FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Saturday, 14 December 2013, 1PM to 4 PM, AT 1003
|
|
- Alan Butler
- 5 years ago
- Views:
Transcription
1 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
2 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
3 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 = 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 kg 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
4 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 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 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
5 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 " ( ) m % ' 5.85 ( ) = 6.8m (.5 point) 5
6 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 ( ) = 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 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
7 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 in N = 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 in N = 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
8 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 m/ and at angle θ = 5.0. Alpha Particle ma m 1 = 4u = "7 kg Oxygen ma m = 16u = "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 co v f co5 ( point) m Uing v f = , v m 1i = ( 0.358)v 1 f [1] y-component 0 = m 1 v 1 f in 1 + m v f in = 4v 1 f in v f in 5 ( point) v 1 f = 4 in5 in69 v m f = m Subtituting back into [1] give v 1i = ( 0.358) m 105 v 1i = m b) Calculate the change in kiic energy, K. I the colliion elatic? Briefly explain. m ( ) Initial K i = 1 m v 1 1 = "7 kg K f = 1 m 1v 1 f + 1 m v f % ' ( = "15 J m ( ) Final = "7 kg % = "15 J ' ( m ( ) "6 kg % ' ( 8
9 K = K f " K i = "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 > (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 > 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
10 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
11 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
12 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 kg + 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 = kgim " 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
13 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 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
14 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
15 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
FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Saturday, 14 December 2013, 1PM to 4 PM, AT 1003
FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Saturday, 14 December 2013, 1PM to 4 PM, AT 1003 NAME: STUDENT ID: INSTRUCTION 1. This exam booklet has 14 pages. Make sure none are missing 2. There is
More informationt α z t sin60 0, where you should be able to deduce that the angle between! r and! F 1
PART III Problem Problem1 A computer dik tart rotating from ret at contant angular acceleration. If it take 0.750 to complete it econd revolution: a) How long doe it take to complete the firt complete
More informationFALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Thursday, 11 December 2014, 6 PM to 9 PM, Field House Gym
FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Thursday, 11 December 2014, 6 PM to 9 PM, Field House Gym NAME: STUDENT ID: INSTRUCTION 1. This exam booklet has 13 pages. Make sure none are missing 2.
More informationFALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Monday, 14 December 2015, 6 PM to 9 PM, Field House Gym
FALL TERM EXAM, PHYS 111, INTRODUCTORY PHYSICS I Monday, 14 December 015, 6 PM to 9 PM, Field House Gym NAME: STUDENT ID: INSTRUCTION 1. This exam booklet has 13 pages. Make sure none are missing. There
More informationPhysics Exam 3 Formulas
Phyic 10411 Exam III November 20, 2009 INSTRUCTIONS: Write your NAME on the front of the blue exam booklet. The exam i cloed book, and you may have only pen/pencil and a calculator (no tored equation or
More informationKEY. D. 1.3 kg m. Solution: Using conservation of energy on the swing, mg( h) = 1 2 mv2 v = 2mg( h)
Phy 5 - Fall 206 Extra credit review eion - Verion A KEY Thi i an extra credit review eion. t will be worth 30 point of extra credit. Dicu and work on the problem with your group. You may ue your text
More informationDYNAMICS OF ROTATIONAL MOTION
DYNAMICS OF ROTATIONAL MOTION 10 10.9. IDENTIFY: Apply I. rad/rev SET UP: 0 0. (400 rev/min) 419 rad/ 60 /min EXECUTE: 0 419 rad/ I I (0 kg m ) 11 N m. t 800 EVALUATE: In I, mut be in rad/. 10.. IDENTIFY:
More informationProf. Dr. Ibraheem Nasser Examples_6 October 13, Review (Chapter 6)
Prof. Dr. Ibraheem Naer Example_6 October 13, 017 Review (Chapter 6) cceleration of a loc againt Friction (1) cceleration of a bloc on horizontal urface When body i moving under application of force P,
More informationEF 151 Final Exam, Spring, 2009 Page 2 of 10. EF 151 Final Exam, Spring, 2009 Page 1 of 10. Name: Section: sina ( ) ( )( ) 2. a b c = = cosc.
