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Physics 111: Lecture 10

Bin Chen

NJIT Department , 13/e Phys. 111Young/Freedman (Part I): Phys. 111 (Part II): TranslationalChapter Mechanics 10 Key Equations Rotational Mechanics q of point bodies q motion of Rigid Bodies (extended, finite size) q Translational motion. Fl rFsin F r (magnitude of ) (10.2) Size and shape not q + translation,tan more considered complex possible q rigid bodies: fixed size & q F = ma rF (definition of torque vector) (10.3) å ext shape, matters q dynamics q conservation laws: F = ma I & å ext cm zz (10.7) (rotationalq rotational analog modifications of ’s to law for a ) energy conservation q conservation laws: energy & 11 angular momentumKM 22 I 22cm cm (10.8) (rigid body with both translation and rotation)

cm R condition for without slipping (10.11)

Fa M (10.12) ext cm

zz Icm (10.13)

Wd 2 done by a torque (10.20) z 1

W zz 21 (work done by a constant torque) (10.21)

2 11 22 (10.22) WIdItot zz 2 I 1 1 22

P zz (10.23)

Lrprm ( of a particle) (10.24)

L I (for a rigid body rotating around a axis) (10.28)

Copyright © 2012 Pearson Education, Inc. Page 1 of 2 Chapter 9 Rotation of Rigid Bodies q 9.1 Angular and q 9.2 Rotation with Constant q 9.3 Relating Linear and Angular q 9.4 Energy in Rotational Motion q 9.5 Parallel-Axis Theorem q Moments-of- Calculations Rigid Object q A rigid object is one that is nondeformable n The relative locations of all particles making up the object remain constant n All real objects are deformable to some extent, but the rigid object model is very useful in many situations where the deformation is negligible q This simplification allows analysis of the motion of an extended object and q What is the circumference S ? s s = (2p )r 2p = r q q can be defined as the arc r length s along a circle divided s by the radius r: q

q q is a pure number, but commonly is given the artificial unit, radian (“rad”) q Whenever using rotational equations, you MUST use expressed in Conversions q Comparing degrees and radians 2p (rad) = 360! p (rad) =180! q Converting from degrees to radians π θ (rad ) = θ (degrees) 180° q Converting from radians to degrees 180! 360° q(deg rees) = q(rad) 1rad = = 57.3° p 2π q Converting from revolutions to radians 1 revolution = 2π (rad) = 360 rpm: Conversion q A waterwheel turns at 360 revolutions per hour. Express this figure in radians per second.

A) 3.14 rad/s B) 6.28 rad/s C) 0.314 rad/s D) 0.628 rad/s

"#$%&'()%* 1 = 2. "/0/+ + One Dimensional x q What is motion? Change of position over . q How can we represent position along a straight ? q Position definition: n Defines a starting point: origin (x = 0), x relative to origin n Direction: positive (right or up), negative (left or down) n It depends on time: t = 0 (start clock), x(t=0) does not have to be zero. q Position has units of [Length]: meters.

x = + 2.5 m

x = - 3 m Angular Position q Axis of rotation is the center of the disc q Choose a fixed reference line q Point P is at a fixed r from the origin q As the particle moves, the only coordinate that changes is q q As the particle moves through q, it moves though an s. q The angle q, measured in radians, is called the angular position. q Displacement is a change of position in time. q Displacement: Dx = x f (t f ) - xi (ti ) n f stands for final and i stands for initial. q It is a vector quantity. q It has both magnitude and direction: + or - q It has units of [length]: meters. x1 (t1) = + 2.5 m x2 (t2) = - 2.0 m Δx = -2.0 m - 2.5 m = -4.5 m

x1 (t1) = - 3.0 m x2 (t2) = + 1.0 m Δx = +1.0 m + 3.0 m = +4.0 m q The angular displacement is defined as the angle the object rotates through during some time interval

