Rotation Chapters 8 and 9 Rotational motion can be most easily approached by considering how it is analogous to linear motion. For each concept in rotational.

Rotation Chapters 8 and 9 Rotational motion can be most easily approached by considering how it is analogous to linear motion. For each concept in rotational motion, there exists a linear analog. Combined with a few "bridge relationships“ between rotational and linear motion, a table of analogies presents a powerful tool for conceptual understanding.

Rigid Body Rotation Axis of Rotation
RIGID BODY is an extended object whose size and shape do NOT change as it moves. Axis of Rotation

Rotational Kinematics
Kinematics is a complete description of how motion occurs without consideration of the causes of motion Chapter 8

Some terms to Describe Rotational Motion
Angular Position, q in radians Direction is + when measured CCW from x-axis - when measured CW from x-axis Angular Displacement, Dq, change in angular position Since for a full circle, s=2pr, 1 rev = 2p rad = 360o 1 rad =57.3o Arc length, s q(rad) = s/r

A A. B. C. D. Example Which particle has angular position 5p/2? y y x

Angular Velocity, w in rad/s, is the velocity of rotation. Direction is along axis of rotation given by the right hand rule (+ is CCW, - is CW) w vT vT Angular velocity Tangential velocity Only for rad

Using Equations as Guide to Thinking
TANGENTIAL ANGULAR Speed Speed If A is 1m from the center and D is 4 m from the center and all take 10 sec to complete a revolution, what are the velocities at A and D D C B A 0.63 0.63 vA = ______m/s vD = ______m/s wA = ______rad/s wD = ______rad/s 2.5 0.63 How do the variables relate to each other?

A bike wheel with diameter 80cm rotates at a rate of 45rpm
What is the angular velocity of the bike tire (in rad/s) ? What is the tangential speed of the tire in m/s? At 90rpm, what is the tangential velocity of the tire? w vT 2 x 1.88 =3.76m/s

Angular Acceleration, a in rad/s2, is the rate of change of the angular velocity. Direction is along axis of rotation, parallel or anti-parallel to w Angular acceleration Tangential acceleration Only for rad

Only if angle unit is in radians Translational or Linear variables Rotational variables

w and a are in same direction
SPEEDING UP w and a are in same direction SLOWING DOWN and a are in opposite direction wf wi wf wi rotating CCW (+): w +, a + rotating CCW (+): w +, a - wi wf wi wf rotating CW (-): w -, a - rotating CW (-): w -, a +

The Vector Nature of Angular Variables
Right-Hand Rule Grasp the axis of rotation with your right hand, so that your fingers circle the axis in the same sense as the rotation. Your extended thumb points along the axis in the direction of the angular velocity vector. Angular acceleration arises when the angular velocity changes, and the acceleration vector also points along the axis of rotation. The acceleration vector has the same direction as the change in the angular velocity.

w aT aC w v a a Problem: Assume the particle is speeding up.
a) What is the direction of the instantaneous velocity, v? b) What is the direction of the angular velocity, w? c) What is the direction of the tangential acceleration, aT? d) What is the direction of the angular acceleration a? e) What is the direction of the centripetal acceleration, ac? f) What is the direction of the overall acceleration, a, of the particle? g) What changes if the particle is slowing down? Tangent to circle Out of page Same direction as v, tangent to circle Same direction as w, out of page Into center of circle aT and a would point anti- parallel to w and a w aT aC v a w a

A jet awaiting clearance for takeoff is momentarily stopped on the runway. As seen from the front of one engine, the fan blades are rotating with an angular velocity of -110 rad/s, where the – sign indicates a CW rotation. As the plane takes off, the angular velocity of the blades reaches -330 rad/s in 14 s. Find the angular acceleration of the blades assuming it to be constant. -16 rad/s2

Linear and Rotational Kinematics
Equations for Constant Angular Acceleration Rotational Motion (a = constant) Linear Motion (a = constant)

Applying the Equations of Rotational Kinematics
Draw a picture to represent the system if necessary. 2. Select a coordinate system to analyze your system with the x-axis along the initial angular position. Decide which direction of rotation will be + and which – (convention is + for CCW and – for CW). Keep directions consistent for the problem. 3. List the variables that are given 4. Write down the kinematic equation that will be used to solve the problem. 5. Isolate the unknown variable. 6. Solve for relevant unknowns by putting in numbers. Keep in mind that you may sometimes have more than one unknown and therefore need the same number of equations as unknowns.

