PHYSICS 231 Review problems for midterm 2


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1 PHYSICS 31 Revew problems for mdterm Topc 5: Energy and Work and Power Topc 6: Momentum and Collsons Topc 7: Oscllatons (sprng and pendulum) Topc 8: Rotatonal Moton
2 The nd exam wll be Wednesday October 8. The exam wll take place rght here n BPS We wll have assgned seatng, so show up early. You need to show up 10 mn early to guarantee your seat. If you are sck, you MUST have a note from a doctor on the doctor s letterhead or prescrpton pad. You CANNOT use cell phones durng the exam. You can (should!) brng two hand wrtten equaton sheets. up to two sdes each on a 8.5x11 sheet of paper
3 (A) v f v 0 at (B) x v t o 1 at (C) x v t f 1 at (D) 1 x v v ) t f ( 0 vt (E) x 1 ( v v a f 0 )
4 Energy Potental energy (PE) Energy assocated wth poston. Gravtatonal PE: mgh Energy assocated wth poston n grav. feld. Knetc energy KE: ½mv Energy assocated wth moton Elastc PE: ½kx energy stored n stretched/compressed sprng. Mechancal energy ME: ME = KE + PE Conservatve forces ME ME f = 0 Nonconservatve forces ME ME f = E nc
5 Work (the amount of energy transfer) W=(Fcos) x To measure how fast we transfer the energy we defne: Power: The rate of energy transfer Power = P = (Work/tme) = (W/t) (J/s=Watt) P = (Fcos) x/t = (Fcos) v average
6 Equatons of moton Lnear moton x(t) = x o + v o t+ ½at v(t) = v o + at Angular moton (t) = o + o t+ ½t (t) = o + t Angular moton = Rotatonal Moton!
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8 Momentum and Impulse F = m a Newton s nd law F = m v/t a=v/t F = m (v fnal v ntal )/t Defne p = mv F= (p fnal p ntal )/t p: momentum (kg m/s) F= p/t Defnton: p = Impulse CONSERVATION OF MOMENTUM (for a closed system) m 1 v 1f + m v f = m 1 v 1 + m v or p 1 + p = p 1f + p f addton of vectors example p = p 1f + p f
9 Inelastc Soluton Conservaton of momentum: m 1 v 1 + m v = m 1 v 1f + m v f After the collson m 1 and m form one new object wth mass M = m 1 + m and one velocty v f =v 1f = v f m 1 v 1 + m v = v f (m 1 + m ) v f = [m 1 v 1 + m v ] / (m 1 +m ) KE KE f = E nc Wth m = c m 1 v f = [v 1 + cv ] / (1+c) Specal case m = m 1 (c=1) v f = [v 1 + v ] /
10 Elastc Soluton v 1f = [(m 1 m ) v 1 + m v ] / (m 1 +m ) v f = [(m m 1 ) v + m 1 v 1 ] / (m 1 +m ) KE KE f = 0 For m = c m 1 v 1f = [(1c) v 1 + cv ] / (1+c) v f = [(c1) v + v 1 ] / (1+c) Specal case m = m 1 (c=1) v 1f = v v f = v 1
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12 Sprng Total ME at any dsplacement x: ½mv + ½kx Total ME at max. dsplacement A: ½kA Conservaton of ME: ½kA = ½mv + ½kx So: v=±[(a x )k/m] poston X velocty V acceleraton +A 0 ka/m 0 ±A(k/m) 0 A 0 ka/m
13 A x A velocty v A(k/m) A(k/m) x(t)=acos(t) =(k/m) v(t)=asn(t) tme (s) tme (s) ka/m a ka/m a(t)= Acos(t) tme (s)
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15 pendulum vs sprng the pendulum parameter sprng pendulum restorng force F F=kx F=(mg/L)s angular frequency =(k/m) =(g/l) k m mg / L m _ g L
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17 crcular moton T perod tme for one turn f angular velocty n (rad/s) t T angular acceleraton t r radus s r arclength wth n radans v s t r t r tangental velocty a v t r r t tangental acceleraton a c ( v / r) r centrpetal acceleraton
18 A 1 m/s B 10 m/s 6m 5m An Example The angular velocty of A s rad/s. 1) What s ts tangental velocty? If B s keepng pace wth A, ) What s ts angular velocty? 3) What s ts tangental velocty? 1) v = r = 6 = 1 m/s ) Must be the same: rad/s 3) v = r = 5 = 10 m/s
19 Torque: =Fd Center of Gravty: x CG m x m y CG m m y Translatonal equlbrum: F=ma=0 The center of gravty does not move! Rotatonal equlbrum: =0 The object does not rotate Mechancal equlbrum: F=ma=0 & =0 No movement!
20 Moment of nerta m r 1 r r 3 m 3 m 1 =I Moment of nerta I: I=(m r ) compare wth: F=ma The moment of nerta n rotatons s smlar to the mass n Newton s nd law. I = mr all mass s at the same dstance r
21 r v m Rotatonal knetc energy Consder a object rotatng wth constant velocty. Each pont moves wth velocty v. The total knetc energy s: 1 m v 1 m r 1 m r 1 I KE r =½I Conservaton of energy for rotatng object must nclude: rotatonal and translatonal knetc energy and potental energy [PE+KE t +KE r ] ntal = [PE+KE t +KE r ] fnal
22 Whch one goes fastest Use conservaton of mechancal energy. At top (at rest): [½mv +mgh+½i ] = mgh At bottom: [½mv + ½I ] = [½mv + ½(kmr )v /r ] (used: I=kmr and =v/r) Smplfy: [½mv +½kmv ]=(1+k)½mv object I k Combne: E top =E bottom mgh=(1+k)½mv v=[gh/(1+k)] No mass dependence!! A Cylndrcal Mr 1 shell B Sold cylnder (1/) mr 0.5 C Thn sphercal (/3) mr shell D Sold sphere (/5) mr 0.4
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24 Angular momentum 0 L then 0 f t L t L L I L t I I t I I Conservaton of angular momentum If the net torque equals zero, the angular momentum L does not change I = I f f
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