# FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Monday, 14 December 2015, 6 PM to 9 PM, Field House Gym

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1 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 is an equation sheet on page 13. You may tear the equation sheet off. 3. There are two parts to the exam: Part I has twelve multiple choice questions (1 to 1), where you must circle the one correct answer (A, B, C, D, E). Rough work can be done on the backside of the sheet opposite the question page Part II includes nine full-answer questions (13 to 1). Do all nine questions. All works must be done on the blank space below the questions. If you run out of space, you may write on the backside of the sheet opposite the question page. 4. Non-Programmable calculators are allowed 5. Programmable calculators are NOT ALLOWED. 1

4 7. (.5 points) A golf ball is hit so that it leaves the ground at 60 above the horizontal and feels no air resistance as it travels. Which of the following statement about the subsequent motion of the ball while it is in the air is true? Only one correct answer. A) Its speed is zero at its highest point. B) Its velocity is zero at its highest point. C) Its acceleration is always 9.8 m/s downward. D) Its forward acceleration is 9.8 m/s. E) Its acceleration is zero at its highest point. Answer C) 8. (.5 points) A 0.54-kg block is held in place against the spring by a 65-N horizontal external force. The external force is removed, and the block is projected with a velocity v 1 = 1. m/s upon separation from the spring. The block descends a ramp and has a velocity v = 1.4 m/s at the bottom. The track is frictionless between points A and B. The block enters a rough section at B, extending to E. The coefficient of kinetic friction is The velocity of the block is v 3 = 1.4 m/s at C. The block moves on to D, where it stops. In the figure below the initial compression of the spring, in cm, is closest to: A) 0.64 B) 0.43 C) 1.6 D) 1. E).4 HINT: Using Hooke s law the force that holds the spring obeys the relation kx = 65N To answer the question we only need to consider all data in the vicinity where the box is in contact with the spring. Using the fact that the box start from rest and has a final speed of v 1 = 1. m/s, and using conservation of energy (no friction) we obtain 1 mv 1 = 1 kx, where x is the maximum (initial) compression of the spring. Rearranging we obtain x = mv 1. Also we can use the hint kx = 65N, kx ( ) 0.54kg 1.m / s x = = m = 1.cm ANSWER d) 65N 9. (.5 points) The force on a particle is conservative if and only if: A) the particle moves exactly once around any closed path. B) It is not a frictional force. C) the work done on the particle is independent of the path taken by the particle D) it obeys Newton s second law E) it obeys Newton s third law 10. (.5 points) A 1.0 kg-ball moving toward and perpendicular to a wall at.0 m/s, rebounds from the wall at 1.5 m/s. The change in the momentum of the ball is closest to: A) zero B) 0.5kgim / s away from wall C) 0.5kgim / s toward wall D) 3.5kgim / s away from wall E) 3.5kgim / s toward wall ANSWER: D 11. (.5 points) A rod is pivoted about its center. A 5-N force is applied 4 m from the pivot and another 5-N force is applied m from the pivot as shown. The magnitude of the total torque about the pivot is: 4

5 A) 0 N m B) 5 N m C) 8.7 N m D) 15 N m E) 6 N m For the right force the moment arm is r,r = 4m sin30 = m, giving ccw torque of τ R = 5N r,r = 10N im, where we assume ccw is +. For the left force the moment arm is r,l = m sin30 = 1m, giving ccw torque + of τ L = 5N r,l = 5N im. Net torque is τ net = τ L +τ R = 15N im. ANSWER D 1. (.5 points) A massless meter stick on a horizontal frictionless table top is pivoted at the 80-cm mark. A Force F 1 is applied perpendicularly to the end of the stick at 0 cm as shown. A second force F (not shown) is applied perpendicularly at the 60-cm mark. The forces are in the plane of the table top. If the stick does not move, the force exerted at the pivot on the stick: HINT: It is possible to infer the direction of by the information in the question A) must be zero B) must be in the same direction as and have a magnitude C) must be directed opposite to and have a magnitude D) must be in the same direction as and have a magnitude E) must be directed opposite to and have a magnitude If the stick does not move then it is in static equilibrium. The net torque about the pivot (or any other axes) must be zero. Since F 1 induces a CW torque ( F 1 (1m) ), F (BLUE ARROW) must point down to induce a CCW torque ( F (0.6m) ). The net torque must be zero or F 1 ( 1m)+ F ( 0.6m) = 0 F 1 = 0.6F, and F > F 1. For static equilibrium the net force must be zero. F 1 + F + F P = 0 F p = F 1 F. The pivot force, F p (RED ARROW). Since F > F 1, the direction of F 1 to cancel F 1 F. The magnitude is simply F 1 + F or F F 1. ANSWER: D For B 1.5 point 5

