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WileyPLUS Assignment 3 is available Chapters 6 & 7 Due Wednesday, November 11 at 11 pm Next Week Tutorial and Test 3 Chapters 6 & 7 See web page for tutorial questions Friday, October 30, 2009 1 Equations for rotational motion Analogous to the equations for linear motion. ! = !0 + "t 1 ! = ("0 + ")t 2 1 ! = "0t + #t 2 2 v = v0 + at ω v α a θ x 1 x = (v0 + v)t 2 1 x = v0t + at 2 2 v 2 = v02 + 2ax !2 = !20 + 2"# arc length, l = rθ speed, v = rω π radians = 180o Friday, October 30, 2009 2 8.33/27: A child is running around a stationary merry-go-round at 0.25 rad/s. At the moment he sees his favourite horse, one quarter turn away, the merry-go-round starts to move in the same direction as the child is running, accelerating at 0.01 rad/s2. What is the shortest time it takes the child to catch up with the horse? Child !c = 0.25 rad/s What are the angular positions of child and horse? Horse " = 0.01 rad/s2 Friday, October 30, 2009 3 8.26/21: A spinning top is made to spin by pulling on a string that is wrapped around it. The length of the string is L = 64 cm and the radius of the top is r = 2 cm. The string is pulled so that the angular acceleration of the top is 12 rad/s2. What is the final angular velocity of the top? ω 2 = ω02 + 2αθ (think v2 = vo2 + 2ax) What is θ? The number of turns of string around the top is N = L/(2πr), so θ= N x 2π = (L/2πr) x 2π = L/r = 64/2 = 32 rad ω = 28 rad/s Friday, October 30, 2009 4 8.75/14: A golf ball passes through a windmill, which has 8 blades and rotates at ω = 1.25 rad/s. The opening between successive blades is equal to the width of a blade. A golf ball is of diameter d = 0.045 m. What must be the minimum speed of the golf ball so that it passes through an opening between blades? There are 8 blades and 8 gaps between blades. The angular width of each gap is 2π/16 rad. The golf ball must travel at least a distance equal to its diameter while the blade moves the width of one gap. That is, t = d 2π/16 8dω = →v= v ω π v = 0.143 m/s d v v t = θ/ω Friday, October 30, 2009 5 Chapter 9: Rotational Dynamics Sections 1, 2, 3, 6 only • Action of torques • The two conditions of equilibrium • Centre of gravity • Conservation of angular momentum Friday, October 30, 2009 6 Circular Motion B r ! l r l = r! v = r" v " = #$/#t ! = #!/#t A ! in radians: 1800 = % rad. " in rad/s ! in rad/s2 Friday, October 30, 2009 7 Torques w = width of door Torque = (magnitude of force) ! (lever arm) Lever arm, l = w sin 90◦ Torque, ! = Fl = Fw Friday, October 30, 2009 8 Le ve ra rm Torques ! Pivot, axis of rotation w = width of door Lever arm: l = w sin ! Torque, ! = Fl = Fw sin " Friday, October 30, 2009 9 Torques !=0 w = width of door Torque, ! = Fl = Fw sin " = 0 Friday, October 30, 2009 10 9.3/1: The torque applied to the bolt is: ! = Fl l = 0.28 sin 50◦ m lever arm l Line of action of the force = 45 N ! = (45 N) × (0.28 sin 50◦ m) = 9.65 N.m Friday, October 30, 2009 11 F = 720 N of he e ft n Li n o e c o t i or ac f Torque generated by Achilles tendon about ankle joint: l = (0.036 m) × cos 55◦ = 0.0207 m ! = Fl = (720 N)(0.0207 m) = 14.9 N.m Sign convention: torque is positive when tending to rotate counterclockwise about reference point. So, torque = –14.9 N.m. Friday, October 30, 2009 12 Two ways to calculate torques P d Lever arm, l Line of action of the force F# = component of F perp. to line to P F# 900 $ 900 $ F Point of application of F Torque about point P: 1) Torque = (force) & (lever arm): τ = F l = F × d sin θ τ = F⊥ d = F sin θ × d 2) Torque = F# & d : Choose the method that is the more convenient... Friday, October 30, 2009 13 9.