FALL TERM EXAM, PHYS 111, INTRODUCTORY PHYSICS I Saturday, 14 December 013, 1PM to 4 PM, AT 1003 NAME: STUDENT ID: INSTRUCTION 1. Thi exam booklet ha 14 page. Make ure none are miing. There i an equation heet on page 14. You may tear the equation heet off. 3. There are two part to the exam: Part I ha twelve multiple choice quetion (1 to 1), where you mut circle the one correct anwer (a,b,c,d,e). Rough work can be done on the backide of the heet oppoite the quetion page Part II include nine full-anwer quetion (13 to 1). Do all nine quetion. All work mut be done on the blank pace below the quetion. If you run out of pace, you may write on the backide of the heet oppoite the quetion page. 4. Calculator are allowed 1
PART I: MULTIPLE CHOICE QUESTIONS (quetion 1 to 1) For each quetion circle the one correct anwer (a,b,c,d or e). 1. (.5 point) Which of the following five acceleration veru time graph i correct for an object moving in a traight line at a contant velocity of 0 m/? a) I b) II c) III d) IV e) V ANSWER: A, ince a = 0 for contant v. (.5 point) The coordinate of an object i given a a function of time by x = 4t 3t 3, where x i in meter and t i in econd. It average acceleration over the interval from t = 0 to t = i: a) -13 m/ b) 4 m/ c) 10 m/ d) -10 m/ e) -8 m/ Ue v x = dx / dt = 8t 9t ; t 1 = 0, v 1x = 0 ; t 1 =, v 1x = 0m /. ( ) / ( t t 1 ) = 10m / ANSWER: D Finally a avg,x = v x v 1x 3. (.5 point) A cannon fire a projectile a hown. The dahed line how the trajectory in the abence of gravity; point MNOP correpond to the poition of the projectile at one econd interval. If g = 10 m/, the length X,Y,Z are: A) 5 m, 0 m, 45 m B) 0. m, 0.8 m, 1.8 m C) 10 m, 0 m, 30 m D) 5 m, 10 m, 15 m E) 10 m, 40 m, 90 m The projectile ha initial velocity v 0 = v 0 x î + v 0 y ĵ. Without gravity, the projectile would follow the dotted path. Hence, the difference between the dotted path and the actual path (i.e. X,Y,Z) i ( 1 / )gt A dicued X, Y, Z are given by( 1 / )gt. With g = 10 m/, t = 1, X = 5mi 1 ( ) = 5m ; t =, Y = 5mi ( ) = 0m ; t = 3, X = 5mi ( 3) = 45m. ANSWER: A 4. (.5 point) A horizontal force of 1 N puhe a 0.50-kg book againt a vertical wall. The book i initially at ret. If the coefficient of friction are µ = 0.60 and µ k = 0.50 which of the following i true. A) The block will tart moving and accelerate Friction f = Fµ B) If tarted moving downward, the block will accelerate C) The frictional force i 4.9 N D) The frictional force i 7. N F = 1N E) The normal force i 4.9 N From diagram maximum tatic friction i f max = 1N.6 = 7.N f max > Mg = 4.9N, book doe not move (tatic equilibrium). F y = f Mg = 0 " f = 4.9N ANSWER: C Mg = 4.9 N B o o k
5. (.5 point) Three block (A, B, C), each having the ame ma M, are connected by tring a hown. Block C i pulled to the right by a force that caue the entire ytem to accelerate. Neglecting friction, the force acting on block B i: a) 0 b) F / 3 c) F / d) F / 3 e) F Each block ha ma M, F = 3Ma, or a = F/3M. Ue Newton nd Law of block B, F 3 = Ma = M(F/3M) or F 3 = F/3. ANSWER: B 6. (.5 point) A toy cork gun contain a pring whoe pring contant i 10.0 N/m. The pring i compreed 5.00 cm and then ued to propel a 6.00-g cork. The cork, however, tick to the pring for 1.00 cm beyond it untretched length before eparation occur. The muzzle velocity of thi cork i: A) 1.0 m/ B) 1.41 m/ C).00 m/ D).04 