PHYSICS 231 Lecture 18: equilibrium & revision


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1 PHYSICS 231 Lecture 18: equlbrum & revson Remco Zegers Walkn hour: Thursday 11:3013:30 am Helproom 1
2 gravtaton Only f an object s near the surface of earth one can use: F gravty =mg wth g=9.81 m/s 2 In all other cases: F gravty = GM object M planet /r 2 wth G=6.67E11 Nm 2 /kg 2 Ths wll lead to F=mg but g not equal to 9.8 m/s 2 (see Prevous lecture!) If an object s orbtng the planet: F gravty =ma c =mv 2 /r=mω 2 r wth v: lnear velocty ω=angular vel. Our solar system! So: GM object M planet /r 2 = mv 2 /r=mω 2 r Kepler s 3 rd law: T 2 =K s r 3 K s =2.97E19 s 2 /m 3 r: radus of planet T: perod(tme to make one rotaton) of planet 2
3 Prevously Torque: τ=fd Center of Gravty: x CG = m x m y CG = m m y Translatonal equlbrum: ΣF=ma=0 The center of gravty does not move! Rotatonal equlbrum: Στ=0 The object does not rotate Mechancal equlbrum: ΣF=ma=0 & Στ=0 No movement! 3
4 examples: A lot more n the book! Where s the center of gravty? x y CG CG = = m x m m m y cos(53 ) cos( 53 = sn(53 ) sn( 53 = ) = ) = =
5 Weght of board: w What s the tenson n each of the wres (n terms of w)? T 1 T 2 Translatonal equlbrum ΣF=ma=0 T 1 +T 2 w=0 so T 1 =wt 2 Rotatonal equlbrum 0 w Στ=0 T *w+0.75*T 2 =0 T 2 =0.5/0.75*w=2/3w T 1 =1/3w T 2 =2/3w 5
6 µ s n n w (x=0,y=0) w µ s =0.5 coef of frcton between the wall and the 4.0 meter bar (weght w). What s the mnmum x where you can hang a weght w for whch the bar does not slde? Translatonal equlbrum (Hor.) T Ty ΣF x =ma=0 nt x =ntcos37 o =0 so n=tcos37 o Translatonal equlbrum (vert.) T x ΣF y =ma=0 µ s nww+t y =0 µ s n2w+tsn37 o =0 µ s Tcos37 02w+Tsn37 0 =0 1.00T=2w Rotatonal equlbrum: Στ=0 xw+2w4tsn37 0 =0 so w(x+24.8)=0 x=2.8 m 6
7 Tps for study Look through the lecture sheets and pck out the summares to get a good overvew Make an overvew for yourself (about 1 Letter sze paper) Read the chapters n the book to make sure that your overvew contans all the man ssues. Study the examples gven n the lectures Study the problems n LONCAPA Study the workedout examples n the book Practce the prevous mdterm exams. Practce problems from the book 7
8 Revson: chapter 5 Work: W=Fcos(θ) x Energy transfer Power: P=W/ t Rate of energy transfer Potental energy (PE) Energy assocated wth poston. Gravtatonal PE: mgh Energy assocated wth poston n grav. feld. PE stored n a sprng: 1/2kx 2 x s the compresson of the sprng k s the sprng constant Knetc energy KE: 1/2mv 2 Energy assocated wth moton Conservatve force: Work done does not depend on path Nonconservatve force: Work done does depend on path Mechancal energy ME: ME=KE+PE Conserved f only conservatve forces are present KE +PE =KE f +PE f Not conserved n the presence of nonconservatve forces (KE +PE )(KE f +PE f )=W nc 8
9 m=1 kg example A pendulum s pushed wth ntal velocty 0.1 m/s from a heght of 1 cm. How far does t compress the sprng? (assume m does not rse sgnfcantly after httng the sprng) 1 cm k=100 N/m Conservaton of ME: (mgh+1/2mv 2 +1/2kx 2 ) ntal = (mgh+1/2mv 2 +1/2kx 2 ) fnal 1*9.8* *1* = *100*x 2 so x=0.045 m 9
