Physics 2101 S c e t c i cti n o 3 n 3 March 24th Announcements: Review today Exam tonight at 6 PM, Lockett Lockett--6 Class Website:

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1 Physics 101 Section 3 Mach 4 th Announcements: Review today Exam tonight at 6 PM, Lockett-6 Class Website:

2 Chap. 10: Rotation θ () t θ θ θ ( Δ 1 ) ω (t) d θ (t) dt dω t d t α () t dt dt () θ() s θ Rotational Kinematics: ( ONLY IF α constant) θ ω v ω d v dt a a + a tot t + ω α t ω ω0 + αt a tot a t + 1 a + 0t+ a a ( ) ω ω 0 + α θ θ ω αt ω ω αθ Rotational Kinetic enegy: KE ot 1 Iω Moment of inetia: Togue: τ F i i I I COM + Mh all mass I m dm τ F sinϑ θ 1 Wok done by a togue: W τ netdθ τ ( θ f θi ) Δ KEot I ( ωf ωi ) θ 1

3 Chap. 11: Rotational dynamics τ t nd Law fo otation: net i τ i I α { F net F i ma } 1 1 Rolling bodies: vcom ωr acom αr KE I ω + Mv The fiction is always against the tendency of sliding tot com com Angula momentum: p m ( v ) L I dl ω dt net τ net net τ Iα I τ i dω dt d dt (a paticle) (a igid body) Angula momentum L L if τ Iω I ω consevation: 0; i f net i i f f

4 Summay: Effects of otation Fom befoe With Rotation a 3 g 4 m M a gm m sinθ 1 M w + ( m 1 + m ) Done via foce/toque v 4gd M m M W + m + M Done via enegy v 4 gd 3 Done with both emembe this fo late In each of these cases: tanslation ti was sepaate fom otation ti Pue tanslation + Pue otation

5 SHW#6: Thee masses M, M and 3M ae shown in T 3 T T 4 the figue with two pulley of moment of inetial I T 1 and adius. The coefficient of fiction is μ k and the table is L m long. T 1 Mg Ma Ia ( T T 1 ) Iα T 3 T μ k Mg Ma ( T 4 T 3 ) Ia 3Mg T 4 3Ma Need v ω & a α T gm ( 3 μ k ) a I + 5M T gm T 3 a I T 3 MgM 1 μ k ( μ k )+ M 4 μ k I + 3M I + 3M + 3Mg + 3Mg Mg + μ k Mg 1 μ k a I + 3M

6 SHW#6: Thee masses M, M and 3M ae shown in T 3 T T 4 the figue with two pulley of moment of inetial I T 1 and adius. The coefficient of fiction is μ k and the table is L m long. Let us use Enegy. We can define zeo so E ( stat ) mech 0 Afte block 3 has moved a distance d we have 6Mv Iω Eme ch 0 μkmgd + (3 M M) gd 3 Mv Iv μkmgd + Mgd (use v ω ) Md g ( 1 μk ) v : I Mg 1 μ 3M + v I ad : a k 3M + I v f + v i t d : t I d 3M + I Mgd( 1 μ k ) Same as befoe a Mg 1 μ k I + 3M

7 Rolling down a amp : Enegy consideations An object at est, with mass m and adius, oles fom top of incline plane to bottom. What is v at bottom and a thoughout. ( KE ) final 0 KE ot +tans,com 1 mv COM ΔE mech 0 ΔKE tot ΔU [ ] init ( 0 ) final ( mgh ) init + 1 I COMω mgl sinθ Using v com ω h L sinθ h θ Using v v 0 + al v COM gl sinθ 1+ I com mr Using 1 D kinematics (const accel) a v gsinθ com L 1+ I com mr Speed at Bottom Constant Acceleation

8 Fiction and Rolling The static fiction is always opposing the tendency to slide 1) If a com 0, it has no tendency to slide at point of contact no fictional foce ) If a com > 0 (i.e. thee ae net foces) and no slipping occus, τ 0 povided by static fiction foce 3) If a com > 0, and no slipping occus, τ 0 povided by weight and static fiction foce f s θ