m/ E) 4.00 m/ Ue Work-EnergyW el = K = (1 / )mv f " (1 / )mv i, m = 0.006 kg and v i = 0. v f = m/. ANSWER: C 7. (.5 point) A 700-N man jump out of a window into a fire 10 m below. The tretche m before bringing the man to ret and toing him back into the air. The maximum potential energy of the, compared to it untretched potential energy, i: A) 300 J B) 710 J C) 850 J D) 7000 J E) 8400 J Only conervative force act on the ytem. Taking zero gravitational potential energy to be 1m below the initial height (at window), the initial potential energy i mg(1m) = 8400J, where mg = 700 N. At the bottom, when the ha maximum tretch and the man i not moving, all of the gravitational potential energy i converted into the elatic potential energy of the. ANSWER: E 8. (.5 point) Two boy with mae of 40 kg and 60 kg tand on a horizontal frictionle urface holding the end of a light 10-m long rod. The boy pull themelve together along the rod. When they meet the 40-kg boy will have moved what ditance? a) 4 m b) 5 m c) 6 m d) 10 m e) need to know the force they exert 40kg x 40 = 0 x com x 60 = 0m Begin by calculating the center-of-ma (com), uing the diagram above. The initial poition of the 40 kg boy i arbitrary, and we et it to x 40 = 0. The initial poition of the 60 kg boy i x 60 = 10m. The center-of-ma i: x com = m 40x 40 + m 60 x 60 40kg 0 + 60kg 10m = = 6m. m 40 + m 60 40kg + 60kg Since they are on ice, there i no external force, and the center-of-ma mut remain at x com = 6m. The 40 kg boy mut move 6m. ANSWER C 9. (.5 point) A.5-kg tone i releaed from ret and fall toward Earth. After 4.0, the magnitude of it momentum i: a) 98 kg m/ b) 78 kg m/ c) 39 kg m/ d) 4 kg m/ e) 0 kg m/ Ue v y = 9.8m / ( ) 4 ( ), with k = 10 Ue W el = (1 / )kx f (1 / )kx i N/m, x i = 0.05 m and x f = 0.01 m, W el = 0.01 J 60 kg ( ) = 39.mi 1, p y = mv y = 98kgimi 1. ANSWER: A 3
10. (.5 point) At time t = 0, a wheel ha an angular diplacement of zero radian and an angular velocity of +18 rad/. The wheel ha a contant acceleration of -0.55 rad/. In thi ituation, the time t (after t = 0 ), at which the kiic energy of the wheel i twice the initial value, i cloet to: a) 46 b) 57 c) 69 d) 33 e) 79 Ue K rot,1 = 1 I 1 " K rot, = 1 I = K rot,1 = 1 I 1 or ( ) = " 1 = " 18rad / 1 I = 1 I 1 " = 1 = 1 ( ) = "5.46rad / Since initial angular velocity i poitive and the contant angular acceleration i negative, the angular velocity will decreae to zero and then begin to rotate in the other direction till it reache -5.46 rad/. Ue = 1 + "t t = 5.46 rad 1 = " 0.55rad / t = 79 ANSWER e. 18 rad 11. (.5 point) Which of the following tatement regarding angular momentum i correct? a) A particle moving in a traight line with contant peed necearily ha zero angular momentum. b) If the torque acting on a particle i zero about an arbitrary origin, then the angular momentum of the particle i alo zero about that origin. c) The angular momentum of a moving particle depend on the pecific origin with repect to which the angular momentum i calculated. d) If the peed of a particle i contant, then the angular momentum of the particle about any pecific origin mut alo be contant. e) Conider a pla moving in a circular orbit about a tar. Even if the pla i pinning it i not poible for it total angular momentum to be zero. Anwer c) 1. (.5 point) A light triangular plate OAB i in a horizontal plane. Three force, F 1 = 3 N, F = 1 N, and F 3 = 9 N, act on the plate, which i pivoted about a vertical axe through point O. In Figure below, the magnitude of the torque due to force F 1 about the axi through point O i cloet to: a) 1.1 N m b) 1.4 N m c) 0.90 N m d) 1.8 N m e) 1.6 N m Here ue = r " F that give 1 = rf 1 in60 0, where we can ee that the angle between F 1 and r. 1 = rf 1 in 60 0 = 0.6m " 3.0N " in60 0 = 1.56N m anwer e) PART II: FULL ANSWER QUESTIONS (quetion 13 to 1) 4