Δθ = θ f −θi q SI unit: radian (rad) q A counterclockwise rotation is positive. q A clockwise rotation is negative. Velocity q Velocity is the rate of change of position q Average velocity displacement Δx x f − xi v = = avg Δt Δt q Average distance Savg = total distance/total time q Instantaneous velocity dx x − x v = = f i dt lim t Δt→0 Δ displacement Average and Instantaneous q The average angular velocity, ωavg, of a rotating rigid object is the ratio of the angular displacement to the time interval

θf −θi Δθ ωavg = = tf − ti Δt q The instantaneous angular velocity is defined as the limit of the average velocity as the time interval approaches zero Δθ dθ ω ≡ lim = Δt→0 Δt dt q SI unit: (rad/s) Angular Velocity: + or - ? q Angular velocity positive if rotating in counterclockwise q Angular velocity will be negative if rotating in clockwise q Every point on the rotating rigid object has the same angular velocity Average Acceleration q Changing velocity (non-uniform) means an acceleration is present. q Acceleration is the rate of change of velocity. q Acceleration is a vector quantity. q Acceleration has both magnitude and direction. q Acceleration has a unit of [length/time2]: m/s2. q Definition: Dv v f - vi n Average acceleration aavg = = Dt t f - ti

n Instantaneous acceleration Average Angular Acceleration q The average angular acceleration, a, of an object is defined as the ratio of the change in the angular speed to the time it takes for the object to undergo the change: Instantaneous Angular Acceleration q The instantaneous angular acceleration is defined as the limit of the average angular acceleration as the time goes to 0 Δω dω α ≡ lim = Δt→0 Δt dt q SI Units of angular acceleration: rad/s² q Positive angular acceleration is in the counterclockwise direction. n if an object rotating counterclockwise is speeding up n if an object rotating clockwise is slowing down q Negative angular acceleration is in the clockwise direction. n if an object rotating counterclockwise is slowing down n if an object rotating clockwise is speeding up Rotational Kinematics q A number of parallels exist between the equations for rotational motion and those for .

x f - xi Dx θf −θi Δθ v = = ωavg = = avg t − t Δt t f - ti Dt f i q Under constant angular acceleration, we can describe the motion of the rigid object using a set of kinematic equations n These are similar to the kinematic equations for linear motion

n The rotational equations have the same mathematical form as the linear equations Comparison Between Rotational and Linear Equations Angular Motion q At t = 0, a wheel rotating about a fixed axis at a constant angular acceleration has an angular velocity A.of 2.017 rad/s rad/s.2 Two later it has turned through B. 14 rad/s2 C.5.020 complete rad/s2 revolutions. Find the angular acceleration D.of this23 rad/s wheel?2 E. 13 rad/s2 A. 17 rad/s2 B. 14 rad/s2 C. 20 rad/s2 D. 23 rad/s2 E. 12 rad/s2 Relating Angular and Linear Kinematics q Every point on the rotating object has the same angular motion (angular displacement, angular velocity, angular acceleration) q Every point on the rotating object does not have the same linear motion q Displacement s = θr q Velocity v = ωr q Acceleration a = αr Velocity Comparison q The linear velocity is always tangent to the circular path n Called the tangential velocity q The magnitude is defined by the tangential velocity Ds Dq = r Dq Ds 1 Ds v = = w = or v = rw Dt rDt r Dt r Acceleration Comparison q The tangential acceleration is the derivative of the tangential velocity

Dv = rDw Dv Dw = r = ra Dt Dt

at = ra Velocity and Acceleration Note q All points on the rigid object will have the same angular speed, but not the same tangential speed q All points on the rigid object will have the same angular acceleration, but not the same tangential acceleration q The tangential quantities depend on r, and r is not the same for all points on the object v w = or v = rw a = ra r t Centripetal Acceleration q An object traveling in a circle, even though it moves with a constant speed, will have an acceleration n Therefore, each point on a rotating rigid object will experience a centripetal acceleration v2 (rw)2 a = = = rw 2 r r r Resultant Acceleration q The tangential component of the acceleration is due to changing speed q The centripetal component of the acceleration is due to changing direction q Total acceleration can be found from these components

2 2 2 2 2 4 2 4 a = at + ar = r α + r ω = r α + ω Angular and Linear Quantities q For a rigid rotational CD, which statement below is true for the two points A and B on this CD?

