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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'ﬂlilllfh Iltmuglt
AVEI'HgE . (D — — the angle AH in time .‘l.r. AI. so its average angular
velocity in? T .l'ttlr'Jt. d6
Instantaneous: a) : ?
r‘  Angular and linear velocity Dinanon 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 iwr ,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. lamllu I'FHN'nHF‘ 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. ClamHI 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. Lacm 'A1ILI1I 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' = IiiML: 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 FEESm 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] =ﬁMul Solution:
10 h
v =1?—
? S ...
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