Do all nine quetion on the provided area below the quetion. Show all work. 13. (10 point) The poition of a particle moving in an xy plane i given by r = ( 5t 3 6t)î + ( 5 t 4 ) ĵ, with r in meter and t in econd. a) In unit-vector notation, calculate the poition, r, velocity, v, and acceleration, a, at t = 3. r ( t) = xî + yĵ = ( 5t 3 6t)î + ( 5 t 4 ) ĵ at t = 3, r ( 3) = ( 5( 3) 3 6( 3) )î + ( 5 ( 3 ) )4 ĵ = 117mî 157mĵ v t ( ) = v x î + v y ĵ = dx at t = 3, v 3 ( ) = 15 3 a( t) = a x î + a y ĵ = dv x î + dv y dt dt ( ) ( ) î + dy ĵ = d 5t 3 6t î + d 5 t 4 dt dt dt dt ( ( ) 6)î 8( 3)3 ĵ = 19 m î 16 m ĵ ( ) ĵ = d 15t 6 dt ĵ = ( 15t 6)î 8t 3 ĵ ( point) î + d ( 8t 3 ) ĵ = 30tî dt 4t ĵ ( point) at t = 3, a ( 3) = 30( 3)î 4 ( 3 ) ĵ = 90 m î 16 m ĵ b) What i the angle between the poitive direction of the +x axi and a line tangent to the particle' path at t = 3? Give your anwer in the range of (-180 o ; 180 o ). v ( 3) = ( 15( 3) 6)î 8( 3)3 ĵ = 19 m î 16 m ĵ v x > 0 and v y < 0 4 th "16m quadrant ( 90 0 < " < 0 ). = tan "1 19m = 59, = "59 (3 point) 14. (10 point) In the figure, a tone i projected at a cliff of height h with an initial peed of 46.0 m/ directed at an angle θ 0 = 59.0 above the horizontal. The tone trike at A, 5.85 after launching. a) Find the height, h, of the cliff. Initial velocity: x-component v 0 x = ( 46m / )co59 = 3.7m / ; v 0 y = ( 46m / )in59 = 39.4m / h = v 0 y t 1 gt " = 39.4 m % ' 5.85 " ( ) 1 9.8 m % ' 5.85 ( ) = 6.8m (.5 point) 5
b) Find the peed of the tone jut before impact at A. x-component v x = v 0 x = 3.7 m contant y-component v y = v 0 y gt = 39.4 m " 9.8 m % ' 5.85 peed v = 3.7 m " % + '17.93 m " % = 9.7 m ( ) = 17.93 m ( point) c) Find the maximum height reached, H, above the ground. What i the peed of the tone at the maximum height? At maximum height, v y = 0, ue v y = v 0 y g( y y 0 ) = 0 y y 0 = H, H = v 39.4 m 0 y g = " % 9.8 m " % = 79.m (1.5 point). At maximum height v y = 0, and v x = v 0 x = 3.7 m contant. The peed i 3.7m/. 15. (10 point) A loaded penguin led weighing 65 N ret on a plane inclined at angle θ = 3 to the horizontal (ee the figure). Between the led and the plane, the coefficient of tatic friction i 0.6, and the coefficient of kiic friction i 0.17. a) Draw free-body diagram that include all force on the led. Aume that the magnitude of the force F i enough to accelerate the penguin up the incline. f +y a F +x F N θ F g = mg ( point) θ b) Aume that the led i initially at ret, what i the minimum magnitude of the force F that will accelerate the box up the incline? 6