-/8: One end of a metre stick is pinned to a table, so that the stick can rotate freely around the surface of the tabletop. Two forces, both parallel to the tabletop, are applied in such a way that the net torque is zero. Where along the stick must the 6 N force be applied? x Friday, October 30, 2009 14 9.2/3: The engine applies a torque of τeng = 295 N.m to the wheel of a car, which does not slip against the road surface because the static friction force applies a countertorque. The car is travelling at constant velocity. What is the static friction force? As the wheel is turning at constant rate, the net torque applied to it must be zero. Therefore, 'eng = Fs & r Fs = 'eng/r = 295/0.35 = 843 N O v r = 0.35 m 'eng Fs Friday, October 30, 2009 15 The Two Conditions of Equilibrium Balance is important! Friday, October 30, 2009 16 The Two Conditions of Equilibrium For an object to be in equilibrium: 1) The sum of forces acting on the object must be zero: !�Fi = �F1 + �F2 + . . . = 0 Then the acceleration is zero by Newton’s first law. 2) Balance: the sum of torques about any point must be zero: !"i = "1 + "2 + . . . = 0 Then the angular acceleration is zero. Friday, October 30, 2009 17 Clickers! Are the three forces shown sufficient to keep the rod in equilibrium? A For the rod to be in equilibrium, the sum of the forces and of the torques about any point must be zero. A) It is not possible to adjust the magnitudes of the forces so that the net force is zero. B) It is possible to adjust the magnitudes of the forces so that the net force is zero, but the net torque cannot be zero. C) It is possible to adjust the magnitudes of the forces so that the net force and the net torque are both zero. Friday, October 30, 2009 18 For the rod to be in equilibrium, the sum of the forces and of the torques about any point must be zero. A The sum of forces could be made to be zero by adjusting the magnitudes of the forces. That is, F1 + F2 + F3 = 0 First condition of equilibrium is OK Second condition of equilibrium? Calculate torques about A. F1 and F2 exert no torque about A, but F3 does. $ the net torque cannot be zero unless F3 = 0, but then the first condition of equilibrium fails (unless the other forces are zero too!). $ the rod cannot be in equilibrium if only these forces act. Friday, October 30, 2009 19 Focus on Concepts, Question 6 Five hockey pucks are sliding across frictionless ice. The drawing shows a top view of the pucks and the three forces that act on each one. The forces can have different magnitudes (F, 2F, or 3F), and can be applied at different points on the puck. Only one of the five pucks could be in equilibrium. Which one? A) 1 B) 2 C) 3 D) 4 E) 5 D) 4 The net force is zero and the net torque is zero. Friday, October 30, 2009 20 Conditions of equilibrium: Forces acting on the diving board 1) Forces acting on the board: −F1 + F2 −W = 0 2) Torques about the axis: 0 × F1 + 1.4F2 − 3.9W = 0 That is, F2 = 3.9W = 2.786W 1.4 Friday, October 30, 2009 21 -F1 + F2 - W = 0 (first condition) F2 = 2.786W (second condition) F1 = F2 - W = (2.786 - 1)W = 1.786W If the weight of the diver is W = 530 N, then F1 = 947 N, F2 = 2.786W = 1480 N Friday, October 30, 2009 22 What weight can the muscle lift? Maximum tension in muscle = 1840 N Friday, October 30, 2009 23 (Wa acting at centre of gravity of arm) Forces on arm from shoulder joint Calculate torques about the shoulder joint to eliminate Sx, Sy from the torque equations. Friday, October 30, 2009 24 = 1840 N lM = 0.15 sin 13◦ = 0.0337 m 0.15 m = 31 N = 0.28 m = 0.62 m Torques about shoulder joint (axis): 1840 × 0.0337 − 31 × 0.28 −Wd × 0.62 = 0 Wd = 86 N Friday, October 30, 2009 25 = 1840 N = 31 N = 86 N Forces: x : Sx − M cos 13◦ = 0 Sx = 1840 cos 13◦ = 1793 N y : M sin 13◦ + Sy −Wa −Wd = 0 Sy = −1840 sin 13◦ + 31 + 86 = −297 N (shoulder joint exerts a downward force on the arm) Friday, October 30, 2009 26 Forces acting on ladder 50◦ WF = 875 N WL = 355 N A A Ladder is 8 m long. No friction at wall. Friday, October 30, 2009 27