablettifl Angular and Linear Positionflelocity, and Acceleration linearlluantitr Angular Quantity Position .1: Angular position {i Velooit v = E An uiar velooit to * fl 3' a: g 3’ a: A 1 ti - fl - fl A i l at' - d—m - d—EH EBB BIB. GI! £1 — {If — (“I Ilgll Ell" BEBE Bl" lflfl fl! — d: — cit: Equations for instant tinearltteleratlan Equations fartmtstantlngularhtceleratian a = ilvfl + v) (2.3) a = gnu + a] (loo) 1! = in) + at (2.?) to = an“;J + at (10.?) x = at. + vat + ital (2.1a) a = an + {out + gar: no.3) v2 = to]: + Zeb: - x0] (2.11] to2 = mg: -l- 2:1(3 - flu] (10.9] - Torque ris the rotational analog of force, and results from the application of one or more forces. - Torque is relative to a chosen rotation axis. I.‘t.tt:t.l":f:ti::tti:” - Torque depends on: W — the distance from the rotation axis to the force application point. — the magnitude of the force 17“. — the orientation of the force relative to the displacement F from axis to force application point: r = rFsinE’ The same lotto is applied Lu IJJJTL'rcnl angles. Clam-HI to f}. 1' i.!1' xlnachsL Farther may. 1r become; larger. E {11} FaJ'tllcsl uwuy. T hocnlncii greatext. Ffitrqm decrease}: when F TLIII'IIJLII; i3. Lac-m 'A-1ILI1I lb! if} Rotational Inertia and the Analog of Newton’s Law - Rotational inertia I is the rotational analog of mass. — Rotational inertia depends on mass and its distance from the rotation axis. Rmmingme Mam mass near the it‘s harder . . axisis eas. toe in. - Rotational acceleration, torque, = 1, p ' and rotational inertia combine to give the rotational analog of Newton's second law: H Rotation axis ax“ r=1a Finding Rotational Inertia - For a single point mass m, rotational inertia is the product of mass with the square of the distance R from the rotation axis: 1 : mg? - For a system of discrete masses, the rotational inertia is the sum of the rotational inertias of the individual masses: 2 I: my. II The mass element aim contributes - For continuous matter, the ““0"“i"*”"“”:“’”*"'- rotational inertia is given by an integral over the distribution of matter: I : Irz dm Rotational Inertias of Simple Objects Table 1&1 Hutatiunal lnertias Selle sphere ebeul diameter Flat plain about perpendicular axis I=%Mfl1 I=T'iM{n2+i:-2) P L 11.11:: and alien: eenter Thin lie: or heiiew cylinder 1' = Iii-ML: eheui. its axis I = Mil? {b Heliew spherieui shell e‘eeui diameter r=§ufl D Flat plate about central axis. _J._ I—uual L 'lhhlmdabeutend V I=J3'ML1 ll 2M1 FEES-m Educaihn. I1:. Disk or solid eylinder abet: its axis I = 1- HR: Combining Rotational and Linear Dynamics - In problems involving both linear and rotational motion: — IDENTIFY the objects and forces or torques acting. — DEVELOP your solution with drawings and by writing Newton’s law and its rotational analog. Note physical connections between the objects. — EVALUATE to find the solution. — ASSESS to be sure your answer makes sense. A bucket of mass m drops Free—bud dia rams . . into awell, its rope_ for bucke an cylinder Newtons law, bUCkBt' unrollin from a cylinder of # F = mg—T= ma mass and radius R. mops tension T provides "st e connec Ion - What’s its acceleration? .4 figflgnsallaaflag iyr’idgr' T , . RT = laiR ,3. Here I : 1 MR2 _ T 2 Solve the two equations to a, I get mg m Rotational Energy - A rotating object has kinetic energy Km : fife); associated with its rotational motion alone. — It may also have translational kinetic energy: Km : 5 ME. - ln problems involving energy conservation with rotating objects, both forms of kinetic energy must be considered. — For rolling objects, the two are related: - The relation depends on the rotational inertia. Example: A solid ball rolls down a hill. How f35t i5 it moving at the bottom? Equation for energy conservation 1 2 1 2 Mghz—m +—Im Energy bar 1? J E 1%; l % 2 2 2 1 2 1 2 v 1 graphs 0 u m G '5 r: K1“ :3“: +E[EMR1][E] =fiMu-l Solution: 10 h v =1?— ? S ...
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