A) Same distance travelled in 1s B) Same linear velocity A C) Same centripetal acceleration D) Same linear acceleration E) Same angular velocity B Rotational q An object rotating about z axis with an angular speed, ω, has rotational kinetic energy q Each particle has a kinetic energy of 2 n Ki = ½ mivi q Since the tangential velocity depends on the distance, r, from the axis of rotation, we can substitute

vi = wri Rotational Kinetic Energy, cont q The total rotational kinetic energy of the rigid object is the sum of the of all its particles 1 K K m r 2 2 R = ∑ i = ∑ i i ω i i 2

1 " 2 # 2 1 2 K m r I R = $∑ i i %ω = ω 2 & i ' 2 q Where I is called the of inertia Rotational Kinetic Energy, final q There is an analogy between the kinetic energies associated with linear motion (K = ½ mv 2) and the kinetic energy associated with 2 rotational motion (KR= ½ Iw ) q Rotational kinetic energy is not a new type of energy, the form is different because it is applied to a rotating object q Units of rotational kinetic energy are (J) of Point q For a single particle, the definition of moment of inertia is

n m is the mass of the single particle n r is the rotational radius q SI units of moment of inertia are kg.m2 q Moment of inertia and mass of an object are different quantities q It depends on both the quantity of matter and its distribution (through the r2 term) Moment of Inertia q The figure shows three small spheres that rotate about a vertical axis. The perpendicular distance between the axis and the center of each sphere is given. Rank the three spheres according to their moment of inertia about that axis, greatest first ? 1m 36kg a A) a, b, c 2m 8kg B) b, a, c b C) c, b, a 3m D) all tie 4kg c E) a and c tie, b Rotation axis Moment of Inertia of Point Mass q For a composite particle, the definition of moment of inertia is

n mi is the mass of the ith single particle

n ri is the rotational radius of ith particle q SI units of moment of inertia are kg.m2 q Consider an unusual baton made up of four sphere fastened to the ends of very light rods q Find I about an axis perpendicular to the page and passing through the point O where the rods cross Moment of Inertia of Point Mass q For a composite particle, the definition of moment of inertia is

n mi is the mass of the ith single particle

n ri is the rotational radius of ith particle q SI units of moment of inertia are kg.m2 q Consider an unusual baton made up of four sphere fastened to the ends of very light rods q Find I about axis y Moment of Inertia of Extended Objects q Divided the extended objects into many small volume

elements, each of mass Dmi q We can rewrite the expression for I in terms of Dm I = lim r 2Δm = r 2dm Δmi →0 ∑ i i i ∫ q With the small volume segment assumption, I = ∫ ρr 2dV q If r is constant, the integral can be evaluated with known geometry, otherwise its variation with position must be known Moment of Inertia of a Uniform Rigid Rod q The shaded has a mass n dm = l dx l=M/L q Then the moment of inertia is

L / 2 2 2 M Iy = r dm = x dx ∫ ∫−L / 2 L 1 I = ML2 12 M-I for some other common shapes Parallel-Axis Theorem q In the previous examples, the axis of rotation coincided with the axis of symmetry of the object q For an arbitrary axis, the parallel-axis theorem often simplifies calculations q The theorem states 2 I = ICM + MD n I is about any axis parallel to the axis through the of the object

n ICM is about the axis through the center of mass n D is the distance from the center of mass axis to the arbitrary axis Moment of Inertia of a Uniform Rigid Rod q The moment of inertia about y is

L / 2 2 2 M Iy = r dm = x dx ∫ ∫−L / 2 L 1 I = ML2 12 q The moment of inertia about y is 1 L 1 I = I + MD2 = ML2 + M ( )2 = ML2 y' CM 12 2 3 7133_Ch09_pp278-307.qxd 10/14/10 3:38 PM Page 297

