Lecture 17: Torque & Rotational EquilibriumPowerPoint PresentationSlide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Lecture 17: Torque & Rotational EquilibriumQuestions of YesterdayYou are riding on a Ferris wheel moving at constant speed. 1a) At what point is the net force acting on you the greatest?a) the top b) the bottomc) halfway between top and bottomd) the force is the same over the whole motion 1b) Is the net force doing work on you?a) YESb) NO2) If the mass of the moon were doubled, what would happen to its centripetal acceleration? a) it would increaseb) it would decreasec) it would stay the sameForce & Angular AccelerationF = maWhat causes an object to accelerate?Can an object be accelerating if there is no net force acting on the object?What causes ANGULAR acceleration?Force is needed to change the rate of rotation of an objectIs that all?=taT=rFFFFFANGLE of force with respect to the radial direction affects angular accelerationr rPOSITION of force with respect to the axis of rotation affects angular accelerationForce & Angular AccelerationFFFFFThe rate of rotation of an object can’t change unless a FORCE is applied at a certain ANGLE and at a DISTANCE from the axis of rotationr rForce & Angular AccelerationrTorqueFTFr= ?FThe rate of rotation of an object can’t change unless a NET FORCE is applied at a certain ANGLE and at a DISTANCE from the axis of rotationThe rate of rotation of an object can’t change unless the object is acted on by a net TORQUE ( )sinFr= r*FsinTorqueDistance from axis of rotation Force actingon objectAngle between forceand displacement from axis of rotationFFFFDirection of Force Vector Fr rTorque Direction+-+Direction of displacement vector r pointing FROM axis of rotation TO applied forceDirection of Rotation depends onFFFTorque is a VECTOR!!Direction tells you direction of rotation + = Counterclockwise Motion- = Clockwise Motionr rTorque Direction+-+Right-Hand RuleCurl your FINGERS of your RIGHT hand in direction of MOTIONTHUMB points in direction of TORQUE VectorCCW (+) Motion -> Torque = OUT of PAGECW (-) Motion -> Torque = INTO PAGETORQUE Direction is PERPENDICULAR to plane formed by FORCE and DISPLACEMENT VectorsFFFr r++-Right-Hand RuleTORQUE Direction is PERPENDICULAR to plane formed by FORCE and DISPLACEMENT VectorsFFFr rIf you have a FORCE acting on an object and a specific AXIS of rotation…What is the direction of the TORQUE?1) POINT the fingers of your right hand in the direction of r2) CURL your fingers in direction of F3) THUMB points in direction of TorqueDoes the seesaw rotate?What is the TORQUE on the seesaw plank from each person?What is the NET torque on the seesaw plank?r rFgFgWhat is the net Force on the seesaw plank?FPmass = m mass = mFgsTorquer rFgFgFPmass = m mass = mForce of gravity of a HOMOGENEOUS, SYMMETRIC body acts at the center of the objectFgsWhat is the TORQUE on the seesaw plank from each person?What is the NET torque on the seesaw plank?Equilibrium ConditionsAn object at rest or moving at constant linear velocity and/or angular velocity (rotation) is in equilibriumr rFgFgFPmass = m mass = mEquilibrium (a = 0,  = 0) conditionsF = 0 = 0FgsEquilibrium ConditionsIs the system in equilibrium?r/2 rFgFgFPmass = m mass = mEquilibrium (a = 0,  = 0) conditionsF = 0 = 0FgsEquilibrium ConditionsrrFgFgFPmass = m mass = mIs the system in equilibrium now?FgsEquilibrium ConditionsFgFgmass = m mass = mIs the system in equilibrium now?rS = + Fgr - Fgr + FgsrsFgsrrFPEquilibrium ConditionsIs there a point where the pivot could be placed to make the system in equilibrium?FgFgmass = m mass = mIs the system in equilibrium now?r3 = + Fgr1 - Fgr2 + Fgsr3 = 0 Fgsr2r1FPChoosing your Axis of RotationIf we chose a different axis of rotation to calculate our net torque…would the system still be in equilibrium?FgFgmass = m mass = mDo the torques exerted on the object depend on the chosen axis of rotation? What about the net torque?r3Fgsr2r1FPChoosing your Axis of RotationIf we chose a different axis of rotation to calculate our net torque…would the system still be in equilibrium?FgFgmass = m mass = mDo the torques exerted on the object depend on the chosen axis of rotation? What about the net torque?r3Fgsr2r1FPaxis ofrotationChoosing your Axis of RotationFgFgmass = m mass = mr3Fgsr2r1FPaxis ofrotationThe net torque of system is independent of the chosen axis of rotation…choose an axis that makes the problem easiest!! = +FPr1 - Fg (r1 + r2) - Fgsr3 = 0Choosing your Axis of RotationThe net torque of system is independent of the chosen axis of rotation…choose an axis that makes the problem easiest!!FgFgmass = m mass = mr3Fgsr2r1FPaxis ofrotationIf there is an unknown force F, choose your axis to be at the position of that unknown force (r = 0 ->  = 0)Choosing your Axis of RotationA uniform 10.0 N picture frame is supported as shown above.Find the tension in the cords and the magnitude of FFT1T210.0 N60o40.0 cm30.0 cm*Center of Massr rFgFgFPmass = m mass = mForce of gravity of a HOMOGENEOUS, SYMMETRIC body acts at the CENTER (the axis of symmetry) of the objectFgsCenter of Massr rFgFgFPmass = m mass = mCENTER OF MASSThe point on an extended object where a single force Fg = mg can act to represent the force of gravity acting on the entire extended objectFgsC.M.Center of Massr rFgFgFPmass = m mass = mCENTER OF MASSThe rotation induced from Fg positioned at the center of mass is the same as that induced from all the Fg’s acting on the extended objectFgsC.M.Center of MassCENTER OF MASS of people + seesaw systemPosition of axis of rotation where system is in equilibriumFgFgmass = m mass = mrSFgsr2r1FPC.M.In 2-dimensions the center of mass is defined by an x and y coordinate (a single point in x-y space)Center of MassFgFgmass = m mass = mrSFgsr2r1C.M.xCM = m1x1 + m2x2 + m3x3….m1 + m2 + m3….m1y1 + m2y2 + m3y3….m1 + m2 + m3….yCM = ==∑mixi∑mi∑miyi∑miPractice ProblemA 20.0-m, 500-N uniform ladder rest against a frictionless wall, making an angle of 60.0o with the horizontal.Find the horizontal and vertical forces exerted on the base of the ladder by the Earth when an 1000-N firefighter is 5.0 m from the bottom.If the ladder is just on the verge of slipping