10 Savng electrcty A smart student decdes to save energy by connectng hs exercse treadmll to hs laptop battery. If t takes 70 J to move the belt on the treadmll by 1 meter and 50% of the generated energy s stored n the battery, how far must the student run to use hs 100 W laptop for free for 2 hours? Work done by student: W=70*d J Energy gven to the battery 0.5W=35*d J 100 W laptop for 2 hours: 100 J/s*3600*2 s=7.2e+5 J 720 KJ 7.2E+5=35*d so d=(7.2e+5)/35=2.1e+4 m =21km!!! 10
11 Nonconservatve vs conservatve case h 45 0 A block of 1 kg s pushed up a 45 o slope wth an ntal velocty of 10 m/s. How hgh does the block go f: a) there s no frcton b) f the coeffcent of knetc frcton s 0.5. A) Conservaton of ME: (mgh+1/2mv 2 ) ntal = (mgh+1/2mv 2 ) fnal *1*10 2 =1*9.8*h+0. So h=5.1 m B) Energy s lost to frcton: (mgh+1/2mv 2 ) ntal = (mgh+1/2mv 2 ) fnal +W frcton [W=F x=µn x=µmgcos(45 o )h/sn(45 o )=0.5*1*9.8*h=4.9h] *1*10 2 =1*9.8*h h so h=3.4 m 11
12 Chapter 6 Momentum p=mv F= p/ t Impulse (the change n momentum) p= F t Inelastc collsons Elastc collsons Momentum s conserved Some energy s lost n the collson: KE not conserved Perfectly nelastc: the objects stck together. Conservaton of momentum: m 1 v 1 +m 2 v 2 =(m 1 +m 2 )v f Momentum s conserved No energy s lost n the collson: KE conserved Conservaton of momentum: m 1 v 1 +m 2 v 2 =m 1 v 1f +m 2 v 2f Conservaton of KE: ½m 1 v 12 +½m 2 v 22 =½m 1 v 1f2 +½m 2 v 2 2f (v 1 v 2 )=(v 2f v 1f ) 12
13 Inelastc collson 5 kg 0.1 kg V m/s 5 m/s An excellent, but somewhat desperate sharp shooter tres to stop a cannon ball amed drectly at hm by shootng a bullet from hs gun aganst t. Wth what velocty does he need to shoot the bullet to stop the cannon ball assumng that the bullet gets stuck n the ball? How much energy s released? Inelastc: only conservaton of momentum. m 1 v 1 +m 2 v 2 =(m 1 +m 2 )v f so 0.1V5*5=0 v=25/0.1=250 m/s Change n knetc energy: Before: ½m 1 v 12 +½m 2 v 22 = = J After: 0 J Release: J 13
14 h=100 m 1 2 Elastc collson Two balls (m 1 =1 kg, m 2 =2 kg) are released on a slope and collde n the valley. How far does each go back up? Step 1: calculate ther veloctes just before the collson: ball 1: cons. of ME: mgh=0.5mv 2 9.8*100=0.5 v 2 v=44. m/s ball 2: cons. of ME: 2*9.8*100=0.5*2*v 2 v=44. m/s Step 2: Collson, use cons. of P and KE (smplfed). m 1 v 1 +m 2 v 2 =m 1 v 1f +m 2 v 2f so 4488=v 1f +2v 2f (v 1 v 2 )=(v 2f v 1f ) so 88=v 2f v 3v 1f 2f =44 v 2f =15 v 1f =73 Step 3: Back up the ramp: cons. of ME: ball 1: 0.5mv 2 =mgh 0.5*73 2 =9.8h h=272 m ball 2: 0.5mv 2 =mgh 0.5*15 2 =9.8h h=11.5 m 14
15 Bouncng ball A 0.5 kg ball s dropped to the floor from a heght of 2 m. If t bounces back to a heght of 1.8 m, what s the magntude of ts change n momentum? Some energy s lost n the bounce. Just before t hts the ground, ts velocty s: (use conservaton of ME) mgh=1/2mv 2 so v= (2gh)= (2*9.8*2)= 6.26 m/s p=m[6.26(5.93)]=0.5*12.2=6.1 kgm/s After the bounce t goes back up 1.8 m. Just after t bounces back t velocty s: (use conservaton of ME) mgh=1/2mv 2 so v= (2gh)= (2*9.8*1.8)=5.93 m/s Must be negatve!! So m/s 15