7133_Ch09_pp278-307.qxd 10/14/10 3:38 PM Page 297

7133_Ch09_pp278-307.qxd 10/14/10 3:38 PM Page 297

CHAPTER 9 SUMMARY

CHAPTER SUMMARY Rotational kinematics:9 When a rigid body rotates about ¢u du du y dvz vz = lim = (9.3) vz 5 az 5 S dt dt a stationary axis (usually called the z-axis), its position ¢t 0 ¢t dt is describedCHAPTER by an angular coordinate u. The angular 2 At t2 At t1 9 SUMMARY ¢vz dvz d u a lim Du dv velocityRotational v zkinematics: is the time derivativeWhen a rigid of ubody, and rotatesthe angular about z = ¢u = du = 2 du y z vz = ¢limt S 0 ¢t = dt dt (9.3) vz 5 u2az 5 acceleration a is the time derivative of v or the second S dt dt a stationary axisz (usually called the z-axis),z its position t 0 ¢ ¢t dt (9.5), (9.6) u1 derivative of u. (See Examples 9.1 and 9.2.) If the angu- At t At t x is describedRotational by kinematics: an angularWhen coordinate a rigid ubody. The rotates angular about ¢u du 2 du 2 O dvz 1 ¢vz dvz d u (9.3) v 5 y a 5 lar acceleration is constant, then u, v , and a are related vz = lim = z z z z S dt dt Du velocitya stationary vz is the axis time (usually derivative called of the u ,z-axis), and the its angular position az = ¢limt 0 ¢t =dt1 2 = u u S v t a t 2 (9.11) by simple kinematic equations analogous to those for = 0¢t+ 0 0¢z t + 2 dtz dt u2 accelerationis described a isby the an timeangular derivative coordinate of uv. The or theangular second 2 At t2 At t1 z z ¢vz dvz d u u straight-line motion with constant linear acceleration. a (constantlim a only) (9.5), (9.6) Du 1 x derivativevelocity of v uz. is (See the timeExamples derivative 9.1 andof u ,9.2.) and theIf theangular angu- z = z= = 2 t S 0 t dt Ou (See accelerationExample 9.3.) a is the time derivative of v or the second ¢ ¢ dt 2 lar acceleration isz constant, then u, vz , and zaz are related 1 u u - u0 = 2 v0z +1 vz 2t (9.5), (9.6)(9.10) 1 x derivative of u. (See Examples 9.1 and 9.2.) If the angu- 7133_Ch09_pp278-307.qxd 10/14/10 3:38 PM Page 297 by simple kinematic equations analogous to those for u = u0 + v0zt + 2 azt (9.11) O

lar acceleration is constant, then u, v , and a are related straight-line motion with constant linearz acceleration.z (constant1 az only)1 2 2 by simple kinematic equations analogous to those for u = (constantu0 + v0z t a+z only)2 az t (9.11) (See Example 9.3.) vz = v0z + az t (9.7) straight-line motion with constant linear acceleration. 1 (constant(constant a az only) only) (See Example 9.3.) u - u0 = 2 v0zz + vz t (9.10) 1 2 2 vu - u0 v= v02z a+ vuz t u (9.10)(9.12) z =(constant0z2 +1 az z only)-2 0 (constant a only) (constant a z only) vz = v0z +1 azz t 1 2 2 (9.7)