6 PART II: FULL ANSWER QUESTIONS (question 13 to 1) Do all nine questions on the provided area below the questions. Show all works. 13. (10 points) A shot putter runs with a speed of 9.0 m/s. He throws the metal ball with a velocity 15.0 m/s at an angle of 53 above the horizontal (velocity is relative to himself). The ball is released at.0 m above the ground. Neglect air resistance. a) Draw a diagram of the path of the trajectory (or a motion diagram) of the metal ball, which shows the direction of the velocity and acceleration at the following points: i) the instant it leaves the shot putter; ii) its maximum height; iii) the instant when it hits the ground. Calculate the x and y component of the initial velocity (with respect to the ground). 13 (Continued) (0.5 point) diagram ( points) (0.5 point) b) Determine the maximum height of the ball, above the ground. Easiest method v y = v 0y g( y y 0 ) y y 0 = v 0y v y (1 points) g v 0y = 1.0 m/s, y 0 = 0, v y = 0 at maximum height, y = max. height (0.5 point) y = v 0y ( 1m /s) = = 7.35m. g 9.8m /s ( ) The height above ground is.0m m= 9.35 m (0.5 points) Hard method v 0y = 1.0 m/s, y 0 = 0, v y = 0 at maximum height, y = max. height (0.5 point) v y = v 0y gt 0 = v 0y gt t = v 0y g y = y 0 + v 0y t 1 gt = 0 + ( 1.0m /s) 1.s 1.0m /s = =1.s (1 points) 9.8m /s ( ) 1 ( 9.8m /s ) 1.s ( ) = 7.35m The height above ground is.0m m= 9.35 m (0.5 points) c) Find the time, after it was thrown, when the ball hits the ground. Answer is.60 s, but you must show work. How far did the ball travel horizontally? Take t = 0 y 0 = 0,v 0 y = 1.0m / s, and when it hits the ground t =?, y =.0m. y = y 0 + v 0 y t 0.5gt = 1t 4.9t 4.9t 1t = 0 ( points) 6

7 1 ± ( 1 Solution is t = ) 4( 4.9) ( ) = 1.3s ± 1.38s =.61s and 0.15s ( 4.9) physical answer is.6 s. range = x = v 0x t = 18.0m /s ( )(.6s) = 46.8m 14. (10 points) In the diagram below block A has a mass of 8.00 kg and block B has mass 8.00 kg. The coefficient of kinetic friction between block B and the table is µ k = 0.5. Block B is moving right and accelerating to the right with a = 1.50 m/s. a) Draw a free-body diagram of block A. Hence, find the tension in the rope connecting Blocks A and B. A T A m A g ( ) ( ) = 66.4N a T A m A g = m A a T A = m A g a T = 8kg 9.8mi s 1.5mi s b) Draw a free-body diagram of block B. Calculate the friction force on block B. Determine the tension of the rope connecting blocks B and C. f k T B F N = m B g B a m B g Newton s Second Law T A T A T B f k = m B a T B = T A f k m B a = 66.4N 19.6N 8kg 1.5mi s = 34.8N Friction f k = m B gµ k = 8kg( 9.8mi s )0.5 = 19.6N c) Finally, draw a free-body diagram of block C. Determine the mass of block C C T A m C g (3 points) (3 points) ( ) ( ) = 34.8N /( 9.8mi s +1.5mi s ) = 3.08kg a T B m C g = m C a T B = m C g+ a m C = T B / g+ a (4 points) 7