y-comp F y = F N mg co" = 0, mg = 65N, F N = ( 65N )co3 = 59.83N (1.5 point) Maximum tatic friction f max = F N µ = 59.83N 0.6 = 15.56N The minimum force F occur when F F g in 3 f max = 0 " F = 65N in3 + 15.56N = 41N (1.5 point) c) For the force calculated in part b, what i the acceleration of the box? Since F = 41 N i ufficient to accelerate the box + penguin, the friction i kiic. f k = F N µ k = ( 59.83N ) 0.17 = 10.17N 65N x-comp F x = F F g in 3 f k = ma, m = = 6.63kg ( point) 9.8mi a = 41N 65N in3 10.17N = 0.81 m 6.63kg 16. (10 point) A 6.6 kg brick move along an x axi. It acceleration a a function of it poition i hown in the figure below. a) Ue graphical integration to calculate the work done on the brick a it move from x = m to x = 8m. x f Ue W = F x dx " area under F x v. x, from x = x i to x f. For a v x plot, we write F = ma to x i x f get W = m a dx " m [area under a v. x], from x = x i to x f. With x i = m to x f = 8m. x i Area = = minu m 8m 0m 8m 0 m m ( point) area = 1 ( bae) ( height) 1 ( 0to8 bae) ( height) = 1 0to 8m " 0 m 1 m " 5 m m = 75 ( point) W = m area = 6.6kg 75 m = 495J. ( point) b) If the peed of the brick i. m/ when it i at x = m, calculate it peed when it i at x = 8m. Ue work-energy theorem W = K = (1 / )mv f " (1 / )mv i, v i =. m. ( point) 7
495J + 1 mv i = 1 mv f v f = " 495J 6.6kg +. m % ' ( = 1.44 m ( point) 17. (10 point) In the figure, projectile particle 1 i an alpha particle and target particle i an oxygen nucleu. The alpha particle i cattered at angle θ 1 = 69.0 and the oxygen nucleu recoil with peed 1.60 10 5 m/ and at angle θ = 5.0. Alpha Particle ma m 1 = 4u = 6.68 10 "7 kg Oxygen ma m = 16u =.67 10 "6 kg 1u atomic ma unit = 1.67 " 10 7 kg a) Uing conervation of momentum calculate the initial and final peed of the alpha particle. x-component m 1 v 1i = m 1 v 1 f co 1 + m v f co = 4v 1 f co69 + 16v f co5 ( point) m Uing v f = 1.6 10 5, v m 1i = ( 0.358)v 1 f + 3.94 10 5 [1] y-component 0 = m 1 v 1 f in 1 + m v f in = 4v 1 f in 69 + 16v f in 5 ( point) v 1 f = 4 in5 in69 v m f = 5.44 10 5 m Subtituting back into [1] give v 1i = ( 0.358) 5.44 10 5 + 3.94 m 105 v 1i = 5.89 10 5 m b) Calculate the change in kiic energy, K. I the colliion elatic? Briefly explain. m ( ) 5.89 10 5 Initial K i = 1 m v 1 1 = 1 6.68 10"7 kg K f = 1 m 1v 1 f + 1 m v f % ' ( = 1.16 10 "15 J m ( ) 5.44 10 5 Final = 1 6.68 10"7 kg % = 1.31 10 "15 J ' ( m ( ) 1.6 10 5 + 1.67 10"6 kg % ' ( 8
K = K f " K i = 1.5 10 "16 J Colliion i not elatic ince the energy i not conerved Uually for an inelatic colliion kiic energy i lot K < 0. The exception would be an exploion, which i not the cae here. I ue a quetion from an online problem that randomize θ 1 and θ. Only certain range will give K < 0, but the value of the angle in thi quetion give K > 0. 18. (10 point) In the figure below, a lawn roller in the form of a olid cylinder ( I = 1 MR ) i being pulled horizontally by a horizontal force, B, applied to an axle through the center of the roller, a hown in the ketch. The roller ha radiu R = 0.68 meter and ma M = 68 kg. The roller roll without lipping on a rough urface with µ = 0.3, and ha a linear acceleration of a =.5 m. After it roll 1.0 m it fall down an icy lope. Rough urface µ > 0 1.0 m icy mooth lope (no friction) h = 3.0 m f a) For the part when the roller i till on the flat urface. Draw a free-body diagram that include all force acting on the roller. The diagram mut how the direction of the linear and angular acceleration. You mut briefly explain the reaon for the direction of force of friction, a well a the reaon why you ue tatic and not kiic friction. Ue Newton law for tranlation and rotation to find the magnitude of the force B. y n R mg α a B +x Tranlation x-axi B f = ma [1] Rotation, torque = f R = I" From the diagram it i obviou that the lever arm of f about O i R. The force B, mg and n pa through O and contribute no torque. Uing, the no-lip condition, R = a, and I = 1 MR give f = 1 Ma [] ( point) Subtituting [] into [1] B 1 Ma = Ma " B = 3 Ma = 3 68kg ( ) %.5 m ' ( = 55N 9