9 Rolling down a amp: acceleation Acceleation downwads Fiction povides toque Static fiction points upwads Yo Yo Newton s nd Law Linea vesion x ˆ : f s F g sinθ ma com y ˆ : N F g cosθ 0 Tension povides toque Hee θ 90 Angula vesion z ˆ : I com α τ Rf s a R ( mgsinθ ma com ) com I com Al Axle: R R 0 a com αr α a com R Rf s I com gsinθ a com 1+ I com mr g a com 1+ I com mr 0

10 a com a com g sinθ Icom 1+ MR Cylinde Hoop MR I1 I MR g sin θ g sin θ a1 a 1 + I / MR 1 + I / MR 1 g sinθ g sinθ a 1 a 1 + MR /MR 1 + MR / MR gsinθ gsinθ a1 a 1+ 1/ 1+ 1 gsinθ gsinθ a1 (0.67) gsin θ a (0.5) gsinθ 3

11 Pob NON smooth olling motion A bowle thows a bowling ball of adius R along a lane. The ball slides on the lane, with initial speed v com,0 and initial angula speed ω 0 0. The coefficient of kinetic fiction between the ball and the lane is μ k. The kinetic fictional foce f k acting on the ball while poducing a toque that causes an angula acceleation of the ball. When the speed v com has deceased enough and the angula speed ω has inceased enough, the ball stops sliding and then olls smoothly. a) [Afte it stops sliding] What is the v com in tems of ω? Smooth olling means v com Rω b) Duing the sliding, what is the ball s linea acceleation? Fom nd law: (linea) ˆ x : f k ma com ˆ y : N mg 0 But f k μ k N μ k mg So a com f k m μ k g c) Duing the sliding, what is the ball s angula acceleation? Fom nd law: τ Rf k ( ˆ z ) But f k μ k N (angula) Iα( ˆ z ) τ Rf k ( ˆ z ) μ k mg So Iα Rf k R( μ k mg) α Rμ kmg I d) What is the speed of the ball when smooth olling begins? When does v com Rω? v com v 0 + a com t v com v 0 μ k gt Fom kinematics: ω ω 0 + αt t ω α Iω Rμ k mg I v com R v com v 0 μ k g Rμ k mg e) How long does the ball slide? v com t v 0 v com μ k g v 0 1+ ImR

12 Poblem A paticle of mass m slides down the fictionless suface though height h and collides with the unifom vetical od (of mass M and length d), sticking to it. The od pivots about point O though the angle θ befoe momentaily stopping. Find θ. What do we know? Whee ae we tying to get to? What do we know about the amp? What does this give us? Consevation of Mechanical Enegy going down amp! ( + 0) ( KE init + U init ) KE final + U final ( 0 + mgh) 1 mv final This gives us the speed of the mass ight befoe hitting the od: v m,bottom gh Thee s a collision! What kind? What do we know about this collision? What is constant? Only Consevation of Angula Momentum holds! L initial L final (choose a point wisely) ( p m,0 + 0) ( p m, f + I od ω) ( d(mv m,0 )) md ( ω bottom )+( 1 Md 3 )ω bottom This gives the angula speed of the mv m,0 od/mass just afte hitting: ω m gh bottom dm + 1 Md dm + 1 Md 3 3 Now what? Afte the collision, what holds? Now Consevation of Mechanical enegy holds again! Solve fo θ: ( KE init + U init ) ( KE final + U final ) 1 ( I totω + 0) 0 + mgd(1 cosθ + Mg( 1 d)(1 cosθ bottom ) ( ( final ) final ) ) θ cos 1 1 m h dm+ ( 1 M )m + 1 M 3