8 15. (10 points) A 5.0 kg crate slides toward a spring (force constant k = 500N / m ) located 1.0 m away with a speed of 6.7 m/s. The coefficient of kinetic and static friction between all surfaces are µ = 0. 3 and µ = 0. 4, respectively. k s Final speed v F =? v 1 =? Initial speed v 0 = 6.7 m/s x 1.0 m a) Use the work-energy theorem or conservation of energy to find the speed (v 1 ) of the crate when it reaches the spring 1.00 m away from its initial position. Work done by friction W f = K 1 K 0 change in kinetic energy 1 1 f k µ k ( 1m) = mgµ k ( 1m) = mv1 mv0 1 1 mv1 = mv0 mgµ k ( 1m) v1 = v0 gµ k ( 1m) v 1 = ( 6.7m / s) 9.8m / s ( )( 1m) = 6.5m / s ( ) 0.3 b) After the spring has been compressed by x = 0. m as shown above, what is the velocity of the box v F? Use Conservation of energy between point 1 and the final point when it compresses the spring by x = 0. m:w external = ΔE mech + ΔE th where ΔE mech = ΔU + ΔK = ( U F U 1 ) + ( K F K 1 ), ΔE th = f k d, and W ext = 0, giving K 1 + U el el 1 f k d = K F + U F Using K F = 1 mv el el F,U 1 = 0,U F = 1 kx = ( 50N / m) ( 0.m) = 10J f k d = mgµ k x = ( 5.0kg) ( 9.8m / s )( 0.3) ( 0.m) =.94J K 1 = 1 mv = 1 1 ( 5kg )( 6.5m/ s) = 97.7 J (1point) K 1 + U el 1 f k d = K F + U el F 84.76J = 1 ( 5kg)v F v F = 5.83 m ( point) s 16. (10 points) A hockey puck B rest on a frictionless ice surface is struck by a second puck A, which was initially traveling at 40 m/s as shown in the diagram below. After the collision, the pucks are deflected as shown below. Assume the puck have the same mass. a) Find the speed of each puck after the collision 8

9 Use conservation of momentum and assume that the mass of the pucks are the same. 1 point Y-component 0 = mv A sin 30 mv B sin point v A = v B. (1) 0.5 point X-component mv A1 = mv A cos 30 + mv B cos point v A1 = 3 v + 1 A v B. () 0.5 point Substituting equation (1) into equation () and using v A 40.0m / s 40.0m / s = = ( v B ) + 1 v = 1.93v B B. 1 point v B = 0.7m / s ( ) = 9.3m / s. 1 point Using equation (1) v A = v B = 0.7m / s b) Calculate the change in kinetic energy due to the collision. Is this an elastic collision? Briefly justify your answer. ΔK = K K 1 = 1 mv A + 1 mv B 1 mv A1. 1 point ΔK = 1 m( 9.3m / s) + 1 m( 0.7m / s) 1 m( 40.0m / s) = 157m.1 point In an elastic collision, the total mechanical energy is conserved. In part (b) it is shown that the kinetic energy decreases. Hence total mechanical energy is not conserved and the collision is not elastic. point 17. (10 points) In the figure to the left, a stuntman of mass 80.0 kg swings on a rope from a 5.0 m high ledge towards a 70.0 kg villain standing on the ground. Assume stuntman is initially at rest. Also assume that the stuntman velocity is horizontal just before he collides with the villain. A) What is the speed of the stuntman just before he hits the villain? This is found by using conservation of mechanical energy; basically the stuntman s energy at the top must equal his energy just before he hits the villain. grav Take the gravitational PE to be zero at the bottom, U bottom = 0 and U grav top = m s gh (m s = 80.0kg, h = 5.0 m) at the top. 1 Conservation of energy m sgh = msvs, where v s is the stuntman s speed just before collision. m s gh = 1 m v s s v s = gh = ( 9.8m / s )( 5.0m) = 9.9m / s (4 points) B) What is the horizontal component of the velocity of the stuntman just before he hits the villain? Hint: Look carefully at the diagram. Obviously, at the bottom just before the collision the stuntman s velocity is horizontal, so the horizontal component of the velocity is 9.9m/s. ( points) C) Just after they (stuntman + villain) collide, and become entangled, what is their speed as they slide on the floor? Assume the floor is frictionless.