Since we aume the wheel roll without lipping, the urface doe not move with repect to each other, and the force of friction i tatic, f. Friction f i the only force that can produce a torque on the wheel. Hence it mut be in the indicated direction to induce a cw angular acceleration. b) Uing the diagram from part a) determine the minimum coefficient of tatic friction, µ, in order for the roller to roll without lipping. Your anwer mut be maller than 0.3, which i the value for the urface of the problem. Uing equation [] f = 1 Ma = 1 ( 68kg).5 m " % = 85N Uing diagram of part a) the maximum value of tatic friction i f mgµ. Hence the minimum coefficient i found by equation the two value: 85N f = 85N = mgµ µ = 68kg " 9.8 m = 0.18 ( point) c) Find the linear and angular peed of the roller at the edge of the lope. Aume roller tart from ret without any rotational peed. Aume that initially, v 0 = 0 and 0 = 0. Linear v = ax v = ".5 m " 1m =.4 m Angular = v R =.4 m 0.68m = 3.3 rad d) Ue conervation of energy to find the linear (v) and angular peed (ω) when it reache the bottom of the lope h = 3.0 m below. Aume that after it reache the edge, there i no force acting on it (i.e. B = 0). The rotation energy will not change ince there no longer any friction, there will be no torque on the wheel to alter it rotational peed. Hence we ue conervation of energy, K 1 + U 1,grav = K + U,grav, with (1) at the top and () at the bottom, and the kiic energy include only the linear peed. Take zero PE to be at the bottom U,grav = 0, we get 1 mv + mgh = 1 mv F v F = v + gh =.4 m " % ' " 9.8 m % ' ( 3m) = 8.0 m the final angular peed i F = = 3.3 rad, ince the angular part i unaltered. ( point) 19. (10 point) In the figure below, box A and B are connected by a rope-pulley ytem. Box A move to the right, and the rope move over the pulley without lipping, and the pulley ha a clockwie angular velocity,. The data are hown in the figure. v T A T A ω ma of pulley i M P = 3.0 kg f m A = 10.0 kg T B I = 1 M PR olid cylindrical pulley 10
Coefficient of Friction µ k = 0.5 T B R = 0. m, radiu of cylinder m B =4.0kg a) Draw a free body diagram of the boxe howing all the force acting on it. Draw a free body diagram of the pulley howing all the force (including thoe due to the hinge) acting on it. The diagram hould include the direction of linear acceleration, a, of m A and m B, and the angular acceleration α of the pulley. The ytem will decelerate, o that the acceleration i left for A and up for box B. a n v a T B f k A T A B m A g m B g Now we draw a diagram for the wheel below T A α + Take counterclockwie (ccw) a poitive Note that the angular acceleration, α, i ccw. R ω Note that angular velocity ω i cw or negative. R T B F Hinge (blue arrow) i the force of the hinge on the wheel that ultimately upport the wheel. b) Ue Newton law for tranlation and rotation to find the linear acceleration (magnitude and direction) a of m A and m B, and the angular acceleration (magnitude and ene of rotation), α, of