Textbook Reference: Wilson, Buffa, Lou: Chapter 8 Glencoe Physics: Chapter 8

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1 AP Physics Rotational Motion Introduction: Which moves with greater speed on a merry-go-round - a horse near the center or one near the outside? Your answer probably depends on whether you are considering the translational or rotational motion of the horses. Have you ever linked arms with friends at a skating rink while making a turn? If you have, you probably noticed that the person on the inside moved very little while the person on the outside had to run to keep up. The outside person traveled a greater distance per period of time and therefore had the greater translational speed. During the same period of time all skaters rotated through the same angle per period of time and had the same rotational speed. In our previous study of motion we discussed translational motion - that is the motion of bodies moving as a whole without regard to rotation. In this unit we will extend our ideas of motion to include the rotation of a rigid body about a fixed axis. If the axis is inside the body we tend to say the body rotates about its axis. If the axis is outside the body, we say the body revolves about an axis. An example of this would be the earth which daily rotates about its axis and yearly revolves around the sun. An object rotating about an axis tends to remain rotating about the same axis unless acted upon by a net external influence. This property of a body to resist changes in its rotational state is called rotational inertia. The rotational inertia of a body depends on the amount of mass the body possesses and on the distribution of that mass with respect to the axis of rotation. The greater the distance of the bulk of the mass from the axis of rotation - the greater the rotational inertia. A long pendulum has a greater rotational inertia than a short one. The period of a pendulum is directly proportional to the square root of the length of the pendulum. It takes more time to change the rotational inertia of a long pendulum as it swings back and forth. People and animals with long legs tend to walk with slower strides than those with short legs for the same reason. Have you ever tried running with your legs straight? Performance Objectives: Upon completion of the readings and activities of the unit and when asked to respond either orally or on a written test, you will be able to: state relationships between linear and angular variables. recognize that the rotational kinematics formulas are analogous to the translational ones. Use these formulas to solve problems involving rotating bodies. define rotational inertia or moment of inertia. Calculate the rotational inertia for a point mass, a system of point masses, and rigid bodies. calculate the kinetic energy of a rotating body. define torque. Calculate the net torque acting on a body. state Newton s second law for rotation. Recognize that the rotational dynamics formulas are analogous to the translational ones. Use these formulas to solve problems involving rotating bodies. use the work-kinetic energy theorem for rotation to solve problems. Textbook Reference: Wilson, Buffa, Lou: Chapter 8 Glencoe Physics: Chapter 8 "To every thing -- turn, turn, turn there is a season -- turn, turn, turn and a time for every purpose under heaven." -- The Byrds (with a little help from Ecclesiastes) Recall: From the definition of a radian (arc length/radius) θ = s/r, where s is the arc length, r is the radius and θ is the angle measure in radians. The following quantities are called the bridges between linear and angular measurements: s = rθ v = rω a T = rα a R = v 2 /r = rω 2 Definitions and Conversions: 1. What angle in radians is swept out by an arc 3.0 m in length, on the circumference of a circle whose radius is 2.0 m? 1.5 rad