10 Use conservation of momentum of the horizontal component before and after the stuntman + villain collision. m s v s = ( m s + m v )v f, m s = 70.0 kg and v f the final speed of the entangled pair. ( )( 9.9m / s) v f = m v s s = 80.0kg m s + m v 80.0kg kg = 5.8m / s (4 points) 18. (10 points) In the diagram below, a solid uniform ball ( I = 5 MR, R = 0.15 m) rolls without slipping with a linear speed of 5.0 m/s. It moves up the hill, and goes over a 8 meter high cliff. a) Calculate the linear, v T, and rotational, ω T, speed at the top of the hill. First use conservation of total mechanical energy between position 1 and. U g mv Iω 1 = U g + 1 mv + 1 Iω 1 mv Iω 1 = mgh + 1 mv + 1 Iω Since I = 5 mr 1 Iω 1 = 1 5 mr ω 1 = 1 5 m ( Rω 1). Using the condition for rolling without slipping v 1 = Rω 1 gives 1 Iω 1 = 1 5 mv 1, and similarly 1 Iω = 1 5 mv. 1 mv Iω 1 = mgh + 1 mv + 1 Iω 1 mv mv 1 = mgh + 1 mv mv 1 mv mv 1 = mgh + 1 mv mv 7 10 mv 1 = mgh mv ( points) v = v 1 10 gh = ( 5m / s 7 ) 10 7 ω = v /R = 10rad / s ( 9.8m / s )( 8.0m) = 15.6m / s b) Calculate the linear, v B, and rotational, ω B, speed when it lands on the ground. Now use conservation of total mechanical energy between position and 3. U g + 1 mv + 1 Iω = U g mv Iω 3 = mgh + 1 mv + 1 Iω = 1 mv Iω 3 Now note that as the ball roll up the hill, from position to 3, without slipping, the friction between the ball and ground induces a torque that changes its angular velocity. But as it leaves the cliff to go from position to 3 there is no friction to change the angular velocity: ω = ω 3. This gives mgh + 1 mv + 1 Iω = 1 mv Iω 3 mgh + 1 mv = 1 mv 3 ( points) v 3 = v + gh = ( 15.6m / s) + ( 9.8m / s )( 8.0m) = 7.9m / s ω 3 = v 3 /R = 186rad / s 10

11 C) In part b, if you did the question correctly, the final linear speed of the ball should be greater than the initial speed (5 m/s). Does the fact that this speed is greater than the initial linear speed means that the ball gains energy? Explain your answer in one or two sentences. It is noted that the linear speed at position 3 ( v 3 = 7.9m / s ) is greater than at position 1 ( v = 5m / s ), even though they are at the same height. This is explained by the fact that the kinetic energy includes rotational component. When the ball roll up the hill from 1 to it gains PE of mgh and it loses mgh in KE, which since there is friction both the linear and rotational component lose energy so that v 1 > v and ω 1 > ω. When it falls off the cliff from to 3 it regains mgh in KE, but only the rotational component cannot change so that ω = ω 3, meaning that the linear component v < v 3 regains the whole mgh so that v 1 < v 3, but also ω = ω 3 < ω 1, so the rotational energy at position 3 is less than at 1. However it is noted that the total kinetic energy (rotational + linear) are the same at position 1 and 3. ( points) 19. (10 points) In the figure below, box A and B are connected by a rope-pulley system. Box A moves to the right, and the rope moves over the pulley without slipping, and the pulley has a clockwise angular velocity, ω. The data are shown in the figure. v T A T A ω mass of pulley is M P = 3.0 kg f m T B I = 1 A = 10.0 kg M P R solid cylindrical pulley Coefficient of Friction µ k = 0.5 T B R = 0. m, radius of cylinder m B =4.0kg a) Draw a free body diagram of the boxes showing all the forces acting on it. Draw a free body diagram of the pulley showing all the forces (including those due to the hinge) acting on it. The diagrams should include the direction of linear acceleration, a, of m A and m B, and the angular acceleration α of the pulley. The system will decelerate, so that the acceleration is 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 counterclockwise (ccw) as positive Note that the angular acceleration, α, is ccw. R ω Note that angular velocity ω is cw or negative. R T B F Hinge (blue arrow) is the force of the hinge on the wheel that ultimately supports the wheel. b) Use Newton s law for translation and rotation to find the linear acceleration (magnitude and direction) a of m A and m B, and the angular acceleration (magnitude and sense of 11