Example: The blades of a ceiling fan start from rest and, after two revolutions, have an angular speed of 0.50 rev/s. The angular acceleration of the blades is constant. What is the angular acceleration of the fan? What is the angular speed after eight revolutions?

The blade of a lawn mower is rotating at an angular speed of 17 rev/s
The blade of a lawn mower is rotating at an angular speed of 17 rev/s. The tangential speed of the outer edge of the blade is 32 m/s. What is the radius of the blade?

NONuniform Circular Motion
Magnitude of the velocity (speed) is NOT constant and the direction of the velocity continuously changes as object moves around in a circle aT and a due to change of velocity magnitude or speed. They are due to forces acting in the line of motion. ac is due to change in direction of velocity. It is due centripetal force. aT = ra v2 v1 ac a Dq r w

Magnitude of the velocity (speed) is NOT constant and the direction of the velocity continuously changes as object moves around in a circle v1 ac r aT = ra Dq j a v2 Acceleration and net force vectors do not point to the center in nonuniform circular motion. They have both radial (centripetal) and tangential components. w

Example: A 220kg speedboat is making a circular turn (radius 32m) around a buoy. During the turn, the engine applies a net tangential force of 550N to the boat. The initial tangential speed of the boat is 5m/s. a) Find aT. b) After 2s into turn, find ac. aT F v0 ac0 r Dq ac aT v ac=3.1 m/s2

Rotational Dynamics Chapter 9
Dynamics is a complete description of WHY motion occurs

Rotation of a Rigid Body
RIGID BODY is an extended object whose size and shape do NOT change as it moves. All points in object move along different paths around rotation axis. One point – center of mass – follows the path of a point particle. Object must be treated as an EXTENDED object All points in object move along parallel paths. Object can be treated as a point.

2F F F F F F Rotational Dynamics q Linear Motion Angular Motion
What causes rotational motion to change? What causes angular acceleration? 2F Linear Motion FORCE, F, causes acceleration F F F F F=Fsinq q F No torque No torque Angular Motion TORQUE, t, causes angular acceleration

Distance to axis of rotation
Rotational Dynamics What causes rotational motion to change? What causes angular acceleration? Linear Motion - FORCE, F, causes acceleration Angular Motion - TORQUE, t, causes angular acceleration Units of Torque: N m Distance to axis of rotation Angle between r and F Magnitude of Force TORQUE is a VECTOR + when it produces CCW rotation about the axis - when it produces CW rotation about the axis

A wheel turns freely on an axle at the center
A wheel turns freely on an axle at the center. Which one of the forces shown will provide the largest positive torque? Negative torque? E A r r/2 2F F A B C D E

Problem: Luis uses a 20cm long wrench to tighten a nut
Problem: Luis uses a 20cm long wrench to tighten a nut. The wrench handle is tilted 30o above the horizontal and Luis pulls straight down on the end with a force of 100 N. How much torque does Luis exert on the nut? F// r=20cm q F=Fsinq 30o F

(Translational AND Rotational)
Static Equilibrium (Translational AND Rotational)

Torque and Static Equilibrium
Object at rest is in static equilibrium Linear Motion Rotational Motion

Static Equilibrium of a Rigid Body
A rigid object at rest is in static equilibrium – it has zero translational acceleration and zero angular acceleration. The net torque about ANY AXIS is zero if a rigid object is in rotational equilibrium.

Applying Equilibrium Conditions to a Rigid Body
Select object. 2. Draw a Free Body Diagram. We are now dealing with extended objects and the position of the forces are important. 3. Choose a convenient x- and y- axis and resolve forces into their x- and y-components 4. Apply Newton’s 2nd Law for equilibrium to each direction: Fx = 0 and Fy = 0 . 5. Select a convenient axis of rotation or pivot point. The rotation axis is arbitrary since object is in equilibrium with respect to ANY axis 6. Identify the point where each force acts and calculate the torque produced about the axis of rotation. Set t = 0. \ 6. Solve equations in 4 and 6 to determine unknowns