vz =(constantv0z + az taz only) (9.7) Relating linear and angular kinematics: The angular v =2 (constantrv 2 az only) (9.13) y atan 5 ra Chap. 9 Summary vz = v0z + 2az u - u0 (9.12) v CHAPTER speed v of a rigid body is the magnitude of its angular 2 d v 2 dv v 5 rv SUMMARY vatanz =(constant=v0z += 2 ra z only)u -= ur0a (9.12)(9.14) S 9 velocity. The rate of change of v is a = dv dt. For a dt zdt Linear a (constant az only)1 2 P particle in the body a distance r from the rotation axis, 2 Relatingv linear1 and 2angular kinematicsacceleration a 5 v2r Rotational Kinematics > S 2 rad the speed v and the components of the acceleration a arad = = v r (9.15) of point P r s r dv Rotational kinematics: When a rigid body rotates aboutareRelating related linear to v and and angular¢ au. (Seed ukinematics: Examples 9.4The and angular 9.5.) dvu = ryv z (9.13) y uatan 5 ra Relatingvz = linearlim and angular= kinematics: The(9.3) angular vz 5 v rv az 5 (9.13) y atan 5 ra x S dt = dt v a stationary axis (usually called the z-axis), its positionspeed v of a rigidt 0 bodyt is dtthe magnitude of its angular dv dv O speed v of¢ a rigid¢ body is the magnitude of its angular dv dv v v 5 rv At attan At t r ra (9.14) vS 5 rv is described by an angular coordinate u. The angularvelocity. The rate of change of v is 2a dv dt. For a a2 = =r1 =ra (9.14) S a velocity. The rate¢v ofz changedvz of vd isu a= dv dt. For a tan = dt = dt= Linear a P = Dudt dt Linear P velocity vz is the time derivative of u, and the angularparticle ainz the= lim body a distance= r=from the rotation axis, 2 acceleration 2 particle in thet S 0 body a distance r from2 the rotation axis, 2 acceleration 2 ¢ ¢t dt dt S vv u22 2 2 a 5 avradr 5 v r acceleration az is the time derivative of vz or the secondMoment of inertia and rotational kinetic energy:>> The S aI m 1 r 1 v2mr2 r 2 Á (9.15) Axisof of point P rad the speedthe speed v and v andthe thecomponents components of of (9.5),the the acceleration acceleration (9.6) aa aradrad= == u ==+v r + (9.15) of point P rv r s derivative of u. (See Examples 9.1 and 9.2.) If the angu-moment of inertia I of a body about a given axis is a rr 1 x rotation s are relatedare related to v to and v and a. (Seea. (See Examples Examples 9.4 9.4 and and 9.5.) 9.5.) O 2 u mu2 = a m i r i (9.16) r2 x x lar acceleration is constant, then u, vz , and az are relatedmeasure of its rotational inertia: The greater the value 1 2 Momenti of inertia and O O 2 I 5 by simple kinematic equations analogous to those forof I, the umore= u 0difficult+ v0zt it+ is2 atoz tchange the(9.11) state of the m Smiri 1 2 1 i K Iv (9.17) straight-line motion with constant linear acceleration.body’s rotation. The moment of inertia can be expressed rotational= 2 kinetic energy r1 1 (constant az only) K 5 Iv2 (See Example 9.3.) as a sum over the particles m i that make up the body, 2 2 2 Moment of inertia and rotational kinetic energy: The I m 1 r 1 2 m 2 r 2 2 Á Axis of = + + eachMoment of which of inertia is at and1 its rotationalown perpendicular kinetic energy: distanceThe r I = m 1r 1 + m 2r 2 + Á Axis of v momentu - uof0 inertia= 2 v I0zof+ a bodyvz t about a given(9.10) axis is ia rotation vr3 moment of inertia I of a body about a given axis is a 2 rotation m2 m3 r from the axis. The rotational kinetic energy of a rigid = a m ir i 2 (9.16) 2 m2 measure of its rotational inertia: The greater the value m r (9.16) r (constant a only) = ai i i 2 bodymeasure rotating of its about rotational1 a fixedz inertia: axis2 dependsThe greater on the the angular value I 5 2 of I, the more difficult it is to change the state of the i m Smiri 2 speedof I, the v more and the difficult moment it isof to inertia change I for the that state rotation of the 1 2 1 i I 5 Smiri vz v0z az t (9.7) K Iv (9.17) m1 body’s rotation.