16 θ f ϖ = t ω = f lm t 0 θ t θ t = θ t Chapter 7 Average angular velocty (rad/s) Instantaneous Angular velocty ω f α = t α = f lm t 0 ω t ω t = ω t Average angular acceleraton (rad/s 2 ) Instantaneous angular acceleraton 2π rad π/360 rad 1 rad 360/2π deg Be aware that sometmes rev/s or rev/mn s asked 16
17 Angular vs lnear/tangental ω = v r lnear r α = a r angular See e.g. the bke example (lecture 16) 17
18 Angular moton Rotatonal moton θ(t)= θ(0)+ω(0)t+½αt 2 ω(t)= ω(0)+αt Centrpetal acceleraton a c =v 2 /r drected to the center of the crcular moton Also v=ωr, so a c =ω 2 r ΣF to center =ma c =mv 2 /r Ths acceleraton s caused by known force (gravtaton, frcton, tenson ) Make sure you understand how to use Kepler s 3rd law and the general defnton of gravtatonal PE. 18
19 A ball of mass 2 kg s attached to a strng of 1m and whrled around a smooth horzontal table. If the tenson n the strng exceeds 200 N, t wll break. What wll be the speed at the moment the strng breaks? ΣF to center =ma c =mv 2 /r T=2*v 2 /1=200 N v=10 m/s Whrlng ball v 19
20 Consder... a chld playng on a swng. As she reaches the lowest pont n her swng, whch of the followng s true? A) The tenson n the robe s equal to her weght B) The tenson n the robe s equal to her mass tmes her acceleraton C) Her acceleraton s downward and equal to g (9.8 m/s 2 ) D) Her acceleraton s zero E) Her acceleraton s equal to her velocty squared dvded by the length of the swng. ΣF to center =Tmg=ma c =mv 2 /L Lowest pont, so no lnear acceleraton!!! 20
21 Two objects... Are rotatng. One starts wth an ntal lnear velocty of 1 m/s and rotates wth a radus of 2 m. The other starts from rest and undergoes a constant angular acceleraton over a crcle wth radus 3 m. What should ts angular acceleraton be, so that t overtakes (for the frst tme) the frst object after 10 revolutons? ω 1 =v 1 /r 1 =1/2=0.5 rad/s θ 1 (t)= θ 1 (0)+ω 1 (0)t+½α 1 t 2 =0.5t θ 1 (t)=10*2π=0.5t so t=40π θ 2 (t)= θ 2 (0)+ω 2 (0)t+½α 2 t 2 =½α 2 t 2 20π=½α 2 (40π) 2 so α 2 =8E03 rad/s 21
22 Tcosθ T θ θ Tsnθ mg Concal moton If the mass of the swngng object s 1 kg, and θ=30 o what should the velocty of the object be so that does not snk or rse? The length of the robe s 2 m. Vertcal drecton: ΣF=ma Tcosθmg=0 So T=mg/cosθ=1*9.8/0.866=11.3 N Horzontal drecton: ΣF=ma c Tsnθ=ma c 11.3*0.5=ma c =1v 2 /r v 2 =11.3*0.5*2 so v=3.4 m/s 22
23 Chapter 8 Summary: see begnnng of ths lecture! 23
24 ?N 30 o Openng a hatch door. A person s tryng to open a 1 meter, 20 kg, trap door by pullng a rope attached to ts non rotatng end at an angle of 30 o. Wth what force should he at least pull? 1 m Center of gravty of the door: halfway the door s length: 0.5m Στ=m door gd CG +F pull,perpendcular l 0=20*9.8*0.5+F pull sn(30 o )*1 F pull =196 N 24
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