Physics 111: Mechanics Lecture 11

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Physcs 111: Mechancs Lecture 11 Bn Chen NJIT Physcs Department

Textbook Chapter 10: Dynamcs of Rotatonal Moton q 10.1 Torque q 10. Torque and Angular Acceleraton for a Rgd Body q 10.3 Rgd-Body Rotaton about a Movng Axs q 10.4 Work and Power n Rotatonal Moton (partally covered n prevous lecture) q 10.5 Angular Momentum q 10.6 Conservaton of Angular Momentum q 10.7* Gyroscopes and Precesson

Dynamcs of Rotaton

Force vs. Torque q Forces cause acceleratons q What cause angular acceleratons? q A door s free to rotate about an axs through O q There are three factors that determne the effectveness of the force n loosenng the tght bolt: n n n The magntude of the force The poston of the applcaton of the force The angle at whch the force s appled

Torque Defnton q Torque, t, s the tendency of a force to rotate an object about some axs q Let F be a force actng on an object, and let r be a poston vector from a rotatonal center to the pont of applcaton of the force, wth F perpendcular to r. The magntude of the torque s gven by

q q Cross Product The cross product of two vectors says somethng about how perpendcular they are. Magntude: C A B B snq q B A A snq n n n n q s smaller angle between the vectors Cross product of any parallel vectors zero Cross product s maxmum for perpendcular vectors Cross products of Cartesan unt vectors: ˆ ˆj kˆ; ˆ ˆ 0; ˆ kˆ - ˆ; j ˆj ˆj 0; ˆj kˆ ˆ kˆ kˆ 0 z j y j k k x

Cross Product q q Drecton: C perpendcular to both A and B (rght-hand rule) n n n n n Place A and B tal to tal Rght hand, not left hand Four fngers are ponted along the frst vector A sweep from frst vector A nto second vector B through the smaller angle between them Your outstretched thumb ponts the drecton of C Frst practce A B B A? A B B A?

Torque Unts and Drecton q The SI unts of torque are N. m q Torque s a vector quantty q Torque magntude s gven by q Torque wll have drecton n n If the turnng tendency of the force s counterclockwse, the torque wll be postve If the turnng tendency s clockwse, the torque wll be negatve

Understand snf q The component of the force (F cosf ) has no tendency to produce a rotaton q The component of the force (F snf ) causes t to rotate q The moment arm, l, s the perpendcular dstance from the axs of rotaton to a lne drawn along the drecton of the force l r sn f

Net Torque q The force F 1wll tend to cause a counterclockwse rotaton about O q The force F wll tend to cause a clockwse rotaton about O q St t 1 + t + t 3 F 1 l 1 F l q If St ¹ 0, starts rotatng q If St 0, rotaton rate does not change

Battle of the Revolvng Door q A man and a boy are tryng to use a revolvng door. The man enters the door on the rght, pushng wth 00 N of force drected perpendcular to the door and 0.60 m from the hub, whle the boy exerts a force of 100 N perpendcular to the door, 1.5 m to the left of the hub. Fnally, the door wll A) Rotate n counterclockwse B) Rotate n clockwse C) Stay at rest D) Not enough nformaton s gven Boy Man

The Swngng Door q Two forces F 1 and F are appled to the door, as shown n fgure. Suppose a wedge s placed 1.5 m from the hnges on the other sde of the door. What mnmum force F 3 must the wedge exert so that the force appled won t open the door? Assume F 1 150 N, F 300 N, θ 30 1.5 m t t1 + t + t 3 0 F 3 F t1 F1 r1 snq1 150 sn 0 θ.0 m F 1 t 3 F r t -F r 3 snq 300 sn30 3 snq -1.5F 0 + 300-1.5F 0 F3 3 3 3 0 00 Nm 300Nm

Newton s Second Law for a Rotatng Object q When a rgd object s subject to a net torque ( 0), t undergoes an angular acceleraton q The angular acceleraton s drectly proportonal to the net torque q The angular acceleraton s nversely proportonal to the moment of nerta of the object q The relatonshp s analogous to

Newton nd Law n Rotaton q The two rgd objects shown n fgure have the same mass, radus, and ntal angular speed. If the same brakng torque s appled to each, whch takes longer to stop? A) Sold cylnder w -w at B) Thn cylnder shell C) More nformaton s needed f I 1 MR 1 I MR D) Same tme requred for sold and thn cylnders

Strategy to use the Newton s nd Law Draw or sketch system. Adopt coordnates, ndcate rotaton axes, lst the known and unknown quanttes, Draw free body dagrams of key parts. Show forces at ther ponts of applcaton. fnd torques about a (common) axs May need to apply Second Law twce to each part Ø Translaton: Ø Rotaton: F t net net å F ma t Ia Make sure there are enough (N) equatons; there may be constrant equatons (extra condtons connectng unknowns) Smplfy and solve the set of (smultaneous) equatons. Fnd unknown quanttes and check answers å Note: can have F net.eq. 0 but t net.ne. 0

The Fallng Object q A sold, frctonless cylndrcal reel of mass M.5 kg and radus R 0. m s used to draw water from a well. A bucket of mass m 1. kg s attached to a cord that s wrapped around the cylnder. q (a) Fnd the tenson T n the cord and acceleraton a of the bucket. q (b) If the bucket starts from rest at the top of the well and falls for 3.0 s before httng the water, how far does t fall?

