Rotation of Rigid Bodies

Chapter 9 Rotation of Rigid Bodies

PowerPoint® Lectures for University Physics, Twelfth Edition – Hugh D. Young and Roger A. Freedman

Lectures by James Pazun

CopyrightCopyright © © 2008 2008 Pearson Pearson Education Education Inc., Inc., publishing publishing as as Pearson Pearson Addison-Wesley Addison-Wesley Goals for Chapter 9 • To study rotational kinematics • To relate linear to angular kinematics • To define moments of inertia and determine rotational kinetic energy • To calculate the moment of inertia

• As you listen to a CD, watch a DVD, ride an airplane, or do anything that involves objects that turn in circles, you’re already involved in the process of rotational dynamics and rotational kinematics. • “Real-world” rotations can be very complicated because of stretching and twisting of the rotating object. We can make a substantial start in our understanding by studying the ideal situation of rigid bodies rotating.

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Homework 7, 15, 25 Æ Specific to 9.1, 9.2 and 9.3 35, 39, 45, 53 Æ Specific to 9.4 and 9.5 79, 85, 89 Æ General problems Read 361 to 370 Æ in chapter 10

The test on this material will be after we cover Ch 11

• A car’s analog speedometer gives us a very good example to begin defining rotational motion. • Consider the clockwise (or counterclockwise) motion of a rigid, fixed-length speedometer needle about a fixed pivot point.

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Angular motions in revolutions, degrees, and radians • One complete cycle of 360° is one revolution. • One complete revolution is 2π radians. • Relating the two, 360° = 2 π radians or 1 radian = 57.3°.

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Angular displacement is the angle being swept out • Like a second hand sweeping around a clock, a radius vector will travel through a displacement of degrees, radians, or revolutions. • We denote angular displacement as θ (theta). It is the angular equivalent of x or y in earlier chapters.

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Angular velocity Angular velocity is denoted by the symbol ω (omega). Angular velocity is measured in radians per second (SI standard) as well as other measures such as r.p.m. (revolutions per second). Δθ = θ2 −θ1 ω θ 2θ− 1 Δ av−z = = t2 − t1 θΔt ω dθ = z dt What does the subscript z refer to? How does it differ from the subscript x for linear velocity? Bike wheel demo:

How does ω differ for different r? vt?

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Calculating Angular Velocity The flywheel in a car engine under test has an angular position given by: θ = (2.0 rad/s3)t3 The diameter of the flywheel is 0.36 m.

Find the angle θ, in radians and in degrees, at times t1 = 2.0 s and t2 = 5.0 s. Find the distance that a particle on the rim moves during that time interval. Find the average angular velocity, in rad/s and in rev/min between t1 = 2.0 s and t2 = 5.0 s.

Find the instantaneous angular velocity at time t = t2 = 5.0 s.

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Angular velocity is a vector • You can visualize the position of the vector by sweeping out the angle with the fingers of your right hand. The position of your thumb will be the position of the angular velocity vector. This is called the “right-hand rule.”

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Angular acceleration The angular acceleration is the change of angular velocity divided by the time interval during which the change occurred. ω −ω Δω α = 2 1 = av−z t − t Δt α ω 2 1 d d 2θ = = z dt dt 2

Use the symbol α (alpha) to denote radians per second2.

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Calculating Angular Acceleration The instantaneous angular velocity of the flywheel in the previous problem was: 3 2 ωZ = (6.0 rad/s )t Find the average angular acceleration between times t1=2.0 s and t2 = 5.0 s. Find the instantaneous angular acceleration at time t2= 5.0 s.

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley We have four fundamental equations for angular kinematics Check out the derivation of the constant angular acceleration equations on 333 to 334.

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley The circular motion of a DVD You have just finished watching a movie on DVD and the disc is slowing to a stop. The angular velocity of the disc at t = 0 is 27.5 rad/s and its angular acceleration is a constant – 10.0 rad/s2. A line PQ on the surface of the disc lies along the +x-axis at t = 0. What is the disc’s angular velocity at t = 0.300 s? What angle does the line PQ make with the + x-axis at this time?

