Lecture 17: Torque & Rotational Equilibrium

Questions of Yesterday You 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 bottom c) halfway between top and bottom d) the force is the same over the whole motion 1b) Is the net force doing work on you? a) YES b) NO 2) If the mass of the moon were doubled, what would happen to its centripetal acceleration? a) it would increase b) it would decrease c) it would stay the same

Force & Angular Acceleration  F = ma What 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 object Is that all?  =   t aT=raT=r

F F F FF ANGLE of force with respect to the radial direction affects angular acceleration rr POSITION of force with respect to the axis of rotation affects angular acceleration Force & Angular Acceleration

F F F FF The 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 rotation rr Force & Angular Acceleration

r Torque FTFT FrFr  = ? F The 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 rotation The rate of rotation of an object can’t change unless the object is acted on by a net TORQUE (  ) sin  Fr  = r*Fsin  Torque Distance from axis of rotation Force acting on object Angle between force and displacement from axis of rotation  F

F F F Direction of Force Vector F rr Torque Direction Direction of displacement vector r pointing FROM axis of rotation TO applied force Direction of Rotation depends on

F F F Torque is a VECTOR!! Direction tells you direction of rotation + = Counterclockwise Motion - = Clockwise Motion rr Torque Direction + - +

Right-Hand Rule Curl your FINGERS of your RIGHT hand in direction of MOTION THUMB points in direction of TORQUE Vector CCW (+) Motion -> Torque = OUT of PAGE CW (-) Motion -> Torque = INTO PAGE TORQUE Direction is PERPENDICULAR to plane formed by FORCE and DISPLACEMENT Vectors F F F rr + + -

Right-Hand Rule TORQUE Direction is PERPENDICULAR to plane formed by FORCE and DISPLACEMENT Vectors F F F rr If 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 r 2) CURL your fingers in direction of F 3) THUMB points in direction of 

Torque Does the seesaw rotate? What is the TORQUE on the seesaw plank from each person? What is the NET torque on the seesaw plank? rr FgFg FgFg What is the net Force on the seesaw plank? FPFP mass = m F gs

Torque rr FgFg FgFg FPFP mass = m Force of gravity of a HOMOGENEOUS, SYMMETRIC body acts at the center of the object F gs What is the TORQUE on the seesaw plank from each person? What is the NET torque on the seesaw plank?

Equilibrium Conditions An object at rest or moving at constant linear velocity and/or angular velocity (rotation) is in equilibrium rr FgFg FgFg FPFP mass = m Equilibrium (a = 0,  = 0) conditions  F = 0   = 0 F gs

Equilibrium Conditions Is the system in equilibrium? r/2r FgFg FgFg FPFP mass = m Equilibrium (a = 0,  = 0) conditions  F = 0   = 0 F gs

Equilibrium Conditions rr FgFg FgFg FPFP mass = m Is the system in equilibrium now? F gs

Equilibrium Conditions FgFg FgFg mass = m Is the system in equilibrium now? rSrS   = + F g r - F g r + F gs r s F gs rr FPFP

Equilibrium Conditions Is there a point where the pivot could be placed to make the system in equilibrium? FgFg FgFg mass = m Is the system in equilibrium now? r3r3   = + F g r 1 - F g r 2 + F gs r 3 = 0 F gs r2r2 r1r1 FPFP

Choosing your Axis of Rotation If we chose a different axis of rotation to calculate our net torque…would the system still be in equilibrium? FgFg FgFg mass = m Do the torques exerted on the object depend on the chosen axis of rotation? What about the net torque? r3r3 F gs r2r2 r1r1 FPFP

Choosing your Axis of Rotation If we chose a different axis of rotation to calculate our net torque…would the system still be in equilibrium? FgFg FgFg mass = m Do the torques exerted on the object depend on the chosen axis of rotation? What about the net torque? r3r3 F gs r2r2 r1r1 FPFP axis of rotation

Choosing your Axis of Rotation FgFg FgFg mass = m r3r3 F gs r2r2 r1r1 FPFP axis of rotation The net torque of system is independent of the chosen axis of rotation…choose an axis that makes the problem easiest!!   = +F P r 1 - F g (r 1 + r 2 ) - F gs r 3 = 0

Choosing your Axis of Rotation The net torque of system is independent of the chosen axis of rotation…choose an axis that makes the problem easiest!! FgFg FgFg mass = m r3r3 F gs r2r2 r1r1 FPFP axis of rotation If 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 Rotation A uniform 10.0 N picture frame is supported as shown above. Find the tension in the cords and the magnitude of F F T1T1 T2T N 60 o 40.0 cm 30.0 cm *

Center of Mass rr FgFg FgFg FPFP mass = m Force of gravity of a HOMOGENEOUS, SYMMETRIC body acts at the CENTER (the axis of symmetry) of the object F gs

Center of Mass rr FgFg FgFg FPFP mass = m CENTER OF MASS The point on an extended object where a single force F g = mg can act to represent the force of gravity acting on the entire extended object F gs C.M.

Center of Mass rr FgFg FgFg FPFP mass = m CENTER OF MASS The rotation induced from F g positioned at the center of mass is the same as that induced from all the F g ’s acting on the extended object F gs C.M.

Center of Mass CENTER OF MASS of people + seesaw system Position of axis of rotation where system is in equilibrium FgFg FgFg mass = m rSrS F gs r2r2 r1r1 FPFP C.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 Mass FgFg FgFg mass = m rSrS F gs r2r2 r1r1 C.M. x CM = m 1 x 1 + m 2 x 2 + m 3 x 3 …. m 1 + m 2 + m 3 …. m 1 y 1 + m 2 y 2 + m 3 y 3 …. m 1 + m 2 + m 3 …. y CM = = = ∑m i x i ∑m i ∑m i y i ∑m i

Practice Problem A 20.0-m, 500-N uniform ladder rest against a frictionless wall, making an angle of 60.0 o 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 when the firefighter is 10.0 m up, what is the coefficient of static friction between ladder and ground?

Practice Problem A hungry 500-N bear walks out on a beam in an attempt to retrieve some goodies (50 N) hanging at the end of the beam. The beam is attached to a wall by a hinge and supported from the other end by a cable. The beam is uniform, weighs 200 N and is 10 m long. -Draw a free-body diagram of the beam -When the bear is at x = 1.00 m, find the tension in the wire and the components of the reaction force at the hinge -If the wire can withstand a maximum tension of 1000 N what is the maximum distance the bear can walk before the wire breaks?

Questions of the Day 1) If an object is rotating at a constant angular speed which statement is true? a) the system is in equilibrium b) the net force on the object is ZERO c) the net torque on the object is ZERO d) all of the above 2) Student 1 (mass = m) sits on the left end of massless seesaw of length L and Student 2 (mass = 2m) sits at the right end. Where must the pivot be placed so the system is in equilibrium? a) L/2 b) L/3 from the right (from Student 2) c) L/3 from the left (from Student 1) d) the system cant be in equilibrium