EF 5 Final Exam, Spring, 9 Page of EF 5 Final Exam, Spring, 9 Page of Name: Section: Guideline: Aume 3 ignificant figure for all given number unle otherwie tated Show all of your work no work, no credit
More informationPHYSICS 221, FALL 2011 EXAM #2 SOLUTIONS WEDNESDAY, NOVEMBER 2, 2011
PHYSICS 1, FALL 011 EXAM SOLUTIONS WEDNESDAY, NOVEMBER, 011 Note: The unit vectors in the +x, +y, and +z directions of a right-handed Cartesian coordinate system are î, ĵ, and ˆk, respectively. In this
More informationME 141. Engineering Mechanics
ME 141 Engineering Mechanic Lecture 14: Plane motion of rigid bodie: Force and acceleration Ahmad Shahedi Shakil Lecturer, Dept. of Mechanical Engg, BUET E-mail: hakil@me.buet.ac.bd, hakil6791@gmail.com
More informationMath 273 Solutions to Review Problems for Exam 1
Math 7 Solution to Review Problem for Exam True or Fale? Circle ONE anwer for each Hint: For effective tudy, explain why if true and give a counterexample if fale (a) T or F : If a b and b c, then a c
More informationPhysics 2. Angular Momentum. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB
Phyic Angular Momentum For Campu earning Angular Momentum Thi i the rotational equivalent of linear momentum. t quantifie the momentum of a rotating object, or ytem of object. To get the angular momentum,
More informationPhysics 6A. Angular Momentum. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB
Phyic 6A Angular Momentum For Campu earning Angular Momentum Thi i the rotational equivalent of linear momentum. t quantifie the momentum of a rotating object, or ytem of object. f we imply tranlate the
More informationSeat: PHYS 1500 (Fall 2006) Exam #2, V1. After : p y = m 1 v 1y + m 2 v 2y = 20 kg m/s + 2 kg v 2y. v 2x = 1 m/s v 2y = 9 m/s (V 1)
Seat: PHYS 1500 (Fall 006) Exa #, V1 Nae: 5 pt 1. Two object are oving horizontally with no external force on the. The 1 kg object ove to the right with a peed of 1 /. The kg object ove to the left with
More information3pt3pt 3pt3pt0pt 1.5pt3pt3pt Honors Physics Impulse-Momentum Theorem. Name: Answer Key Mr. Leonard
3pt3pt 3pt3pt0pt 1.5pt3pt3pt Honor Phyic Impule-Momentum Theorem Spring, 2017 Intruction: Complete the following workheet. Show all of you work. Name: Anwer Key Mr. Leonard 1. A 0.500 kg ball i dropped
More informationa = f s,max /m = s g. 4. We first analyze the forces on the pig of mass m. The incline angle is.
Chapter 6 1. The greatet deceleration (of magnitude a) i provided by the maximum friction force (Eq. 6-1, with = mg in thi cae). Uing ewton econd law, we find a = f,max /m = g. Eq. -16 then give the hortet
More informationPHYSICS LAB Experiment 5 Fall 2004 FRICTION
FRICTION In thi experiment we will meaure the effect of friction on the motion of a body in contact with a particular urface. When a body lide or roll over another, it motion i oppoed by the force of friction
More informationRolling, Torque & Angular Momentum
PHYS 101 Previous Exam Problems CHAPTER 11 Rolling, Torque & Angular Momentum Rolling motion Torque Angular momentum Conservation of angular momentum 1. A uniform hoop (ring) is rolling smoothly from the
More informationPhysics Sp Exam #4 Name:
Phyic 160-0 Sp. 017 Ea #4 Nae: 1) A coputer hard dik tart ro ret. It peed up with contant angular acceleration until it ha an angular peed o 700 rp. I it coplete 150 revolution while peeding up, what i
More information= v 0 x. / t = 1.75m / s 2.25s = 0.778m / s 2 nd law taking left as positive. net. F x ! F
Multiple choice Problem 1 A 5.-N bos sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at one instant the bos sliding to the right at 1.75 m/s
More informationDiscover the answer to this question in this chapter.
