# Physics 207: Lecture 16, Pg 1 Lecture 16Goals: Chapter 12 Chapter 12  Extend the particle model to rigid-bodies  Understand the equilibrium of an extended.

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Physics 207: Lecture 16, Pg 1 Lecture 16Goals: Chapter 12 Chapter 12  Extend the particle model to rigid-bodies  Understand the equilibrium of an extended object.  Analyze rolling motion  Understand rotation about a fixed axis.  Employ “conservation of angular momentum” concept Assignment: l HW7 due March 25 th l After Spring Break Tuesday: Catch up

Physics 207: Lecture 16, Pg 2 Rotational Dynamics: A child’s toy, a physics playground or a student’s nightmare l A merry-go-round is spinning and we run and jump on it. What does it do? What principles would apply? l We are standing on the rim and our “friends” spin it faster. What happens to us? l We are standing on the rim a walk towards the center. Does anything change?

Physics 207: Lecture 16, Pg 3 Rotational Variables l Rotation about a fixed axis:  Consider a disk rotating about an axis through its center: l How do we describe the motion: (Analogous to the linear case )  

Physics 207: Lecture 16, Pg 4 Rotational Variables... l Recall: At a point a distance R away from the axis of rotation, the tangential motion:  x =  R  v =  R  a =  R   R v =  R x 

Physics 207: Lecture 16, Pg 5 Comparison to 1-D kinematics AngularLinear And for a point at a distance R from the rotation axis: x = R  v =  R  a T =  R Here a T refers to tangential acceleration

Physics 207: Lecture 16, Pg 6 Exercise Rotational Definitions A. The wheel is spinning counter-clockwise and slowing down. B. The wheel is spinning counter-clockwise and speeding up. C. The wheel is spinning clockwise and slowing down. D. The wheel is spinning clockwise and speeding up A friend at a party (perhaps a little tipsy) sees a disk spinning and says “Ooh, look! There’s a wheel with a negative  and positive  !” l Which of the following is a true statement about the wheel? 

Physics 207: Lecture 16, Pg 7 Example: Wheel And Rope l A wheel with radius r = 0.4 m rotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such that it has a constant acceleration a T = 4 m/s 2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2  radians) aTaT r

Physics 207: Lecture 16, Pg 8 Example: Wheel And Rope l A wheel with radius r = 0.4 m rotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such that it has a constant acceleration a T = 4 m/s 2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2  radians) Revolutions = R = (    and a T =  r      t + ½  t    R = (    + ½  a T /r  t    R = (0.5 x 10 x 100) / 6.28 aTaT r

Physics 207: Lecture 16, Pg 9 System of Particles (Distributed Mass): l Until now, we have considered the behavior of very simple systems (one or two masses). l But real objects have distributed mass ! l For example, consider a simple rotating disk and 2 equal mass m plugs at distances r and 2r. l Compare the velocities and kinetic energies at these two points. 1 2 

Physics 207: Lecture 16, Pg 10 System of Particles (Distributed Mass): l Twice the radius, four times the kinetic energy l The rotation axis matters too! 1 K= ½ m v 2 = ½ m (  r) 2 2 K= ½ m (2v) 2 = ½ m (  2r) 2 

Physics 207: Lecture 16, Pg 11 A special point for rotation System of Particles: Center of Mass (CM) l If an object is not held then it will rotate about the center of mass. l Center of mass: Where the system is balanced !  Building a mobile is an exercise in finding centers of mass. m1m1 m2m2 + m1m1 m2m2 + mobile

Physics 207: Lecture 16, Pg 12 System of Particles: Center of Mass l How do we describe the “position” of a system made up of many parts ? l Define the Center of Mass (average position):  For a collection of N individual point like particles whose masses and positions we know: (In this case, N = 2) y x r2r2 r1r1 m1m1 m2m2 R CM

Physics 207: Lecture 16, Pg 13 Sample calculation: l Consider the following mass distribution: (24,0) (0,0) (12,12) m 2m m R CM = (12,6) X CM = (m x 0 + 2m x 12 + m x 24 )/4m meters Y CM = (m x 0 + 2m x 12 + m x 0 )/4m meters X CM = 12 meters Y CM = 6 meters

Physics 207: Lecture 16, Pg 14 System of Particles: Center of Mass l For a continuous solid, convert sums to an integral. y x dm r where dm is an infinitesimal mass element.

Physics 207: Lecture 16, Pg 15 Connection with motion... l So for a rigid object which rotates about its center of mass and whose CM is moving: For a point p rotating:  V CM

Physics 207: Lecture 16, Pg 16 Rotation & Kinetic Energy l Consider the simple rotating system shown below. (Assume the masses are attached to the rotation axis by massless rigid rods). l The kinetic energy of this system will be the sum of the kinetic energy of each piece: K = ½  m 1 v 1  + ½  m 2 v 2  + ½  m 3 v 3  + ½  m 4 v 4  rr1rr1 rr2rr2 rr3rr3 rr4rr4 m4m4 m1m1 m2m2 m3m3 