2 2. What angle in radians is swept out by an arc of length cm on the circumference of a circle of diameter cm? What is the angle in degrees? 1.57 rad The angle between two radii of a circle of radius 2.00 m is 0.60 rad. What length of arc is intercepted on the circumference of the circle by the two radii? 1.2 m 4. What is the angular velocity in radians per second of a flywheel spinning at the rate of 7230 revolutions per minute? 757 rad/sec 5. If a wheel spins with an angular velocity of 625 rad/s, what is its frequency in revolutions per minute? 5968 rpm 6. Compute the angular velocity in rad/s, of the crankshaft of an automobile engine that is rotating at 4800 rev/min. 503 rad/sec Rotational Kinematics: Rotational motion is described with kinematic formulas just like the translational motion formulas. To get the rotational kinematic formulas, substitute the rotational variables. 7. A flywheel accelerates uniformly from rest to an angular velocity of 94 radians per second in 6.0 seconds. What is the angular acceleration of the flywheel in radians per second squared? 16 rad/s 2 8. a) Calculate the angular acceleration in radians per second squared of a wheel that starts from rest and attains an angular velocity of 545 revolutions per minute in 1.00 minutes. b) What is the angular displacement in radians of the wheel during the first minutes? c) During the second minutes? 0.95 rad/s rad rad 9. Find the angular displacement in radians during the second 20.0 second interval of a wheel that accelerates from rest to 725 revolutions per minute in 1.50 minutes? 506 radians 10. A fly wheel requires 3.0 seconds to rotate through 234 rad. Its angular velocity at the end of this time is 108 rad/s. Find a) the angular velocity at the beginning of the 3 second interval; b) the constant angular acceleration. 48 rad/s 20.0 rad/s a) A cylinder 0.15 m in diameter rotates in a lathe at 750 rpm. What is the tangential velocity of a point on the surface of the cylinder? b) The proper tangential velocity for machining cast iron is about 0.60 m/s. At how many rpm should a piece of stock 0.05 m in diameter be rotated in a lathe? 5.89 m/s 229 rev/min 12. Find the required angular velocity of an ultracentrifuge in rpm for the radial acceleration of a point 1.00 cm from the axis to equal 300,000g (that is 300,000 times the acceleration of gravity) x 10 5 rev/min 13. A wheel rotates with a constant angular velocity of 10 rad/s. a) Compute the radial acceleration of a point 0.5 m from the axis using the relation, a = rω 2. b) Find the tangential velocity of the point, and compute its radial acceleration from the relation, a c = v 2 /r. 50 m/s 2 5m/s 50 m/s 2 Rotational Inertia is the resistance of a rotating body to changes in its angular velocity. According to Newton's First Law a body tends to resist a change in its motion. The amount of inertia a body possesses is directly related to the mass. For rotational motion, an analogous situation exists. However, rotational inertia depends on the mass and on the distribution of the mass about the axis of rotation. This quantity that relates mass and position of the mass relative to the axis of rotation is called the moment of inertia and has units of kg-m 2. The symbol for moment of inertia is I. For a point mass m a distance r from the axis of rotation, the moment of inertia will be I = mr 2. For bodies made up of several small masses just add all the moments of inertia together. For bodies which

3 are not composed of discrete point masses but are continuous distributions of matter, the methods of calculus must be used to find the moment of inertia. The moments of inertia for a few simple but important rigid bodies of uniform composition are listed in the chart below. 14. Small blocks, each of mass 2.0 kg, are clamped at the ends and at the center of a light rod 1.2 m long. Compute the moment of inertia of the system about an axis passing through a point one-third of the length from one end of the rod if the moment of inertia of the light rod can be neglected kg-m Four small spheres, each of mass kg, are arranged in a square m on a side and connected by light rods of negligible mass. Find the moment of inertia of the system about an axis a) perpendicular to the plane of the square through the center kg-m 2 b) bisecting two opposite sides of the square kg-m What is the rotational inertia of a solid ball 0.50 min radius that weighs 80.0 N if it is rotated about a diameter? kg-m What is the rotational inertia of a thick ring that is rotating about an axis perpendicular to the plane of the ring passing through its center? The ring has a mass of 1.20 kg and a diameter of 45.0 cm. The hole in the ring is 15.0 cm wide kg-m A wagon wheel is constructed as shown in the figure. The radius of the wheel is m and the rim has a mass of 1.20 kg. Each of the eight spokes, which lie along a diameter and are m long has a mass of kg. What is the moment of inertia of the wheel about an axis through its center and perpendicular to the plane of the wheel? kg-m 2

4 Kinetic Energy of Rotation: Because a rotating rigid body consists of particles in motion, it has kinetic energy. This kinetic energy is computed using the moment of inertia of the body and the angular velocity. KE = ½Iω The rotor of an electric motor has a rotational inertia of 45 kg-m2. What is its kinetic energy if it turns at 1500 revolutions per minute? 555 kj 20. A grinding wheel in the shape of a solid disk is m in diameter and has a mass of 3.00 kg. The wheel is rotating at 3600 rev/min about an axis through its center. a) What is its kinetic energy? 1066 J b) How far would it have to drop in free fall to acquire the same kinetic energy? 36.3 m 21. The flywheel of a gasoline engine is required to give up J of kinetic energy while its angular speed decreases from 660 rev/min to 540 rev/min. What moment of inertia is required for the wheel? kg-m A phonograph turntable has a kinetic energy of J when turning at 78 rpm. What is the moment of inertia of the turntable about the rotation axis? 23. Energy is to be stored in a large flywheel in the shape of a disk with radius of 1.20 m and a mass of 80.0 kg. To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 5000 m/s 2. What is the maximum kinetic energy that can be stored in the flywheel? 1.20 x 10 5 J Torque: To change the translational inertia of a body you have to apply a net external force. To change the rotational inertia of a body you have to apply a torque (rhymes with fork). If you studied torque in previous science courses it was probably defined as the product of the force and the length of the torque arm. The torque arm (sometimes called lever arm) is the perpendicular distance between the line of action of the force and the axis of rotation. In order to solve problems involving torque, you need to understand how torque is calculated and then be able to calculate the net torque acting on a body. τ = r x F = rf sin θ 24. Calculate the torque (magnitude and direction) about point 0 due to the force F in each of the situations sketched in the figure. In each case the object to which the force is applied has length of 4.00 m, and the force F =20.0 N. a) 80.0 m-n ccw b) 69.3 m-n ccw c) 40.0 m-n ccw d) 34.6 m-n cw e) 0 f) Calculate the resultant torque about point O for the two forces applied in the figure below. 28 m-n cw