the pulley. The no-lip condition i ueful. Aume the rope doe not lip on the pulley. Note thatt A T B. Alo you may want to aume that Box B accelerate up. BOX A BOX B y-com F y = n m A g = 0 " n = m A g y-com F y = T B m B g = m B a [1] friction f k = m A gµ k x-axi F y = T A f k = m A a " T A m A gµ k = m A a [] We now calculate the torque on the wheel: = T A R " T B R = I. Uing I = 1 M PR, and the no-lip condition a = R. Now do [] [1] T A m A gµ k T B + m B g = m A a m B a, which give T A T B + m A a + m B a = m A gµ k m B g. Subtituting [3] T A T B = 1 Ma, 1 M a + m a + m a = m gµ m g P A B A k B " 1 M + m + m % P A B ' a = m A gµ k m B g 11
m Solving for a, a = A gµ k m B g " 1 M % P + m A + m B ' 10kg ( 9.8 m = ( 0.5 4kg ( 9.8 m 1 3kg + 10.0kg + 4.0kg a = 0.63 m, and uing the no lip condition = a R = 0.63 m 0.m = 3. rad 0. (10 point) In figure below, a carouel ha a radiu of 3.0 m and a moment of inertia of I C = 8000kg m, for rotation about axi perpendicular to the it center. The carouel i rotating unpowered and without friction with an angular velocity of 1. rad/. An 80-kg man run with a velocity of 5.0 m/, on a line tangent to the rim of the carouel, overtaking it. The man run onto the carouel and grab hold of a pole on the rim. +y +x Direction perpendicular to x-y plane indicate +z out of the page indicate z into page a) Before the colliion, what i the magnitude of the angular momentum of the rotating carouel, L C, with repect to the center of the carouel? What i the direction of L C? Direction (+x, +y, +z, -x, -y, -z) are a indicated in the above figure. Direction (+x, +y, +z, -x, -y, -z) are a indicated in the above figure. The magnitude of angular momentum i L C = I c c = 8000.0kgim " 1. rad = 9600 kgim The rotation i ccw, o uing the right hand rule, the direction i out of the page in the +z direction. (3.5 point) b) Before the colliion, what i the magnitude of the angular moment of the running 80- kg man, L M, with repect to the center of the carouel? What i the direction of L M? center of carouel R = 3.0 m p = mv = (80 kg) (5.0 m/) = 400 kg-m/ The angular momentum of a particle i defined a L M = R p, where R i the poition vector from the center of the carouel to the particle when it reache the edge of the carouel, and p i the momentum. Since they are perpendicular the magnitude i imply L M = Rp = 3.00m 400 kgim = 100 kgim 1
Uing the right hand rule on the diagram above it i eay to that the direction i out of the page in the +z direction. (3.5 point) c) After the colliion when the man i on the carouel, what i the magnitude of the final angular velocity of the carouel (with the man on it), fc? What i the direction of the final angular velocity fc? Note: I total = I C + mr Firt find the moment of inertia of the carouel with the man on it about the center I total = I C + mr = 8000kgim + 80kg ( 3.0m) = 870kg m. The conervation of angular momentum (only the z-direction i relevant) give: 9600 kgim + 100 kgim L c + L M = I tot fc " fc = 870kgim Direction i +z. (3 point) = 1.4 rad 1. (10 point) A diving board of length 3.00 m i upported at a point (P) 1.00 m from the end, and a diver weighing 50 N tand at the free end. The diving board i of uniform cro ection and weigh 300 N. a) Find the force of upport at point P Hint: Do the torque part firt Net torque about point O mut be zero. Take clockwie to poitive. Torque due to diver (poitive): D = WDlD Torque due to board (poitive), weight aumed to be at it center of ma: B = W l B CM 13