Erwan, whoe ma i 65 kg, goe Bungee jumping. He ha been in free-fall for 0 m when the bungee rope begin to tretch. hat will the maximum tretching of the rope be if the rope act like a pring with a 100 N/m
More informationPHYSICS 211 MIDTERM II 12 May 2004
PHYSIS IDTER II ay 004 Exa i cloed boo, cloed note. Ue only your forula heet. Write all wor and anwer in exa boolet. The bac of page will not be graded unle you o requet on the front of the page. Show
More informationCHAPTER VII FRICTION
CHAPTER VII FRICTION 1- The block brake conit of a pin-connected lever and friction block at B. The coefficient of tatic friction between the wheel and the lever i and a torque of i applied to the wheel.
More informationPHYSICSBOWL March 29 April 14, 2017
PHYSICSBOWL 2017 March 29 April 14, 2017 40 QUESTIONS 45 MINUTES The ponor of the 2017 PhyicBowl, including the American Aociation of Phyic Teacher, are providing ome of the prize to recognize outtanding
More informationRotational Kinematics and Dynamics. UCVTS AIT Physics
Rotational Kinematics and Dynamics UCVTS AIT Physics Angular Position Axis of rotation is the center of the disc Choose a fixed reference line Point P is at a fixed distance r from the origin Angular Position,
More informationKinematics (special case) Dynamics gravity, tension, elastic, normal, friction. Energy: kinetic, potential gravity, spring + work (friction)
Kinematics (special case) a = constant 1D motion 2D projectile Uniform circular Dynamics gravity, tension, elastic, normal, friction Motion with a = constant Newton s Laws F = m a F 12 = F 21 Time & Position
More informationPhysics 218: Exam 1. Class of 2:20pm. February 14th, You have the full class period to complete the exam.
Phyic 218: Exam 1 Cla of 2:20pm February 14th, 2012. Rule of the exam: 1. You have the full cla period to complete the exam. 2. Formulae are provided on the lat page. You may NOT ue any other formula heet.
More informationHalliday/Resnick/Walker 7e Chapter 6
HRW 7e Chapter 6 Page of Halliday/Renick/Walker 7e Chapter 6 3. We do not conider the poibility that the bureau might tip, and treat thi a a purely horizontal motion problem (with the peron puh F in the
More informationAssessment Schedule 2017 Scholarship Physics (93103)
Scholarhip Phyic (93103) 201 page 1 of 5 Aement Schedule 201 Scholarhip Phyic (93103) Evidence Statement Q Evidence 1-4 mark 5-6 mark -8 mark ONE (a)(i) Due to the motion of the ource, there are compreion
More informationChapter 12: Rotation of Rigid Bodies. Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics
Chapter 1: Rotation of Rigid Bodies Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics Translational vs Rotational / / 1/ m x v dx dt a dv dt F ma p mv KE mv Work Fd P Fv / / 1/ I
More informationName: Answer Key Date: Regents Physics. Energy
Nae: Anwer Key Date: Regent Phyic Tet # 9 Review Energy 1. Ue GUESS ethod and indicate all vector direction.. Ter to know: work, power, energy, conervation of energy, work-energy theore, elatic potential
More informationPHYSICS 221, FALL 2010 EXAM #1 Solutions WEDNESDAY, SEPTEMBER 29, 2010
PHYSICS 1, FALL 010 EXAM 1 Solutions WEDNESDAY, SEPTEMBER 9, 010 Note: The unit vectors in the +x, +y, and +z directions of a right-handed Cartesian coordinate system are î, ĵ, and ˆk, respectively. In
More informationAP Physics Charge Wrap up
AP Phyic Charge Wrap up Quite a few complicated euation for you to play with in thi unit. Here them babie i: F 1 4 0 1 r Thi i good old Coulomb law. You ue it to calculate the force exerted 1 by two charge
More information= 16.7 m. Using constant acceleration kinematics then yields a = v v E The expression for the resistance of a resistor is given as R = ρl 4 )
016 PhyicBowl Solution # An # An # An # An # An 1 C 11 C 1 B 31 E 41 D A 1 B E 3 D 4 B 3 D 13 A 3 C 33 B 43 C 4 D 14 E 4 B 34 C 44 E 5 B 15 B 5 A 35 A 45 D 6 D 16 C 6 C 36 B 46 A 7 E 17 A 7 D 37 E 47 C