13 Sample Poblem * 1 9 Fou thin, unifom ods, each with mass M and length d, ae attached igidly to a vetical axle to fom a tunstile (ts). The tunstile otates ccw with an initial angula velocity ω i. A mud ball (mb) of mass m 1/3 M and initial speed v i is thown along the path shown and sticks to one end of one od. What is the final angula velocity of the ball tunstile system? ( angula vesion) L m v 1) Is KE tot conseved. To find ω f of the combined system, what is conseved? ) Is E mech conseved. 3) Is P conseved. Using Consevation of angula momentum about axis: 4) Is conseved. Li L f L + L L + L about axle ts, i mb, i ts, f mb, f [ I ω ] + L I L, ω +, ts i ts, i mb i ts f ts, f mb f L L I ω & L I ω ts, i ts i ts, f ts f ts 4 I Md + M d Md v NO! NO! NO! YES! 4 4 Md ω + L Md ω + L,, 3 i ts, i mb i 3 f ts, f mb f Lmb, f Imbωf ( md ) ωf Md ω 1 3 f Md ω + L Md ω + Md ω i ts, i mb, i 3 f ts, f 3 f mb, f L mv sinϑ d M v sin Mdv 1 1 mb, i i 3 i 6 i Md ω + Mdv Md ω + Md ω Md ω i ts, i 6 i mb, i 3 f ts, f 3 f mb, f 3 f 4 ( 1 ω ) Md Mdv v ω ω + ω i i 4 i f 5 5 i 10 f Md d 3

14 Chap. 1-13: 13: Equilibium, Gavitation Static equilibium: Gavitation law: F τ net net F 1 G m 1m 1 0 (balance of foces) 0 (balance of togue) ˆ 1 F 1,net Vitual otation fo togues F 1 + F 13 + F F 1n M F mag ; ag G Consevative foce n i1 F 1i Gavitational potential enegy: mm 1 U () G ; Δ U fi W Fg d U total 1 mm i j G (many objects) i< j ij Keple s Laws: da dt L m nd constant ( -Law) T 4π GM 3 d (3 - Law)

15 Poblem 37: A unifom plank, with a length L of 6.10 m and a weight of mg 445 N, ests on the gound and against fictionless olle at the top of a wall of height h3.05 m. The plank emains in equilibium fo any value of θ 70 but slips if θ <70. Find the coefficient of static fiction between the plank and the gound. What do we know: θ 70 is citical angle: s s F ix 0: fs N1 sinθ 0 (1) F i 0 F iy 0: N + N1cos θ mg 0 () Choose a convenient VIRTURAL otation axis h L τ i 0 N1 mg cosθ 0 (3) sinθ f μ N y N 1 N x f s mg

16 Ch. 13 # 11: Two sphees of mass m and a thid of mass M fom an equilateal tiangle, and a foth of mass m 4 is at the cente of the tiangle. If the net foce on the cente is zeo what is M? If m 4 is doubled what is the answe? The angle at each cone is 60 0, so the distance to the cente mass is l cos 300 l 3 All the x foces cancel, the downwad foce on m 4 is m m F y (net) Gm 4m sin Gm 4m l The downwad foce must equal the upwad foce by M, so 3 Gm 4m Gm 4M l l 3 m M

17 Ch. 13 #46: Fou masses M ae located on each cone of a squae with side d. If d is educed by ½ what is the change in the potential enegy? 1 U total mm i G i< j ij j (many objects) mm mm mm mm mm mm G m m m m m m G( ) d d d d d d 4m m m G ( + ) (4 + ) d d d If d d ; U (4 ) m total + Double in magnitude! d

18 Ch. 13 #64: Two satellites of the same mass ae obiting the eath at a adius. But they ae going in opposite diections and collide. (a) Find the total mechanical enegy fo A and B and the eath befoe the collision. (b) If the collision is completely inelastic (mass m) what is the mechanical enegy ight afte the collision? (c) Just afte the collision is the weckage falling diectly towad Eath s cente o obiting aound the Eath? (a) Fom equation we see that the enegy of each satellite is GM E m - GM Em Total enegy: EA + EB - ( b ) The speed of the weckage is zeo--ight. GM E E ( m) (c) It will fall staight down!!!!

19 Ch. 13 # 93: Atiple sta system is shown in the pictue. The two obiting stas ae always at opposite ends of the diamete of the obit. Find the peiod? The magnitude of the net gavitational foce on one of the smalle stas (mass m) is GMm Gmm Gm m + M + 4 This supplies the centipetal foce needed fo the motion of the sta. Gm m v M m + 4 π v T T π 3/ [ ] G M + m /4