SEESAW or PLANK PROBLEM: A woman whose weight is 530 N is poised at the right end of a diving board with a length of 3.90 m. The board has negligible weight and is bolted down at the left end, while being supported 1.40 m away by a fulcrum. Find the forces F1 and F2 that the bolt and the fulcrum, respectively, exert on the board F2 1.40m 3.90m F1 W Choose Axis (r=0)

Center of Gravity The center of mass of an object or system is the point at which all the WEIGHT can be assumed to reside. Sometimes the system is an assortment of particles and sometimes it is a solid object. Mathematically, you can think of the center of mass as a “weighted average”. If gravitational field is uniform throughout object, then CM and CG are the same

STABLE UNSTABLE Center of Gravity, TORQUE and Equilibrium FN FN W W
In extended objects or a group of objects, force of gravity acts at the CG so the weight of an object can produce a torque CG plays important role in determining whether an object or group of objects remains in equilibrium. WHEN THE CG is in line with a pivot point, then the object or system is in equilibrium. CG W FN CG W FN STABLE UNSTABLE

STABLE STABLE Center of Gravity, TORQUE and Equilibrium FN FN W Wb W
In extended objects, force of gravity acts at the CG so the weight of an object can produce a torque CG plays important role in determining whether an object or group of objects remains in equilibrium. WHEN THE CG of a system is in line with a pivot point, then the system is in equilibrium. CG of the system of boy and seesaw is supported at fulcrum CG W FN CG W FN CG CG Wb STABLE STABLE

Some Common Equilibrium Problems
Plank or See Saw problems 2. Shelf or Crane problems: beam sticking out of the wall 3. Ladder Problems

LADDER PROBLEM: An 8.00-m ladder of weight WL = 355 N leans against a smooth vertical wall. The term “smooth” means that the wall can exert only a normal force directed perpendicular to the wall and cannot exert a frictional force parallel to it. A firefighter, whose weight is WF = 875 N, stands 6.30 m from the bottom of the ladder. Assume that the ladder’s weight acts at the ladder’s center and neglect the hose’s weight. Find the forces that the wall and the ground exert on the ladder.

FN1 WF WL fS FN2 rF=6.3m rL=4.0m Pivot r=0

SHELF PROBLEM: A safe of mass 430kg hangs by a rope from the end of a boom that is 3.0 m long. The boom consists of a hinged beam and a horizontal cable that connects the beam to a wall. The beam is uniform and has a mass of 85 kg; the mass of the cable and rope are negligible. What is the tension in the cable? WS WB T1 T2 // r=3.0m 37o FHy rB=1.5m FHx Pivot r=0

SHELF PROBLEM: What is the tension in the cable
SHELF PROBLEM: What is the tension in the cable? Find the vertical and horizontal hinge forces 37o FHx FHy rB=1.5m r=3.0m WS WB T1 T2 // Pivot r=0

Which of these objects is in static equilibrium?
D A. B. C. D.

A square piece of plywood on a horizontal tabletop is subjected to the two horizontal forces shown at right. Where should a third force of magnitude 5 N be applied to put the piece of plywood into equilibrium? A

Problem: A cat walks along a uniform plank that is 4
Problem: A cat walks along a uniform plank that is 4.00 m long and has a mass of 7.00 kg. The plank is supported by two sawhorses, one m from the left end of the board and the other 1.50 m from its right end. When the cat reaches the right end, the plank just begins to tip. What is the mass of the cat?

Rolling Motion Combination of rotation and translation
vCM vT = wR vT vCM + pure ROTATION pure TRANSLATION When an object rolls without slipping, there is a relationship between the rotational and translational motion. When there is slidding or slipping, there is NO relationship between the rotational and translational motions. R

Rolling Motion (without slipping)
Rotational speed Linear speed Rotational acceleration Linear acceleration DxCM = s = rq

Rolling Motion Combination of rotation and translation
Rolling and Slipping CM Travels more than 1 circumference for every full rotation No relationship between the translational motion (vCM) and rotational motion (Rw) dCM > Rq vCM > Rw aCM > Ra Rolling without Slipping CM Travels 1 circumference for every full rotation There is a relationship between the translational motion (vCM) and rotational motion (Rw) dCM = Rq vCM = Rw aCM = Ra

ROLLING WITHOUT SLIPPING Velocity at any point on a rolling object
Rw=vT vT vCM= vT vCM vCM = -vT 2vCM vCM + = Velocity at point in pure ROTATION (in CM ref frame) Velocity at point in pure TRANSLATION (relative to ground) Velocity at point in COMBINED ROLLING MOTION (relative to ground)

C v=2vCM vT vCM vCM vCM v vT (In CM frame) A. B. C. D. E. P
A wheel rolls without slipping. Which is the correct velocity vector for point P on the wheel? C v=2vCM vCM P vT (In CM frame) vCM vCM v vT (In CM frame) A B C D E.