Newton nd Law for Rotaton q Draw free body dagrams of each object q Only the cylnder s rotatng, so apply S t I a q The bucket s fallng, but not rotatng, so apply S F m a q Remember that a a r and solve the resultng equatons a mg r

Cord wrapped around dsk, hangng weght Cord does not slp or stretch à constrant Dsk s rotatonal nerta slows acceleratons Let m 1. kg, M.5 kg, r 0. m For mass m: T y T mg FBD for dsk, wth axs at o : N Mg å F y ma mg - T T m (g - a) Unknowns: T, a åt a 0 Tr I + Tr 1 Mr So far: Equatons, 3 unknowns àneed a constrant: Substtute and solve: mgr mar a - Mr Mr 1 Ia m(g - a)r Mr m a( 1 + ) M I Unknowns: a, a mg Mr a a mg a + ar support force at axs O has zero torque r from no slppng assumpton mg ( 4 rad/s ) r(m + M/)

Cord wrapped around dsk, hangng weght Cord does not slp or stretch à constrant Dsk s rotatonal nerta slows acceleratons Let m 1. kg, M.5 kg, r 0. m For mass m: T y mg å F y ma mg - T T m (g - a) Unknowns: T, a a mg ( 4 rad/s ) r(m + M/) a mg (m+ M/) ( 4.8 m/s ) a r support force at axs O has zero torque T m ( g - a) 1.(9.8-4.8) 6N mg x f - x f 1 vt + at 0 + 1 4.8 3 1.6m

Momentum of Rotaton

Angular Momentum q Same basc technques that were used n lnear moton can be appled to rotatonal moton. n F becomes t n m becomes I n a becomes a n v becomes ω n x becomes θ q Lnear momentum defned as q What f mass of center of object s not movng, but t s rotatng? q Angular momentum

Angular Momentum of a Rgd Body q Angular momentum of a rotatng rgd object n L has the same drecton as w n L s postve when object rotates n CCW n L s negatve when object rotates n CW q Angular momentum SI unt: kg m /s L w q Calculate L of a 10 kg dsc when w 30 rad/s, R 9 cm 0.09 m q L Iw and I MR / for dsc q L 1/MR w ½(10)(0.09) (30) 1.96 kg m /s

Angular Momentum of a partcle q Angular momentum of a partcle q Angular momentum of a partcle n r s the partcle s nstantaneous poston vector n p s ts nstantaneous lnear momentum n Only tangental momentum component contrbute n r and p tal to tal form a plane, L s perpendcular to ths plane

Angular Momentum of a Partcle n Unform Crcular Moton Example: A partcle moves n the xy plane n a crcular path of radus r. Fnd the magntude and drecton of ts angular momentum relatve to an axs through O when ts velocty s v. q The angular momentum vector ponts out of the dagram q The magntude s L rp snq mvr sn (90 o ) mvr q A partcle n unform crcular moton has a constant angular momentum about an axs through the center of ts path O

Angular momentum III q Angular momentum of a system of partcles n angular momenta add as vectors n be careful of sgn of each angular momentum p r L L... L L L all all n 1 net å å + + + " p r - p r L 1 1 net ^ ^ + for ths case: 1 1 1 p r p r L L L net + +

Calculatng angular momentum for partcles Two objects are movng as shown n the fgure. What s ther total angular momentum about pont O? L net L 1 + L r1 p1 + r p m L net r mv snq - r mv snq r mv 1 1 1 - r 1 mv.8 3.1 3.6-1.5 6.5. 31.5-1.45 9.8 kgm / s 1 m 1

Lnear Momentum and Force q Lnear moton: apply force to a mass q The force causes the lnear momentum to change q The net force actng on a body s the tme rate of change of ts lnear momentum

Angular Momentum and Torque q Net torque actng on an object s equal to the tme rate of change of the object s angular momentum q Usng the defnton of angular momemtum

Angular Momentum and Torque q Rotatonal moton: apply torque to a rgd body q The torque causes the angular momentum to change q The net torque actng on a body s the tme rate of change of ts angular momentum q S t and L to be measured about the same orgn q The orgn should not be acceleratng, should be an nertal frame

Isolated System q Isolated system: net external torque actng on a system s ZERO n n Scenaro #1: no external forces Scenaro #: net external force actng on a system s ZERO

Conservaton of Angular Momentum q where denotes ntal state, f s fnal state q L s conserved separately for x, y, z drecton q For an solated system consstng of partcles, q For an solated system s deformable

q One of the classc scenes n Swan Lake q Physcs explaned

Isolated System τ net L 0 about z - axs Þ L constant å Iω ntal å fnal I f ω f Moment of nerta changes

How fast does the ballerna spn? The ballerna s ntally rotatng wth angular speed 1. radan/s wth her arms/legs out-stretched. The moment of nerta s 6.0 kg m. Now she pull n her arms and legs and the moment of nertal reduces to.0 kg m. (a) what s the resultng angular speed of the ballerna? (b) what s the rato of the new knetc energy to the orgnal knetc energy? L s constant whle moment of nerta changes

Larger I 6 kg-m Smaller w 1. rad/s Smaller I f kg-m Larger w f? rad/s L s constant whle moment of nerta changes, Zero external torque...about a fxed axs Þ L L fnal I w L ntal I f w f L Soluton (a): w f I w I f 6 1. 3.6 rad/s Soluton (b): K K f 1 1 I f w I w f I I f w f ( ) w KE has ncreased f I I f I ( I f ) I I f 3

Translaton Force F Lnear " Momentum p mv Knetc Energy K 1 mv SUMMARY Rotaton Torque t r F Angular Momentum # % &' Knetc Energy K 1 Iw Lnear Momentum Second Law P Systems and Rgd Bodes åp M v cm dp F net dt Angular Momentum L å L å I w for rgd bodes about common fxed axs " dl sys tnet Second Law dt Momentum conservaton - for closed, solated systems P sys constant Apply separately to x, y, z axes L sys constant