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## rOTATIONAL MOTION

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**What Is Rotational Motion?**What does a yo-yo have in common with a merry-go-round? How can we describe this type of motion?**Linear Motion vs. Rotational Motion**• Linear motion involves an object moving from one point to another in a straight line. • Rotational motion involves an object rotating about an axis. • Examples include a merry-go-round, the rotating earth, a spinning skater, a top, and a turning wheel. • What causes rotational motion? • Does Newton’s second law apply?**For every type of linear quantity we have a rotational**quantity that does much the same thing : Linear Quantities Speed Force Mass Momentum Distance Rotational Quantities Rotational (Angular) Speed Torque Rotational Inertia Angular Momentum Angle**On a merry-go-round, a rider near the edge travels a**greater distance in 1 revolution than one near the center. • The outside rider is therefore traveling with a greater linear speed. • Relationship between linear and rotational velocity V=rω**A merry-go-round is accelerated at a constant rate of 0.005**rev/s2, starting from rest.What is its rotational velocity at the end of 1 min? 0.005 rev/s 0.03 rev/s 0.05 rev/s 0.30 rev/s = 0.005 rev/s2 0 = 0 t= 60 s = 0 + t = 0 + (0.005 rev/s2)(60 s) = 0.30 rev/s**How many revolutions does the merry-go-round make in 1**minute? 1.5 rev 3.0 rev 9.0 rev 18.0 rev = 0.005 rev/s2 0 = 0 t= 60 s, = 0.30 rev/s = 0t + 1/2 t2 = 0 + 1/2 (0.005 rev/s2)(60 s)2 = 9 rev**Torque and Balance**• What causes the merry-go-round to rotate in the first place? • What determines whether an object will rotate? • If an unbalanced force causes linear motion, what causes rotational motion?**Torque and Balance**• When is a balance balanced? • Consider a thin but rigid beam supported by a fulcrum or pivot point. • If equal weights are placed at equal distances from the fulcrum, the beam will not tend to rotate: it will be balanced.**To balance a weight twice as large as a smaller weight, the**smaller weight must be placed twice as far from the fulcrum as the larger weight. • Both the weight and the distance from the fulcrum are important. • The product of the force and the distance from the fulcrum is called the torque. • Torque describes the tendency of a weight to produce a rotation.**The distance from the fulcrum to the point of application of**the force must be measured in a direction perpendicular to the line of action of the force. • This distance is called the lever arm or moment arm. • A longer lever arm produces a greater torque. • For a force F and a lever arm l, the resulting torque is:**A 50-N force is applied at the end of a wrench handle that**is 24 cm long. The force is applied in a direction perpendicular to the handle as shown. What is the torque applied to the nut by the wrench? 6 N·cm 12 N·cm 26 N·cm 120 N·cm 0.24 m 50 N = 12 N·cm**What would the torque be if the force were applied half way**up the handle instead of at the end? 6 N·m 12 N·m 26 N·m 120 N·m 0.12 m 50 N = 6 N·m**When the applied force is not perpendicular to the crowbar,**for example, the lever arm is found by drawing the perpendicular line from the fulcrum to the line of action of the force. • We call torques that produce counterclockwise rotation positive, and torques that produce clockwise rotation negative.**Two forces are applied to a merry-go-round with a radius of**1.2 m as shown. What is the torque about the axle of the merry-go-round due to the 80-N force? +9.6 N·m -36 N·m +96 N·m -36 N·m 1.2 m 80 N = +96 N·m (counterclockwise)**What is the torque about the axle of the merry-go-round due**to the 50-N force? +60 N·m -60 N·m +120 N·m -120 N·m -(1.2 m 50 N) = -60 N·m (clockwise)**What is the net torque acting on the merry-go-round?**+36 N·m -36 N·m +96 N·m -60 N·m +126 N·m • 96 N·m (counterclockwise) • 60 N·m (clockwise) • = +36 N·m (counterclockwise)**We want to balance a 3-N weight against a 5-N weight on a**beam. The 5-N weight is placed 20 cm to the right of a fulcrum. What is the torque produced by the 5-N weight? • +1 N·m • -1 N·m • +4 N·m • +4 N·m F = 5 N = - Fl l= 20 cm = 0.2 m = - (5 N)(0.2 m) = -1 N·m**How far do we have to place the 3-N weight from the fulcrum**to balance the system? • 2 cm • 27 cm • 33 cm • 53 cm F = 3 N l = / F = +1 N·m = (+1 N·m) / (3 N) = 0.33 m = 33 cm**The center of gravity of an object is the point about which**the weight of the object itself exerts no torque. • We can locate the center of gravity by finding the point where it balances on a fulcrum. • For a more complex object, we locate the center of gravity by suspending the object from two different points, drawing a line straight down from the point of suspension in each case, and locating the point of intersection of the two lines.**Rotational Inertia and Newton’s Second Law**• In linear motion, net force and mass determine the acceleration of an object. • For rotational motion, torque determines the rotationalacceleration. • The rotational counterpart to mass is rotational inertia or moment of inertia. • Just as mass represents the resistance to a change in linear motion, rotational inertia is the resistance of an object to change in its rotational motion. • Rotational inertia is related to the mass of the object. • It also depends on how the mass is distributed about the axis of rotation.**Simplest example:a mass at the end of a light rod**• A force is applied to the mass in a direction perpendicular to the rod. • The rod and mass will begin to rotate about the fixed axis at the other end of the rod. • The farther the mass is from the axis, the faster it moves for a given rotational velocity.**Simplest example: a mass at the end of a light rod**• To produce the same rotational acceleration, a mass at the end of the rod must receive a larger linear acceleration than one nearer the axis. • It is harder to get the system rotating when the mass is at the end of the rod than when it is nearer to the axis.