More informations much time does it take for the dog to run a distance of 10.0m
ATTENTION: All Diviion I tudent, START HERE. All Diviion II tudent kip the firt 0 quetion, begin on #.. Of the following, which quantity i a vector? Energy (B) Ma Average peed (D) Temperature (E) Linear
More informationWebreview Torque and Rotation Practice Test
Please do not write on test. ID A Webreview - 8.2 Torque and Rotation Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A 0.30-m-radius automobile
More informationPHYSICS 221, FALL 2009 EXAM #1 SOLUTIONS WEDNESDAY, SEPTEMBER 30, 2009
PHYSICS 221, FALL 2009 EXAM #1 SOLUTIONS WEDNESDAY, SEPTEMBER 30, 2009 Note: The unit vectors in the +x, +y, and +z directions of a right-handed Cartesian coordinate system are î, ĵ, and ˆk, respectively.
More informationChapter 9-10 Test Review
Chapter 9-10 Test Review Chapter Summary 9.2. The Second Condition for Equilibrium Explain torque and the factors on which it depends. Describe the role of torque in rotational mechanics. 10.1. Angular
More informationPhysics 1 Second Midterm Exam (AM) 2/25/2010
Physics Second Midterm Eam (AM) /5/00. (This problem is worth 40 points.) A roller coaster car of m travels around a vertical loop of radius R. There is no friction and no air resistance. At the top of
More informationPhysics UCSB TR 2:00-3:15 lecture Final Exam Wednesday 3/17/2010
Physics @ UCSB TR :00-3:5 lecture Final Eam Wednesday 3/7/00 Print your last name: Print your first name: Print your perm no.: INSTRUCTIONS: DO NOT START THE EXAM until you are given instructions to do
More informationSolution Only gravity is doing work. Since gravity is a conservative force mechanical energy is conserved:
8) roller coaster starts with a speed of 8.0 m/s at a point 45 m above the bottom of a dip (see figure). Neglecting friction, what will be the speed of the roller coaster at the top of the next slope,
More informationME 375 EXAM #1 Tuesday February 21, 2006
ME 375 EXAM #1 Tueday February 1, 006 Diviion Adam 11:30 / Savran :30 (circle one) Name Intruction (1) Thi i a cloed book examination, but you are allowed one 8.5x11 crib heet. () You have one hour to
More informationChapter 8. Rotational Equilibrium and Rotational Dynamics
Chapter 8 Rotational Equilibrium and Rotational Dynamics Wrench Demo Torque Torque, τ, is the tendency of a force to rotate an object about some axis τ = Fd F is the force d is the lever arm (or moment
More informationMomentum. Momentum and Energy. Momentum and Impulse. Momentum. Impulse. Impulse Increasing Momentum
Momentum and Energy Chapter 3, page 59-80 Review quetion: 1,3,4,7, 8, 11, 1, 14-17, 0, 1 Momentum Momentum i inertia in motion Ma x velocity Ha both magnitude and direction Large ma or high peed can give
More informationGeneral Physics (PHY 2130)
General Physics (PHY 130) Lecture 0 Rotational dynamics equilibrium nd Newton s Law for rotational motion rolling Exam II review http://www.physics.wayne.edu/~apetrov/phy130/ Lightning Review Last lecture:
More informationConservation of Energy
Conervative Force Conervation of Energ force i conervative if the work done b the force from r to r, but depend on initial and final poition onl Conervative Non-conervative Section #4.5 #4.6 Conervation
More informationChapter 8 - Rotational Dynamics and Equilibrium REVIEW
Pagpalain ka! (Good luck, in Filipino) Date Chapter 8 - Rotational Dynamics and Equilibrium REVIEW TRUE/FALSE. Write 'T' if the statement is true and 'F' if the statement is false. 1) When a rigid body
More informationPHY 171 Practice Test 3 Solutions Fall 2013
PHY 171 Practice et 3 Solution Fall 013 Q1: [4] In a rare eparatene, And a peculiar quietne, hing One and hing wo Lie at ret, relative to the ground And their wacky hairdo. If hing One freeze in Oxford,
More information= s = 3.33 s s. 0.3 π 4.6 m = rev = π 4.4 m. (3.69 m/s)2 = = s = π 4.8 m. (5.53 m/s)2 = 5.