Example An Accelerating Car: An automobile starts from rest and for 20
Example An Accelerating Car: An automobile starts from rest and for 20.0 s has a constant linear acceleration of m/s2 to the right. During this period, the tires do not slip. The radius of the tires is m. At the end of the 20.0-s interval, what is the angle through which each wheel has rotated?

Rotational Dynamics and
Newtons 2nd Law What is the RELATIONSHIP between TORQUE and ANGULAR ACCELERATION?

NONuniform Circular Motion and Newtons 2nd Law
FT causes tangential acceleration; it causes a change in the speed or magnitude of the velocity. Fr changes only the direction of velocity. It does not affect the speed; it does no work. It produces no torque. w FT F m Fr r

NONuniform Circular Motion and Newtons 2nd Law
FT F (Only if angle unit is radians) Fr r This is the relationship between torque and angular acceleration for a single particle moving in circular motion. To look at rotational dynamics, must expand idea to an extended rigid object.

Rotational Motion and Newtons 2nd Law
ALL pts on rigid body rotate at same a t1 = r1FT1 = r1m1aT1 = m1r12a w t2 = r2FT2 = r2m2aT2 = m2r22a t3 = r3FT3 = r3m3aT3 = m3r32a F2 m1 r1 If we did this for ALL particles in the object, the net torque on the entire object would be F1 r2 m2 r3 F3 m3 Rotational Inertia, I

Rotational Motion and Newtons 2nd Law
ALL pts on rigid body rotate at same a w Rotational Inertia, I F2 m1 Newtons 2nd Law for Rotation of a Rigid Body r1 F1 r2 m2 r3 (angle unit must be in radians) F3 m3 Rotational equivalent of FORCE Rotational equivalent of MASS

Rotational Inertia (Moment of Inertia)
ri is the perpendicular distance between the mass and the rotation axis. (Units are kg m2) Moment of inertia is the rotational equivalent of mass. Rotational Inertia of an object depends on the mass AND on how the mass is distributed about an axis of rotation. It is constant for a particular rigid body and a particular axis. Both involve same mass. Left is easier to rotate since the mass is distributed closer to the rotation axis.

ri is the perpendicular distance between the mass and the rotation axis. (Units are kg m2) Which rod has a greater Moment of Inertia (is harder to rotate)? I depends on where the axis of rotations is Both involve same mass. Top is easier to rotate since the mass is distributed closer to the rotation axis. Smaller I (easier to rotate) Larger I (harder to rotate)

Problem: Two 3kg masses are at the ends of a 4m bar with negligible mass. What is the moment of inertia of the object when w 2m 3kg a) when the masses rotate around the center of the rod 3kg w 4m 3kg b) when the masses rotate around one end of the rod

Three small spheres that rotate about a vertical axis are shown. The perpendicular distance between the axis of rotation and center of mass of each sphere is given. Rank the 3 spheres from greatest to least rotational inertia. w m1=36kg 1 m m2=9kg 2 m m3=4kg 3 m

ri is the perpendicular distance between the mass and the rotation axis. (Units are kg m2) For a solid object

Moments of Inertia of Common Shapes

Linear Dynamics Rotational Dynamics
Net Force (N): Fnet Net Torque (N m): tnet Mass (kg): m Moment of Inertia (kg m2): I Acceleration (m/s2): a Angular Acceleration (rad/s): a Newtons 2nd Law Angular acceleration is caused by torques. The larger the moment of inertia, the smaller the angular acceleration. Acceleration is caused by forces. The larger the mass, the smaller the acceleration.