5 26. Calculate the net torque (magnitude and direction) on the beam shown in the figure to the right about a) an axis through O, perpendicular to the figure m-n ccw b) an axis through C, perpendicular to the figure m-n ccw 27. A small ball of mass 0.75 kg is attached to one end of a 1.25 meter long massless rod, and the other end of the rod is hung from a pivot. When the resulting pendulum is30 from the vertical, what is the magnitude of the torque about the pivot? 4.6 m-n 28. Find the net torque on the wheel in the figure about the axle through O if a = 10 cm and b = 25 cm m-n cw Rotational Dynamics: In studying translational dynamics we made use of Newton's Second Law, which related the acceleration of a body and the forces applied to the body. An analogous relationship exists between angular acceleration and a quantity we call a torque. Qualitatively speaking, torque is the tendency of a force to cause a rotation of the body on which it acts. Mathematically speaking, torque is defined as the cross product of the moment arm and the applied force. The moment arm is the perpendicular distance between the force applied and the axis of rotation. The unit for torque is a meter-newton. The symbol for torque is the lower-case Greek letter tau, τ. 29. A net force of 10.0 N is applied tangentially to the rim of a wheel having a 0.25 m radius. If the rotational inertia of the wheel is kg m 2, what is its angular acceleration? 5 rad/s A solid ball is rotated by applying a force of 4.7 N tangentially to it. The ball has a radius of 14 cm and a mass of 4.0 kg. What is the angular acceleration of the ball? 21 rad/s A fly wheel in the shape of a thin ring has a mass of 30.0 kg and a diameter of 0.96 m. A torque of 13 m-n is applied tangentially to the wheel. How long will it take for the flywheel to attain an angular velocity of 10.0 rad/s? 5.3 sec 32. A cord is wrapped around the rim of a flywheel 0.5 m in radius, and a steady pull of 50.0 N is exerted on the cord. The wheel is mounted with frictionless bearings on a horizontal shaft through its center. The moment of inertia of the wheel is 4.0 kg-m 2. Compute the angular acceleration of the wheel rad/s 2 Angular Momentum and Angular Impulse: The angular momentum of a rigid body about a fixed axis is defined as: L = Iω or for a single particle The product of the torque and the time interval during which it acts is called the angular impulse, J θ. The angular impulse acting on the body causes a change in the angular momentum of the body about the same axis. Jθ = ΔL = τ Δt

6 Conservation of angular momentum states that when the net external torque on a system is zero, the angular momentum of the system remains constant. This principle of conservation of angular momentum ranks with the principles of conservation of linear momentum and conservation of energy as one of the most fundamental of physical laws. 33. Calculate the angular momentum of a uniform sphere of radius 0.20 m and mass 4.0 kg if it is rotating about an axis along a diameter at (a) 6.0 rad/s and (b) 5.0 rev/s. 2.0 kg-m 2 /sec 34. A solid wooden door 1.0 m wide and 2.0 m high is hinged along one side and has a total mass of 50.0 kg. Initially open and at rest, the door is struck at its center with a hammer. During the blow, an average force of N acts for 0.01 seconds. Find the angular velocity of the door after the impact. 35. A man of mass 70.0 kg is standing on the rim of a large disk that is rotating at 0.5 rev/s about an axis through its center. The disk has mass kg and radius 4.0 m. Calculate the total angular momentum of the man-plusdisk system.

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