Seat: PHYS 500 (Fall 0) Exa #, V 5 pt. Fro book Mult Choice 8.6 A tudent lie on a very light, rigid board with a cale under each end. Her feet are directly over one cale and her body i poitioned a hown.
More informationChapter 12: Rotation of Rigid Bodies. Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics
Chapter 12: Rotation of Rigid Bodies Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics Translational vs Rotational 2 / / 1/ 2 m x v dx dt a dv dt F ma p mv KE mv Work Fd P Fv 2 /
More informationName (please print): UW ID# score last first
Name (please print): UW ID# score last first Question I. (20 pts) Projectile motion A ball of mass 0.3 kg is thrown at an angle of 30 o above the horizontal. Ignore air resistance. It hits the ground 100
More information( kg) (410 m/s) 0 m/s J. W mv mv m v v. 4 mv
PHYS : Solution to Chapter 6 Home ork. RASONING a. The work done by the gravitational orce i given by quation 6. a = (F co θ). The gravitational orce point downward, oppoite to the upward vertical diplacement
More informationWrite your name legibly on the top right hand corner of this paper
NAME Phys 631 Summer 2007 Quiz 2 Tuesday July 24, 2007 Instructor R. A. Lindgren 9:00 am 12:00 am Write your name legibly on the top right hand corner of this paper No Books or Notes allowed Calculator
More informationis acting on a body of mass m = 3.0 kg and changes its velocity from an initial
PHYS 101 second major Exam Term 102 (Zero Version) Q1. A 15.0-kg block is pulled over a rough, horizontal surface by a constant force of 70.0 N acting at an angle of 20.0 above the horizontal. The block
More informationConditions for equilibrium (both translational and rotational): 0 and 0
Leon : Equilibriu, Newton econd law, Rolling, Angular Moentu (Section 8.3- Lat tie we began dicuing rotational dynaic. We howed that the rotational inertia depend on the hape o the object and the location
More informationPHYSICS 218 Exam 3 Fall, 2013
PHYSICS 218 Exam 3 Fall, 2013 Wednesday, November 20, 2013 Please read the information on the cover page BUT DO NOT OPEN the exam until instructed to do so! Name: Signature: Student ID: E-mail: Section
More informationtwo equations that govern the motion of the fluid through some medium, like a pipe. These two equations are the
Fluid and Fluid Mechanic Fluid in motion Dynamic Equation of Continuity After having worked on fluid at ret we turn to a moving fluid To decribe a moving fluid we develop two equation that govern the motion
More informationFinal Comprehensive Exam Physical Mechanics Friday December 15, Total 100 Points Time to complete the test: 120 minutes
Final Comprehenive Exam Phyical Mechanic Friday December 15, 000 Total 100 Point Time to complete the tet: 10 minute Pleae Read the Quetion Carefully and Be Sure to Anwer All Part! In cae that you have
More informationTwo-Dimensional Rotational Kinematics
Two-Dimensional Rotational Kinematics Rigid Bodies A rigid body is an extended object in which the distance between any two points in the object is constant in time. Springs or human bodies are non-rigid
More informationElastic Collisions Definition Examples Work and Energy Definition of work Examples. Physics 201: Lecture 10, Pg 1
Phyic 131: Lecture Today Agenda Elatic Colliion Definition i i Example Work and Energy Definition of work Example Phyic 201: Lecture 10, Pg 1 Elatic Colliion During an inelatic colliion of two object,
More informationy(t) = y 0 t! 1 2 gt 2. With y(t final ) = 0, we can solve this for v 0 : v 0 A ĵ. With A! ĵ =!2 and A! = (2) 2 + (!