Physics 207: Lecture 16, Pg 17 Rotation & Kinetic Energy Notice that v 1 =  r 1, v 2 =  r 2, v 3 =  r 3, v 4 =  r 4 l So we can rewrite the summation: We recognize the quantity, moment of inertia or I, and write: rr1rr1 rr2rr2 rr3rr3 rr4rr4 m4m4 m1m1 m2m2 m3m3 

Physics 207: Lecture 16, Pg 18 Calculating Moment of Inertia where r is the distance from the mass to the axis of rotation. Example: Calculate the moment of inertia of four point masses (m) on the corners of a square whose sides have length L, about a perpendicular axis through the center of the square: mm mm L

Physics 207: Lecture 16, Pg 19 Calculating Moment of Inertia... For a single object, I depends on the rotation axis! Example: I 1 = 4 m R 2 = 4 m (2 1/2 L / 2) 2 L I = 2mL 2 I 2 = mL 2 mm mm I 1 = 2mL 2

Physics 207: Lecture 16, Pg 20 Home Exercise Moment of Inertia A triangular shape is made from identical balls and identical rigid, massless rods as shown. The moment of inertia about the a, b, and c axes is I a, I b, and I c respectively.  Which of the following is correct: ( A) ( A) I a > I b > I c (B) (B) I a > I c > I b (C) (C) I b > I a > I c a b c

Physics 207: Lecture 16, Pg 21 Home Exercise Moment of Inertia I a = 2 m (2L) 2 I b = 3 m L 2 I c = m (2L) 2 l Which of the following is correct: ( A) ( A) I a > I b > I c (B) I a > I c > I b (C) (C) I b > I a > I c a b c L L

Physics 207: Lecture 16, Pg 22 Moments of Inertia Solid disk or cylinder of mass M and radius R, about perpendicular axis through its center. I = ½ M R 2 Some examples of I for solid objects: R L r dr r dm l For a continuous solid object we have to add up the mr 2 contribution for every infinitesimal mass element dm.  An integral is required to find I : Use the table…

Physics 207: Lecture 16, Pg 23 Moments of Inertia Some examples of I for solid objects: Thin hoop (or cylinder) of mass M and radius R, about an axis through it center, perpendicular to the plane of the hoop is just MR 2 R Thin hoop of mass M and radius R, about an axis through a diameter. R

Physics 207: Lecture 16, Pg 24 Exercise Rotational Kinetic Energy A. ¼ B. ½ C. 1 D. 2 E. 4 l We have two balls of the same mass. Ball 1 is attached to a 0.1 m long rope. It spins around at 2 revolutions per second. Ball 2 is on a 0.2 m long rope. It spins around at 2 revolutions per second. l What is the ratio of the kinetic energy of Ball 2 to that of Ball 1 ? Ball 1 Ball 2

Physics 207: Lecture 16, Pg 25 Exercise Rotational Kinetic Energy K 2 /K 1 = ½ m  r 2 2 / ½ m  r 1 2 =  2 /  2 = 4 l What is the ratio of the kinetic energy of Ball 2 to that of Ball 1 ? (A) 1/4 (B) 1/2 (C) 1 (D) 2 (E) 4 Ball 1 Ball 2

Physics 207: Lecture 16, Pg 26 Rotation & Kinetic Energy... l The kinetic energy of a rotating system looks similar to that of a point particle: Point Particle Rotating System v is “linear” velocity m is the mass.  is angular velocity I is the moment of inertia about the rotation axis.

Physics 207: Lecture 16, Pg 27 Moment of Inertia and Rotational Energy Notice that the moment of inertia I depends on the distribution of mass in the system.  The further the mass is from the rotation axis, the bigger the moment of inertia. l For a given object, the moment of inertia depends on where we choose the rotation axis (unlike the center of mass). In rotational dynamics, the moment of inertia I appears in the same way that mass m does in linear dynamics ! lSo where

Physics 207: Lecture 16, Pg 28 Exercise Work & Energy Strings are wrapped around the circumference of two solid disks and pulled with identical forces, F, for the same linear distance, d. Disk 1 has a bigger radius, but both are identical material (i.e. their density  = M / V is the same). Both disks rotate freely around axes though their centers, and start at rest.  Which disk has the biggest angular velocity after the drop? W  F d = ½ I  2 ( A) ( A) Disk 1 (B) (B) Disk 2 (C) (C) Same FF 11 22 start finish d

Physics 207: Lecture 16, Pg 29 Exercise Work & Energy Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same linear distance. Disk 1 has a bigger radius, but both are identical material (i.e. their density  = M/V is the same). Both disks rotate freely around axes though their centers, and start at rest.  Which disk has the biggest angular velocity after the drop? W = F d = ½ I 1  1 2 = ½ I 2  2 2 and  1 = (I 2 / I 1 ) ½  2 and I 2 < I 1 ( A) ( A) Disk 1 (B) (B) Disk 2 (C) (C) Same FF 11 22 start finish d

Physics 207: Lecture 16, Pg 30 Lecture 16 Assignment: l HW7 due March 25 th l For the next Tuesday: Catch up

Physics 207: Lecture 16, Pg 31 Lecture 16 Assignment: l HW7 due March 25 th l After Spring Break Tuesday: Catch up

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