Using Newton’s 2nd Law for Rotation
In the Scottish game of caber toss, contestants toss a heavy uniform pole, landing it on its end. A 5.9m pole with a mass of 79kg has just landed on its end. It is tipped 25o from the vertical and is starting to rotate about the end that touches the ground. A) Find the angular acceleration. B) What is the tangential acceleration of the left end of the log? 25o Pivot r=0 FN Fg

Ropes and Pulleys Linear Motion Angular Motion T3 T2 T1 w a
Frictionless, massless pulley Pulley does NOT rotate Rope slides over frictionless pulley Rope tension does NOT change when it passes over a frictionless, massless pulley Pulley has mass and friction Pulley rotates If pulley TURNS WITHOUT THE ROPE SLIPPING ON IT, then vrope=vT pulley=Rw arope = aTpulley=Ra Rope tension CHANGES when it passes over a rotating pulley with friction. T3 T2 w R T1 / a Pulley Rotation At rim, vT = Rw aT = Ra Rope Translation vrope = Rw arope = Ra

vobject = Rw aobject = Ra
Ropes and Pulleys T3 T2 w R a Motion constraints for an object connected to a pulley of radius R by a nonslipping rope vobject = Rw aobject = Ra Note that these are magnitudes only

Pulley Block F1 Mg FT mg +t a M R a + m
A light string is wrapped around a uniform pulley (disk) of mass M=2.5kg and radius R=20cm. Suspended from the end of the string is a block of mass of m=1.2kg. Find the acceleration of the falling block, the angular acceleration of the pulley and the tension in the string. The string does not slip and there is no friction in the axle. If the block starts at rest, how many revolutions does the pulley in 2s? Block Pulley mg FT Mg F1 / // +t a M R + a m

A light string is wrapped around a uniform pulley (disk) of mass M=2
A light string is wrapped around a uniform pulley (disk) of mass M=2.5kg and radius R=20cm. Suspended from the end of the string is a block of mass of m=1.2kg. Find the acceleration of the falling block, the angular acceleration of the pulley and the tension in the string. The string does not slip and there is no friction in the axle. If the block starts at rest, how many revolutions does the pulley in 2s? mg FT Mg F1 / // M R Pulley m

Rotational Work and Energy

Work When a force acts on an object over a distance, it is said that work was done upon the object. Work tells us how much a force transfers energy to a system. Work done by a constant force that points in the same direction as the displacement

Rotational Work Constant force pulls rope out a distance, s
FT Constant force pulls rope out a distance, s s r FT q Wheel rotates through the angle q = s/r Work done by a force Work done by a torque Unit: J must be in rad

Rotational Kinetic Energy
ALL pts on rigid body rotate at same w K1 = ½m1v1T2 = ½m1r12w2 w K2 = ½m2v2T2 = ½m2r22w2 K3 = ½m3v3T2 = ½m3r32w2 vT2 m1 vT1 r1 If we did this for ALL particles in the object, the total rotational kinetic energy would be r2 m2 r3 m3 vT3 Rotational Inertia, I

Kinetic Energy Translational Kinetic Energy Rotational Kinetic Energy
(angle unit must be in radians) Rotational Kinetic Energy Rotational equivalent of MASS

Power Translational Rotational

s h w Roll without slip vCM=Rw
A solid disk and a hoop with mass M and radius R start from rest and roll without slipping down an incline. What is the velocity at the bottom in terms of h? w Disk Hoop Sphere Spherical Shell h s q System is isolated, WNC=0 since FN and fS do no work on object Disk Roll without slip vCM=Rw Disk Hoop

Find the acceleration of an object rolling down an inclined plane
Find the acceleration of an object rolling down an inclined plane. The object rolls without slipping. It has a mass of M and radius R. +a,t Torque around Rotation axis at CG FN fS // q Fgsinq // +a Fgcosq Fg q Disk Hoop Sphere Spherical Shell

Which wins the RACE? 4 different spherical objects with the same mass and radius are released from rest and roll down an inclined plane. The objects roll without slipping. Rank the objects according to the order that they reach the finish line C < A < D < B Spherical Shell Disk Hoop Sphere Spherical Shell Sphere D C Hoop B Disk A

A single particle moving in circular motion has rotational kinetic energy and angular momentum and can experience a torque if a force acts tangentially. r w F FT Fr vT (angle unit is radians) Moment of inertia of single particle moving in a circle of radius r is mr2 (all mass is traveling at r)