1. The angle between the vector! A = 3î! 2 ĵ! 5 ˆk and the positive y axis, in degrees, is closest to: A) 19 B) 71 C) 90 D) 109 E) 161 The dot product between the vector! A = 3î! 2 ĵ! 5 ˆk and the unit
More informationPhysics 53 Summer Final Exam. Solutions
Final Exam Solutions In questions or problems not requiring numerical answers, express the answers in terms of the symbols given, and standard constants such as g. If numbers are required, use g = 10 m/s
More informationChapter 10. Rotation
Chapter 10 Rotation Rotation Rotational Kinematics: Angular velocity and Angular Acceleration Rotational Kinetic Energy Moment of Inertia Newton s nd Law for Rotation Applications MFMcGraw-PHY 45 Chap_10Ha-Rotation-Revised
More informationPhysics 6A. Practice Midterm #2 solutions
Phyic 6A Practice Midter # olution 1. A locootive engine of a M i attached to 5 train car, each of a M. The engine produce a contant force that ove the train forward at acceleration a. If 3 of the car
More informationChapter 8 continued. Rotational Dynamics
Chapter 8 continued Rotational Dynamics 8.4 Rotational Work and Energy Work to accelerate a mass rotating it by angle φ F W = F(cosθ)x x = s = rφ = Frφ Fr = τ (torque) = τφ r φ s F to s θ = 0 DEFINITION
More informationTable of Contents. Pg. # Momentum & Impulse (Bozemanscience Videos) Lab 2 Determination of Rotational Inertia 1 1/11/16
Table of Contents g. # 1 1/11/16 Momentum & Impulse (Bozemanscience Videos) 2 1/13/16 Conservation of Momentum 3 1/19/16 Elastic and Inelastic Collisions 4 1/19/16 Lab 1 Momentum 5 1/26/16 Rotational tatics
More informationCHAPTER 8 TEST REVIEW MARKSCHEME
AP PHYSICS Name: Period: Date: 50 Multiple Choice 45 Single Response 5 Multi-Response Free Response 3 Short Free Response 2 Long Free Response MULTIPLE CHOICE DEVIL PHYSICS BADDEST CLASS ON CAMPUS AP EXAM
More informationTranslational vs Rotational. m x. Connection Δ = = = = = = Δ = = = = = = Δ =Δ = = = = = 2 / 1/2. Work
Translational vs Rotational / / 1/ Δ m x v dx dt a dv dt F ma p mv KE mv Work Fd / / 1/ θ ω θ α ω τ α ω ω τθ Δ I d dt d dt I L I KE I Work / θ ω α τ Δ Δ c t s r v r a v r a r Fr L pr Connection Translational
More informationRotation. PHYS 101 Previous Exam Problems CHAPTER
PHYS 101 Previous Exam Problems CHAPTER 10 Rotation Rotational kinematics Rotational inertia (moment of inertia) Kinetic energy Torque Newton s 2 nd law Work, power & energy conservation 1. Assume that
More informationLinear Motion, Speed & Velocity
Add Important Linear Motion, Speed & Velocity Page: 136 Linear Motion, Speed & Velocity NGSS Standard: N/A MA Curriculum Framework (006): 1.1, 1. AP Phyic 1 Learning Objective: 3.A.1.1, 3.A.1.3 Knowledge/Undertanding
More informationPhysics 6A. Practice Midterm #2 solutions. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB
Phyic 6A Practice Midter # olution or apu Learning Aitance Service at USB . A locootive engine of a M i attached to 5 train car, each of a M. The engine produce a contant force that ove the train forward
More informationHandout 7: Torque, angular momentum, rotational kinetic energy and rolling motion. Torque and angular momentum
Handout 7: Torque, angular momentum, rotational kinetic energy and rolling motion Torque and angular momentum In Figure, in order to turn a rod about a fixed hinge at one end, a force F is applied at a
