Presentation on theme: "Torque Looking back Looking ahead Newton’s second law."— Presentation transcript:
1Torque Looking back Looking ahead Newton’s second law. The mathematics of circular motion.Looking aheadApply rigid-body model to extended object.Understand the rotation of a rigid body around a fixed axis.1
2Torque Definition ¿ = +- r Ftangential = r F sin Á Rotational effect of force. Tells how effective force is at twisting or rotating an object.¿ = +- r Ftangential = r F sin ÁUnits N mSign, CCW rotation is positiveDepends on 3 properties:1) Magnitude of force F.2) Its distance r from pivot point.3) Angle Á.
3Concept QuizThe picture below shows three different ways of using a wrench to loosen a stuck nut. Assume the applied force F is the same in each case.In which of the cases is the torque on the nut the biggest?A. Case 1 B. Case 2 C. Case 3CORRECT10
4Concept QuizThe picture below shows three different ways of using a wrench to loosen a stuck nut. Assume the applied force F is the same in each case.In which of the cases is the torque on the nut the smallest?A. Case 1 B. Case 2 C. Case 3CORRECT12
5Torque Example and Quiz A person raises one leg to an angle of 30 degrees. An ankle weight (89 N) attached a distance of 0.84 m from her hip. What is to torque due to this weight?1) Draw Diagram2) t = F r sin q= F r sin(90 – 30)If she raises her leg higher, the torque due to the weight willA) IncreaseB) SameC) Decrease30F=89 Nr = 0.84 m= 65 N m17
6Equilibrium QuizA rod is lying on a table and has two equal but opposite forces acting on it. What is the net force on the rod?A) Up B) Down C) ZeroWill the rod move? A) Yes B) NoY direction: S Fy = may+F – F = 0yxFYes, it rotates!F21
7Equilibrium Conditions for Equilibrium S F = 0 Translational EQ S t = Rotational EQCan choose any axis of rotation…. Choose Wisely!Steel from podium25
8Equilibrium ExampleA meter stick is suspended at the center. If a 1 kg weight is placed at x=0. Where do you need to place a 2 kg weight to balance it?A) x = 25 B) x=50 C) x=75 D) x=100E) 1 kg can’t balance a 2 kg weight.Steel from podium9.8 N17.6 N50 cmdS t = 09.8 (0.5) – (17.6)d = 0d = 25Balance Demo25
9Static Equilibrium and Center of Mass pivotdGravitational Force Weight = mgActs as force at center of massTorque about pivot due to gravity:¿ = WrsinÁ =mgdObject not in static equilibriumW=mg
10Static Equilibrium Not in equilibrium Equilibrium Center of masspivotdW=mgTorque about pivot 0Center of masspivotTorque about pivot = 0Not in equilibriumEquilibriumA method to find center of mass of an irregular object
11Equilibrium Quiz and Example The picture below shows two people lifting a heavy log. Which of the two people is supporting the greatest weight?1. The person on the left is supporting the greatest weight2. The person on the right is supporting the greatest weight3. They are supporting the same weightCORRECTLook at torque about center:FR L – FL L/2 = 0FR= ½ FLmgFLFRL/2L34
12ExampleA 75 kg painter stands at the center of a 50 kg 3 meter plank. The supports are 1 meter in from each edge. Calculate the force on support A.1 meter1 meterABFAFBMgmg1) Draw FBD2) SF = 03) Choose pivot4) St = 01 meter0.5meterPainter homework problem. Get demo here have man stand at several places calculate force on each support. Right above support, middleFA + FB – mg – Mg = 0-FA (1) sin(90)+ FB (0) sin(90) + mg (0.5)sin(90) + Mg(0.5) sin(90) = 0FA = 0.5 mg Mg = Newtons40
13Concept QuizIf the painter moves to the right, the force exerted by support AA) Increases B) Unchanged C) Decreases1 meter1 meterABFAFBMgmgPainter homework problem. Get demo here have man stand at several places calculate force on each support. Right above support, middle¿A also decreases¿M decreases43
14ExampleHow far to the right of support B can the painter stand before the plank tips?1 meter1 meterABJust before board tips, force from A becomes zeroFBMgmg1) Draw FBD2) SF = 03) Choose pivot4) St = 0Painter homework problem. Get demo here have man stand at several places calculate force on each support. Right above support, middleFB – mg – Mg = 00.5meterxFB (0) sin(90) + mg (0.5)sin(90) – Mg(x) sin(90) = 00.5 m = x M47
15Summary Torque = Force that causes rotation Equilibrium t = F r sin q S F = 0S t = 0Can choose any axis.
16Torque and rotational motion Torque is the quantity that changes rotational motion.For rotations, it is the object that replaces force as the cause of angular acceleration.
17Torque and Rotational Motion Linear motion: F = ma (Newton’s 2nd law linear motion)If force is tangential to a circular path:F=Ft and a = at=r®Rotational motion:¿ = Ft r = m atr=(mr2)® =I®(Newton’s 2nd law rotational motion)Extended object:I = mr2Ftr
18Moment of inertia I = i miri2 Net torque leads to angular acceleration, but we need the moment of inertia to tell us how large an acceleration results.Moment of inertia in rotational problems takes the place that mass has in translational problems.Depends upon the mass and location of the mass from the axis of rotation.Large moment of inertia requires large torque to accelerate object (just like large mass requires large force)
19Concept QuizConsider a rod of uniform density with an axis of rotation through its center and an identical rod with the axis of rotation through one end. Which has the larger moment of inertia?A IC > IEB IC < IEC IC = IE
20Concept Quiz A: IA > IB B: IA < IB C: IA = IB Consider two masses, each of size 2m at the ends of a light rod of length L with the axis of rotation through the center of the rod. The rod is doubled in length and the masses are halved. What happens to I? A: IA > IBB: IA < IBC: IA = IB
21Concept TestConsider a solid disk with an axis of rotation through the center (perpendicular to the diagram).Two holes are cut out near the center and the material is placed near the rim. Which has the larger I? A IA > IBB IA < IBC IA = IB
22Concept TestA mass m hangs from string wrapped around a pulley of radius R. The pulley has a moment of inertia I and its pivot is frictionless. Because of gravity the mass falls and the pulley rotates.The magnitude of the torque on the pulley is..A: greater than mgRB: equal to mgRC: less than mgR (Hint: Is the tension in the string = mg?)
23Acceleration of the mass m? ¿ = Ts R= Icyl ®® = T_sR/ (0.5MR^2)= 2Ts/MRTension of the string from linear motion of block:Ts-mg=mayay = at = - ® R® = -ay/R = -(Ts-mg)/(mR)Ts = -mR ®+mg® = 2Ts/MR = 2(-mR® +mg)R/MR = 2(mR®+mg)/M