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Linear and angular quantities related On a Merry-Go-Round rotating at a constant rate, as a child moves from the center outwards, they move at a faster linear velocity. The child’s angular velocity does not change, only the distance from the center r. At any time the angle θ and the arc length s are related by: θ s = rθ Taking the time derivative of this (r does not depend on time): ds d = r → v = rω dt dt This linear velocity is tangent to the path. Taking the second derivativeω yields tangential (linear) acceleration: dv d = r → a = r dt dt tan α v2 a = = ω 2r rad r Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley An athlete throwing the discus A discus thrower moves the discus in a circle of radius 80.0 cm. At a certain instant the thrower is spinning at an angular speed of 10.0 rad/s and the angular speed is increasing at 50.0 rad/s2. At this instant find the tangential and centripetal components of the acceleration of the discus and the magnitude of the acceleration.

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley An airplane propeller You are asked to design an airplane propeller to turn at 2400 rpm. The forward air speed of the plane is to be 75 m/s and the speed of the tips of the propeller blades through the air must not exceed 270 m/s. What is the maximum radius the propeller can have? With this radius, what is the acceleration of the propeller tip?

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Bicycle pedals and gears How are the angular speeds of the two sprockets below related to the number of teeth on each sprocket?

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Rotational energy • A rotating rigid body consists of a mass in motion, so it has kinetic energy. • Just like linear kinetic energy is ½mv2, the angular energy will be determined by: K = ½ Iω2.

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley 9.4 Energy in Rotational Motion Consider a rigid body as being made up of particles with masses m1, m2, . . . and perpendicular distances from the th axis of rotation r1, r2, . . . In other words the i particle will have a mass of mi and distance ri.

th When the rigid body rotates the speed vi of the i particle is vi = riω. Different particles have different speeds and distances, but the same ω, or the body is not rigid. We can write the kinetic energy of the particle as: 1 2 1 2 2 2 mivi = 2 miri ω

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley 9.4 Energy in Rotationalω Motion The total K can be expressed asω the sum of the individual kinetic energies: 1 2 2 1 2 2 1 2 2 K = 2 m1r1 + 2 m2r2 +…= ∑ 2 miri ω ω i Factoring out the common ω2/2: 1 2 2 2 1 ⎛ 2 ⎞ 2 K = 2 ()m1r1 + m2r2 +… = 2 ⎜∑miri ⎟ω ⎝ i ⎠ The quantity in parentheses is called the moment of inertia of the body. The definition of this quantity follows: 2 2 2 I = m1r1 + m2r2 +…= ∑miri i In terms of the moment of inertia, the kinetic energy would be: 1 2 K = 2 Iω

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Rotational energy changes if parts shift and I changes An engineer is designing a one-piece machine part consisting of three heavy connectors linked by light molded struts. What is the moment of inertia about an axis coinciding with rod BC? What is the moment of inertia of this body about an axis through point A, perpendicular to the plane of the diagram? If the body rotates about this axis with angular speed ω = 4.0 rad/s, what is its kinetic energy?

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Finding the moment of inertia for common shapes Memorize the moments for the bodies below. You will not be given these on the AP Test!!

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Calculating rotational energy A light, flexible, nonstretching cable is wrapped several times around a winch drum, a solid cylinder of mass 50 kg and diameter 0.120 m, which rotates about a stationary horizontal axis held by frictionless bearings. The free end of the cable is pulled with a constant force of magnitude 9.0 N for a distance of 2.0 m. It unwinds without slipping, turning the cylinder as it does so. If the cylinder is initially at rest, fid its final angular speed and the final speed of the cable.

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Variations on Example 9.8 In a lab experiment to test conservation of energy in rotational motion, we wrap a light, flexible cable around a solid cylinder with mass M and radius R. The cylinder rotates with negligible friction about a stationary horizontal axis. We tie the free end of the cable to an object of mass m and release the object with no initial velocity at a distance h above the floor. As the object falls, the cable unwinds without stretching or slipping, turning the cylinder. Find the speed of the falling object and the angular speed of the cylinder just as the object strikes the floor.

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Extended bodies and the Parallel Axis Theorem • An object does not need to rotate through its center of mass. It can rotate around any axis in, on, or out of the body. • Fortunately there is a simple relationship between the

moment of inertia Icm of a body of mass M about an axis through its center of mass and the moment of inertial IP about any other parallel axis that is a distance d from the center of mass, called the parallel-axis theorem. 2 I P = Icm + Md

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley Extended bodies and the Parallel Axis Theorem A part of a mechanical linkage has a mass of 3.6 kg. We measure its moment of inertia about an axis 0.15 m from its 2 center of mass to be Ip = 0.132 kgm . What is the moment of inertia about an axis though the center of mass?

Find the moment of inertia of a thin uniform disk of mass M and radius R about an axis perpendicular to its plane at the edge.

The test on this material will be after we cover Ch 11