More informationPhysics 2212 G Quiz #2 Solutions Spring 2018
Phyic 2212 G Quiz #2 Solution Spring 2018 I. (16 point) A hollow inulating phere ha uniform volume charge denity ρ, inner radiu R, and outer radiu 3R. Find the magnitude of the electric field at a ditance
More informationExam 3 Practice Solutions
Exam 3 Practice Solutions Multiple Choice 1. A thin hoop, a solid disk, and a solid sphere, each with the same mass and radius, are at rest at the top of an inclined plane. If all three are released at
More informationChap. 10: Rotational Motion
Chap. 10: Rotational Motion I. Rotational Kinematics II. Rotational Dynamics - Newton s Law for Rotation III. Angular Momentum Conservation (Chap. 10) 1 Newton s Laws for Rotation n e t I 3 rd part [N
More informationAfternoon Section. Physics 1210 Exam 2 November 8, ! v = d! r dt. a avg. = v2. ) T 2! w = m g! f s. = v at v 2 1.
Name Physics 1210 Exam 2 November 8, 2012 Afternoon Section Please write directly on the exam and attach other sheets of work if necessary. Calculators are allowed. No notes or books may be used. Multiple-choice
More informationWe define angular displacement, θ, and angular velocity, ω. What's a radian?
We define angular displacement, θ, and angular velocity, ω Units: θ = rad ω = rad/s What's a radian? Radian is the ratio between the length of an arc and its radius note: counterclockwise is + clockwise
More informationUniform Acceleration Problems Chapter 2: Linear Motion
Name Date Period Uniform Acceleration Problem Chapter 2: Linear Motion INSTRUCTIONS: For thi homework, you will be drawing a coordinate axi (in math lingo: an x-y board ) to olve kinematic (motion) problem.
More informationSolution to phys101-t112-final Exam
Solution to phys101-t112-final Exam Q1. An 800-N man stands halfway up a 5.0-m long ladder of negligible weight. The base of the ladder is.0m from the wall as shown in Figure 1. Assuming that the wall-ladder
More informationPHYSICS 221 SPRING 2013
PHYSICS 221 SPRING 2013 EXAM 2: April 4, 2013 8:15-10:15pm Name (printed): Recitation Instructor: Section # INSTRUCTIONS: This exam contains 25 multiple-choice questions plus 2 extra credit questions,
More informationChapter 8 continued. Rotational Dynamics
Chapter 8 continued Rotational Dynamics 8.4 Rotational Work and Energy Work to accelerate a mass rotating it by angle φ F W = F(cosθ)x x = rφ = Frφ Fr = τ (torque) = τφ r φ s F to x θ = 0 DEFINITION OF
More informationAP Physics. Harmonic Motion. Multiple Choice. Test E
AP Physics Harmonic Motion Multiple Choice Test E A 0.10-Kg block is attached to a spring, initially unstretched, of force constant k = 40 N m as shown below. The block is released from rest at t = 0 sec.
More informationUniform Circular Motion
Uniform Circular Motion Motion in a circle at constant angular speed. ω: angular velocity (rad/s) Rotation Angle The rotation angle is the ratio of arc length to radius of curvature. For a given angle,
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Two men, Joel and Jerry, push against a wall. Jerry stops after 10 min, while Joel is
More informationChapter 8: Momentum, Impulse, & Collisions. Newton s second law in terms of momentum:
linear momentum: Chapter 8: Momentum, Impulse, & Collisions Newton s second law in terms of momentum: impulse: Under what SPECIFIC condition is linear momentum conserved? (The answer does not involve collisions.)