12 rotation), α, of the pulley. The no-slip condition is useful. Assume the rope does not slip on the pulley. Note thatt A T B. Also you may want to assume that Box B accelerates up. BOX A BOX B net net 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-axis F y net = T A f k = m A a T A m A gµ k = m A a [] We now calculate the net torque on the wheel:τ net = T A R T B R = Iα. Using I = 1 M PR, and the no-slip condition a = Rα. Now do [] [1] T A m A gµ k T B + m B g = m A a m B a, which gives T A T B + m A a + m B a = m A gµ k m B g. Substituting [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 m Solving for a, a = A gµ k m B g 1 M P + m A + m B 10kg 9.8 m = s 0.5 4kg 9.8 m s 1 3kg kg + 4.0kg a = 0.63 m s, and using the no slip condition α = a R = 0.63 m s 0.m = 3. rad s 0. (10 points) In figure below, a carousel has a radius of 3.0 m and a moment of inertia of I C = 8000kg m, for rotation about axis perpendicular to the its center. The carousel is rotating unpowered and without friction with an angular velocity of 1. rad/s. An 80-kg man runs with a velocity of 5.0 m/s, on a line tangent to the rim of the carousel, overtaking it. The man runs onto the carousel and grabs hold of a pole on the rim. +y +x Direction perpendicular to x-y plane indicates +z out of the page indicates z into page USE THIS CONVENTION TO INDICATE THE DIRECTION OF ANGULAR MOMENTUM a) Before the collision, what is the magnitude of the angular momentum of the rotating carousel, L C, with respect to the center of the carousel? What is the direction of L C? Directions (+x, +y, +z, -x, -y, -z) are as indicated in the above figure. 1

13 The magnitude of angular momentum is L C = I c ω c = kgim 1. rad = 9600 kgim s s The rotation is ccw, so using the right hand rule, the direction is out of the the +z direction. (3.5 points) page in b) Before the collision, what is the magnitude of the angular moment of the running 80- kg man, L M, with respect to the center of the carousel? What is the direction of L M? center of carousel R = 3.0 m p = mv = (80 kg) (5.0 m/s) = 400 kg-m/s The angular momentum of a particle is defined as L M = R p, where R is the position vector from the center of the carousel to the particle when it reaches the edge of the carousel, and p is the momentum. Since they are perpendicular the magnitude is simply L M = Rp = 3.00m 400 kgim = 100 kgim s s Using the right hand rule on the diagram above it is easy to that the direction is out of the page in the +z direction. (3.5 points) c) After the collision when the man is on the carousel, what is the magnitude of the final angular velocity of the carousel (with the man on it), ω fc? What is the direction of the final angular velocity ω fc? Note: I total = I C + mr First find the moment of inertia of the carousel with the man on it about the center I total = I C + mr = 8000kgim + 80kg ( 3.0m) = 870kg m. The conservation of angular momentum (only the z-direction is relevant) gives: 9600 kgim kgim L c + L M = I tot ω fc ω fc = s s = 1.4 rad 870kgim s Direction is +z. (3 points) 1. (10 points) In figure below, a shell is shot with an initial velocity v 0 of 19 m/s, at an angle of θ 0 = 5 with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically to the ground. 13