= + The moment of inertia can be expressed = 21 2 r1 1 i K Iv (9.17) 2 axis.body’s (See rotation. Examples(constant The 9.6–9.8.) momenta only) of inertia can be expressed = 2 r1 K 5 Iv as a sum over the particlesz m i that make up the body, 2 K 5 1 Iv2 as a sum over the particles m that make up the body, 2 each of2 which 2 is at its own iperpendicular distance ri r vz = v0z + 2az u - u0 (9.12) The parallel-axis theorem 3 m eachfrom of which the axis. is at The its rotationalown perpendicular kinetic energy distance of a rigid ri r 3 Calculating the(constant moment a of only) inertia: The parallel-axis I I Md2 (9.19) 3 m frombody the axis.rotating The about rotational za fixed kinetic axis depends energy on of the a rigidangular P = cm + 3 theorem relates the moments1 of inertia2 of a rigid body bodyspeed rotating v and about the momenta fixed axisof inertia depends I for onthat the rotation angular d of mass M about two parallel axes: an axis through the cm speedaxis. v and (See the Examples moment 9.6–9.8.) of inertia I for that rotation Relating linear and angular kinematics: The angular center ofv mass= rv (moment of inertia Icm ) and(9.13) a parallel y atan 5 ra Mass M P v I speed v of a rigid body is the magnitude of its angularaxisaxis. a (See distance Examples dvfrom 9.6–9.8.) thedv first axis (moment of inertia v 5 rv cm atan = = r = ra (9.14) S I 5 I 1 Md 2 velocity. The rate of change of v is a = dv dt. For IaP). (See Exampledt 9.9.) dtIf the body has a continuousLinear a 2 P cm Calculating the moment of inertia: The parallel-axis IP = Icm + MdP (9.19) particle in the body a distance r from the rotation axis,,2 the moment of inertia can be calcu-acceleration 2 S theorem relatesv the 2moments of inertia of a rigid body arad 5 2v r d the speed v and the components of the acceleration> latedCalculatinga by aintegration.rad the= moment= (Seev ofr Examplesinertia: The 9.10 parallel-axis(9.15) and 9.11.) of point P IP = Icm + Md (9.19) cm of mass M aboutr two parallel axes: an axis through the r s are related to v and a. (See Examples 9.4 and 9.5.) theorem relates the moments of inertia of a rigid body u d center of mass (moment of inertia Icm ) and a parallel x Mass M I cm P of massaxis Ma distanceabout two d from parallel the first axes: axis an (moment axis through of inertia the O cm 2 center of mass (moment of inertia Icm ) and a parallel MassIP 5 M Icm 1 Md IP). (See Example 9.9.) If the body has a continuous I P axis massa distance distribution, d from the the moment first axis of (momentinertia can of be inertia calcu- cm 2 2 I 5 I 1 Md 2 Moment of inertia and rotational kinetic energy: TheI ). (See IExamplem r 9.9.)m If r the bodyÁ has a continuousAxis of P cm P lated by= integration.1 1 + (See2 2 Examples+ 9.10 and 9.11.) v moment of inertia I of a body about a given axis is a rotation mass distribution, the2 moment of inertia can be calcu- m2 = m i r i (9.16) r2 measure of its rotational inertia: The greater the valuelated by integration.a (See Examples 9.10 and 9.11.) i I 5 2 of I, the more difficult it is to change the state of the m Smiri 1 2 1 i K Iv (9.17) body’s rotation. The moment of inertia can be expressed = 2 r1 K 5 1 Iv2 as a sum over the particles m i that make up the body, 2 each of which is at its own perpendicular distance ri r3 m 297 from the axis. The rotational kinetic energy of a rigid 3 body rotating about a fixed axis depends on the angular speed v and the moment of inertia I for that rotation axis. (See Examples 9.6–9.8.) 297 2 Calculating the moment of inertia: The parallel-axis IP = Icm + Md (9.19) theorem relates the moments of inertia of a rigid body d of mass M about two parallel axes: an axis through the cm 297 center of mass (moment of inertia Icm ) and a parallel Mass M I P axis a distance d from the first axis (moment of inertia cm I 5 I 1 Md 2 IP). (See Example 9.9.) If the body has a continuous P cm mass distribution, the moment of inertia can be calcu- lated by integration. (See Examples 9.10 and 9.11.)

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