Angular Momentum

(angle unit must be in radians)
Momentum Linear momentum Angular Momentum (angle unit must be in radians) Use turntable and solid disk demo. Spinning turntable has angular momentum. There is a net torque that slows it down If I drop this solid disk onto the spinning turntable DROP it … what happens The disk was not rotating initially, had no ang momentum. After, this is a collision, it gained L so what happened to L of turntable? It lost L (no ext torque) Changed I of system L same before and after. W got smaller so I sys got bigger. (if drop further from axis the I even bigger, but hard to calc because I of disk around axis hard to calc) There are many situations where I naturally changes and that results in change in w. Ask for examples I goes up and w goes down and vece versa If there are no external forces on a system of particles, momentum of the system is conserved If there are no external torques on a system of particles, angular momentum of the system is conserved

Conservation of Angular Momentum

Problem: Suppose an empty “carousel" having a radius, R of 1
Problem: Suppose an empty “carousel" having a radius, R of 1.8 m and a mass, M of 50 kg (I = ½mr2) is initially rotating on a frictionless axis counterclockwise at 15 rev/min. before after Simultaneously, four students walk up and place four 5-kg boxes (I = mr2) symmetrically along its outer edge so that each box's center of mass is located 1.50 m from the axis of the carousel. What will become the carousel's new angular velocity at the instant the boxes are in place?

Problem: Suppose an empty “carousel" having a radius, R of 1
Problem: Suppose an empty “carousel" having a radius, R of 1.8 m and a mass, M of 50 kg (I = ½mr2) is initially rotating on a frictionless axis counterclockwise at 15 rev/min. before after Suppose instead that they had placed the boxes so that each box's center of mass was located 50 cm from the axis of the merry-go-round, what would have been its new angular velocity?

Linear and angular momentum are SEPARATELY CONSERVED
One form of momentum does NOT transform into the other form p = 0 L ≠ 0 L = 0 w p=mv Linear momentum of the system is NOT conserved since there is an external force on the system (friction between turntable and surface). Linear momentum gets transferred to the Earth Angular momentum of the system IS conserved since there are no external torques on the system (the axis provides no torque). The angular momentum of the bullet transfers to the turntable.

Angular Momentum of a Point Particle
In general, a single particle moving in any way has angular momentum relative to an axis point, O O r Radius is closest distance to rotation axis, r r q v

Ballistic Pendulum w’ Linear momentum of the system is NOT conserved since the pendulum is fixed to the earth at the axis of rotation and this provides and external force. Linear momentum gets transferred to the Earth Angular momentum of the system IS conserved since there are no external torques on the system (the axis provides no torque). The angular momentum of the ball gets transferred to the pendulum.

M Ballistic Pendulum m, v w’ d
Then if want to know how high it swings or what angle then use conser of energy

Angular Momentum in Astronomy
L=mvR The Angular momentum of the orbiting object is the same at every point on the orbit since the force of gravity produces no external torque. Therefore the orbiting object travels faster when it is closer to the rotation axis and slower when it is further away. This is Kepler’s 3rd Law L=mvR

s, d q v w a t = Ia Linear Dynamics Rotational Dynamics
Displacement s, d q Velocity v w Acceleration a Cause of acceleration Fnet, net force tnet, net torque Inertia m, Mass I, Moment of Inertia Newton’s 2nd Law F = ma t = Ia Work W = F//s WR = tq Kinetic Energy K = ½mv2 K R= ½Iw2 Momentum p = mv L = Iw

A baseball catcher sits on a rotating stool. He reaches out 85.0cm to
catch a 40.0m/s fastball. After catching the ball he spins at a rate of 60.0rpm. His mass is 80.0kg and the mass of the ball is 150g. Find the rotational inertia of the catcher and the stool. 85 cm 0.81 kg m2

Rotation Chapters 8 and 9 Rotational motion can be most easily approached by considering how it is analogous to linear motion. For each concept in rotational.

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Rotation Chapters 8 and 9 Rotational motion can be most easily approached by considering how it is analogous to linear motion. For each concept in rotational.

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Presentation on theme: "Rotation Chapters 8 and 9 Rotational motion can be most easily approached by considering how it is analogous to linear motion. For each concept in rotational."— Presentation transcript:

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