More informationChap10. Rotation of a Rigid Object about a Fixed Axis
Chap10. Rotation of a Rigid Object about a Fixed Axis Level : AP Physics Teacher : Kim 10.1 Angular Displacement, Velocity, and Acceleration - A rigid object rotating about a fixed axis through O perpendicular
More informationRotational Dynamics continued
Chapter 9 Rotational Dynamics continued 9.4 Newton s Second Law for Rotational Motion About a Fixed Axis ROTATIONAL ANALOG OF NEWTON S SECOND LAW FOR A RIGID BODY ROTATING ABOUT A FIXED AXIS I = ( mr 2
More informationChapter 8 continued. Rotational Dynamics
Chapter 8 continued Rotational Dynamics 8.6 The Action of Forces and Torques on Rigid Objects Chapter 8 developed the concepts of angular motion. θ : angles and radian measure for angular variables ω :
More informationExam 1 Solutions. +4q +2q. +2q +2q
PHY6 9-8-6 Exam Solution y 4 3 6 x. A central particle of charge 3 i urrounded by a hexagonal array of other charged particle (>). The length of a ide i, and charge are placed at each corner. (a) [6 point]
More informationPhysics Final Exam Formulas
INSTRUCTIONS: Write your NAME on the front of the blue exam booklet. The exam is closed book, and you may have only pens/pencils and a calculator (no stored equations or programs and no graphing). Show
More informationChapter 9. Rotational Dynamics
Chapter 9 Rotational Dynamics In pure translational motion, all points on an object travel on parallel paths. The most general motion is a combination of translation and rotation. 1) Torque Produces angular
More informationKinetics of a Particle: work and energy
Kinetic of a Particle: work and energy Work ha been done by a force on a particle only when the particle undergoe a diplacement in the direction of the force du F d co du Fdr Unit of work: Joule J= Nm
More informationMechanics II. Which of the following relations among the forces W, k, N, and F must be true?
Mechanics II 1. By applying a force F on a block, a person pulls a block along a rough surface at constant velocity v (see Figure below; directions, but not necessarily magnitudes, are indicated). Which
More informationChapter 12: Rotation of Rigid Bodies. Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics
Chapter 12: Rotation of Rigid Bodies Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics Translational vs Rotational 2 / / 1/ 2 m x v dx dt a dv dt F ma p mv KE mv Work Fd P Fv 2 /
More informationApplication of Newton s Laws. F fr
Application of ewton Law. A hocey puc on a frozen pond i given an initial peed of 0.0/. It lide 5 before coing to ret. Deterine the coefficient of inetic friction ( μ between the puc and ice. The total
More informationNewton s Laws & Inclined Planes
GP: ewton Law & Inclined Plane Phyic Mcutt Date: Period: ewton Law & Inclined Plane The ormal orce, Static and Kinetic rictional orce The normal orce i the perpendicular orce that a urace exert on an object.
More informationGeneral Definition of Torque, final. Lever Arm. General Definition of Torque 7/29/2010. Units of Chapter 10
Units of Chapter 10 Determining Moments of Inertia Rotational Kinetic Energy Rotational Plus Translational Motion; Rolling Why Does a Rolling Sphere Slow Down? General Definition of Torque, final Taking
More informationPHYSICS 221 SPRING 2013
PHYSICS 221 SPRING 2013 EXAM 2: April 4, 2013 8:15-10:15pm Name (printed): Recitation Instructor: Section # INSTRUCTIONS: This exam contains 25 multiple-choice questions plus 2 extra credit questions,
More informationChapter 9. Rotational Dynamics
Chapter 9 Rotational Dynamics In pure translational motion, all points on an object travel on parallel paths. The most general motion is a combination of translation and rotation. 1) Torque Produces angular
More information