Torque due to upport (negative): " Torque at end (zero): F ( 0 ) = 0 S = F l S S = E = " D + " B + " S + " E = WDlD + WBlCM FS l ", zero moment arm. Net Torque: " = 0 F S l = W D l D + W B l CM ( )( 3.00m) + ( 300N )( 1.50m) F S = W l + W l D D B CM = 50N = 010N l S 1.00m Hence the upward force at the upport i F S = 010N (5 point) b) Find the force of upport at point L, at the left-hand end. Since the ytem i at equilibrium the force (of all vertical component) i zero: Taking up to be poitive F = F F W W = 0 " F = F W W y S Hence the downward force at the end i F E = 010N 300N 50N = 1190N (5 point) E B D E S B D 14
Ueful Equation Kinematic x = x 0 + v 0 x t + (1 / )a x t, v x = v 0 x + a x t, v x = v 0 x + a x ( x x 0 ) ; v x = dx / dt, a x = dv x / dt ; v = v x î + v y ĵ + v z ˆk ; a = ax î + a y ĵ + a z ˆk ; average peed avg = (total ditance)/(total ( ) / ( t t 1 ), average acceleration (x-com) time); average velocity (x-com) v avg,x = x x 1 a avg,x = ( v x v 1x ) / ( t t 1 ). Newton Law F = F i = 0 (Object in equilibrium); F = m a (Nonzero force); Weight: F g = mg, g = 9.8m / ; Centripetal acceleration a rad = v r ; Friction f µ F N, f k = µ k F N. Hooke Law F x = kx. Work and Energy W = F d = ( F co )d = F d ; W = K = (1 / )mv f " (1 / )mv i (valid if W i the or total work done on the object);w grav = mg y f y i ( ) (elatic work) W el = (1 / )kx f (1 / )kx i ( ) (gravitational work), Conervation of Mechanical Energy (only conervative force are preent) E mech = U + K W = "U = U U 1 ( ) = "K = K K 1,U 1 + K 1 = U + K,U grav = mgy,u el = (1 / )kx Non-Conervative Force W external = E mech + E th (W ext work done by external force, and we et E int = 0 ), where E th = f k d (thermal energy or negative work done by friction). Uing E mech = U + K = U f " U i ( ) + ( K f " K i ), U f + K f = U i + K i + W ext f k d x f Work due to variable force 1D: W = F x dx " area under F x v. x, from x = x i to x f x i POWER: average P avg = W / t = Fd / t = Fv avg ; intantaneou P = F v = Fvco Momentum: P = m v, J t = Fdt = F av t " t 1 t 1 ( ), J = P = P " P 1. Newton Law in Term of Momentum F = d p / dt. For F = 0, d p / dt = 0 give momentum conervation: P contant. Rotational Kinematic Equation: avg = " " 1 ( ) / ( t t 1 ), avg = (" " 1 ) / ( t t 1 ) ( ) For z = contant, = 0 + "t, = 0 + " 0 t + (1 / )t, = 0 + " 0 Linear and angular variable: = r, v = r, a tan = R (tangential), a rad = v / r = r (radial) Moment of Inertia and Rotational Kiic Energy I = m i r i, K rot = (1/ )I. Center of Ma (COM) r com = m i ri / m i. Torque and Newton Law of Rotating Body: rigid body = Fr ", = " ext i = I, r -moment arm about axi; point = r " F about origin O. Combined Rotation and Tranlation of a Rigid Body K = (1/ )Mv com N i=1 + (1/ )I com, F = M a com, = I com ". Rolling without lipping = R, vcom = R, a com = R. Angular Momentum L = I (olid object) where I i the moment of inertia about the axi of rotation. L = r p " L = mvr, valid for point particle about an origin O. Newton Second Law of rigid body in term of angular momentum = " i ext ( ) = 0 and angular momentum i conerved, L contant. = ( d L / dt). For = 0, d L / dt Equilibrium condition F ezt = F i = 0 about all object, ext = " i = 0 about any point. 15