**Rotational Inertia and Newton’s Second Law**• The resistance to a change in rotational motion depends on: • the mass of the object; • the square of the distance of the mass from the axis of rotation. • For an object with its mass concentrated at a point: • Rotational inertia = mass x square of distance from axis • I = mr2 • The total rotational inertia of an object like a merry-go-round can be found by adding the contributions of all the different parts of the object.**Rotational Inertia and Newton’s Second Law**• Newton’s second law for linear motion: Fnet = ma • Newton’s second law for rotational motion: • The net torque acting on an object about a given axis is equal to the rotational inertia of the object about that axis times the rotational acceleration of the object. net = I • The rotational acceleration produced is equal to the torque divided by the rotational inertia.**Example: a baton with a mass at both ends**• Most of the rotational inertia comes from the masses at the ends. • A torque can be applied at the center of the rod, producing a rotational acceleration and starting the baton to rotate. • If the masses were moved toward the center, the rotational inertia would decrease and the baton would be easier to rotate.**Some objects have more rotational inertia than others**• Objects with mass closer to axis of rotation are easier to rotate, b/c it they have less rotational inertia • If the mass is farther away from the axis, then object will have more rotational inertia, and will therefore be harder to rotate**Two 0.2-kg masses are located at either end of a 1-m long,**very light and rigid rod as shown. What is the rotational inertia of this system about an axis through the center of the rod? • 0.02 kg·m2 • 0.05 kg·m2 • 0.10 kg·m2 • 0.40 kg·m2 I = mr2 = (0.2 kg)(0.5m)2 x 2 = 0.10 kg·m2**Conservation of Angular Momentum**How do spinning skaters or divers change their rotational velocities?**Angular Momentum**• Linear momentum is mass (inertia) times linear velocity: p = mv • Angular momentum is rotational inertia times rotational velocity: L = I • Angular momentum may also be called rotational momentum. • A bowling ball spinning slowly might have the same angular momentum as a baseball spinning much more rapidly, because of the larger rotational inertia I of the bowling ball.**Conservation of Angular Momentum**Angular momentum is conserved if the net external torque acting on the system is zero.**Angular momentum is conserved by changing the angular**velocity • When the masses are brought in closer to the student’s body, his rotational velocity increases to compensate for the decrease in rotational inertia. • He spins faster when the masses are held close to his body, and he spins more slowly when his arms are outstretched.**Kepler’s Second Law**• Kepler’s second law says that the radius line from the sun to the planet sweeps out equal areas in equal times. • The planet moves faster in its elliptical orbit when it is nearer to the sun than when it is farther from the sun.**Kepler’s Second Law**• This is due to conservation of angular momentum. • The gravitational force acting on the planet produces no torque about an axis through the sun because the lever arm is zero: the force’s line of action passes through the sun.**Kepler’s Second Law**• When the planet moves nearer to the sun, its rotational inertia about the sun decreases. • To conserve angular momentum, the rotational velocity of the planet about the sun must increase.**Riding a bicycle and other amazing feats**Why does a bicycle remain upright when it is moving but promptly falls over when not moving?**Angular momentum is a vector**• The direction of the rotational-velocity vector is given by the right-hand rule. • The direction of the angular-momentum vector is the same as the rotational velocity.**Angular momentum and bicycles**• The wheels have angular momentum when the bicycle is moving. • For straight line motion, the direction of the angular-momentum vector is the same for both wheels and is horizontal. • To tip the bike over, the direction of the vector must change, requiring a torque.**Angular momentum and bicycles**• If the bike is not perfectly upright, a gravitational torque acts about the line of contact of the tires with the road. • As the bike begins to fall, it acquires a rotational velocity and angular momentum about this axis. • If the bike tilts to the left, the change in angular momentum points straight back.**Angular momentum and bicycles**• If the bike is standing still, the gravitational torque causes the bike to fall.**Angular momentum and bicycles**• When the bike is moving, the change in angular momentum caused by the gravitational torque adds to the angular momentum already present from the rotating tires. • This causes a change in the direction of the total-angular momentum vector which can be accommodated by turning the wheel.**A student holds a spinning bicycle wheel while sitting on a**stool that is free to rotate. What happens if the wheel is turned upside down? • To conserve angular momentum, the original direction of the angular-momentum vector must be maintained.**A student holds a spinning bicycle wheel while sitting on a**stool that is free to rotate. What happens if the wheel is turned upside down? • The angular momentum of the student and stool, +Ls, adds to that of the (flipped) wheel, -Lw, to yield the direction and magnitude of the original angular momentum +Lw.**A student sits on a stool holding a bicycle wheel with a**rotational velocity of 5 rev/s about a vertical axis. The rotational inertia of the wheel is 2 kg·m2 about its center and the rotational inertia of the student and wheel and platform about the rotational axis of the platform is 6 kg·m2. What is the initial angular momentum of the system? • 10 kg·m2/s upward • 25 kg·m2/s downward • 25 kg·m2/s upward • 50 kg·m2/s downward L = I= (2 kg·m2)(5 rev/s) = 10 kg·m2/s upward from plane of wheel**If the student flips the axis of the wheel, reversing the**direction of its angular-momentum vector, what is the rotational velocity of the student and stool about their axis after the wheel is flipped? • 1.67 rad/s • 3.33 rad/s • 60 rad/s • 120 rad/s = L / I = (20 kg·m2/s) / (6 kg·m2) = 3.33 rad/s in direction of original angular velocity

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