14 a) Determine the horizontal distance from the gun to the top of the trajectory (explosion). First find component of initial velocity: v 0x = v 0 cosθ 0 = 19 m s cos5 = 11.7 m s ; v 0 y = v 0 sinθ 0 = 19m s sin5 = m s At top v y = 0 = v oy gt t = v 0 y / g = 14.97mi s 1 9.8mi s = 1.53s The distance is x explosion = v 0x t = 11.7mi s s = 17.9m. (3 points) b) If the shell does not explode, determine the horizontal distance from the gun to the point where the shell hits the ground. Neglect the height of the gun above the ground. First find the time it hits the ground y = y 0 + v 0 y t 1 gt 0 = 0+ v 0 y t 1 gt t v 0 y 1 gt t = 0,t = v 0 y g = 14.97mi s 1 ( ) = 3.06s 9.8mi s The position is x range = v 0x t = 11.7mi s s = 35.8m (3 points) c) Use the result of part a) and b), and the fact that the center-of-mass (COM) position is not affected by the explosion, to determine the position of the second fragment when it hits the ground. For unexploded shell the COM position is x com = x range = 35.8m Let x position of fragment. Since the COM must remain constant we have x com = x range = 0.5mx +0.5mx explosion = x + x explosion 0.5m+ 0.5m x = x range x explosion = ( 35.8m) 17.9m = 53.7m (4 points) 14

15 Useful Equations Kinematics 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 speed savg = (total distance)/(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 s Laws F net = F i = 0 (Object in equilibrium); F net = m a (Nonzero net force); Weight: F g = mg, g = 9.8m / s ; Centripetal acceleration a rad = v r ; Friction f s µ s F N, f k = µ k F N. Hooke s Law F x = kx. Work and Energy W = F d = ( F cosθ)d = F d ; W net = ΔK = (1 / )mv f (1 / )mv net i (valid if W is the net or total work done on the object);w grav = mg y f y i ( ) (elastic work) W el = (1 / )kx f (1 / )kx i ( ) (gravitational work), Conservation of Mechanical Energy (only conservative forces are present) E mech = U + K W net = ΔU = U U 1 ( ) = ΔK = K K 1,U 1 + K 1 = U + K,U grav = mgy,u el = (1 / )kx Also ΔE mech = ΔU + ΔK = U f U i ( ) + ( K f K i ) = 0 ΔK = ΔU Non-Conservative Forces W external = ΔE mech + ΔE th (W ext work done by external forces, and we set ΔE int = 0 ), where ΔE th = f k d (thermal energy or negative work done by friction). Using ΔE mech = ΔU + ΔK = U f U i Work due to variable force 1D: W = ( ) + ( K f K i ), U f + K f = U i + K i + W ext f k d x f x i F x dx area under F x vs. x, from x = x i to x f Momentum: P = m v, J t = Fdt = F av t t 1 t 1 ( ), J = Δ P = P P 1. Newton s Law in Terms of Momentum F net = d p / dt. For F net = 0, d p / dt = 0 gives momentum conservation: P constant. Rotational Kinematics Equations:ω avg = θ θ 1 ( ) / ( t t 1 ), α avg = ( ω ω 1 ) / ( t t 1 ) ( ) For α z = constant, ω = ω 0 + αt, θ = θ 0 + ω 0 t + (1 / )αt,ω = ω 0 + α θ θ 0 Linear and angular variables: s = rθ, v = rω, a tan = Rα (tangential), a rad = v / r = ω r (radial) Moment of Inertia and Rotational Kinetic Energy I = m i r i, K rot = (1/ )Iω. Center of Mass (COM) r com = m i ri / m i. Torque and Newton s Laws of Rotating Body: rigid bodyτ = Fr, τ net = ext τ i = Iα, r -moment arm about axis; point τ = r F about origin O. Combined Rotation and Translation of a Rigid Body K = (1/ )Mv com + (1/ )I com ω, F net = M a com, τ net = I com α. Rolling without slipping s = Rθ, vcom = Rω, a com = Rα. Angular Momentum L = Iω (solid object) where I is the moment of inertia about the axis of rotation. L = r p L = mvr sinθ, valid for point particle about an origin O. Newton s Second Law of rigid body in terms of angular momentum τ net = τ i ext For τ net = 0, d L / dt ( ) = 0 and angular momentum is conserved, L constant. N i=1 = ( d L / dt). 15

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