Physics 151: Lecture 20, Pg 1 Physics 151: Lecture 20 Today’s Agenda l Topics (Chapter 10) : çRotational KinematicsCh çRotational Energy Ch çMoments of InertiaCh. 10.5

Physics 151: Lecture 20, Pg 2 Rotation l Up until now we have gracefully avoided dealing with the rotation of objects. çWe have studied objects that slide, not roll. çWe have assumed wheels are massless. l Rotation is extremely important, however, and we need to understand it ! l Most of the equations we will develop are simply rotational versions of ones we have already learned when studying linear kinematics and dynamics.

Physics 151: Lecture 20, Pg 3 Recall Kinematic of Circular Motion: R v s tt (x,y)  =  t s = v t s = R  = R  t v = Rv = R x = R cos(  )  = R cos(  t)  y = R sin(  )  = R sin(  t)  = tan -1 (y/x) For uniform circular motion:  is angular velocity x y Animation

Physics 151: Lecture 20, Pg 4 Example: l The angular speed of the minute hand of a clock, in rad/s, is: a.  1800 b.  /60 c.  /30 d.  e. 120  See text: 10.1

Physics 151: Lecture 20, Pg 5 Rotational Variables l Rotation about a fixed axis: çConsider a disk rotating about an axis through its center: l First, recall what we learned about Uniform Circular Motion: (Analogous to )   See text: 10.1

Physics 151: Lecture 20, Pg 6 Rotational Variables... Now suppose  can change as a function of time: l We define the angular acceleration:    See text: 10.1 Consider the case when  is constant.  We can integrate this to find  and  as a function of time: constant

Physics 151: Lecture 20, Pg 7 Example: l The graphs below show angular velocity as a function of time. In which one is the magnitude of the angular acceleration constantly decreasing ? See text: 10.1

Physics 151: Lecture 20, Pg 8 Rotational Variables... l Recall also that for a point a distance R away from the axis of rotation:  x =  R  v =  R And taking the derivative of this we find  a =  R   R v x  See text: 10.2 constant Animation

Physics 151: Lecture 20, Pg 9 Summary (with comparison to 1-D kinematics) AngularLinear And for a point at a distance R from the rotation axis: x = R  v =  R  a =  R See text: 10.3

Physics 151: Lecture 20, Pg 10 Example: Wheel And Rope l A wheel with radius R = 0.4m 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 = 4m/s 2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2  radians) See text: 10.1 a R

Physics 151: Lecture 20, Pg 11 Example: l The turntable of a record player has an angular velocity of 8.0 rad/s when it is turned off. The turntable comes to rest 2.5 s after being turned off. Through how many radians does the turntable rotate after being turned off ? Assume constant angular acceleration. a. 12 rad b. 8.0 rad c. 10 rad d. 16 rad e. 6.8 rad See text: 10.1

Physics 151: Lecture 20, Pg 12 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: rr1rr1 rr2rr2 rr3rr3 rr4rr4 m4m4 m1m1 m2m2 m3m3  Recall text 9.6, systems of particles, CM

Physics 151: Lecture 20, Pg 13 Rotation & Kinetic Energy... So: but v i =  r i rr1rr1 rr2rr2 rr3rr3 rr4rr4 m4m4 m1m1 m2m2 m3m3  vv4vv4 vv1vv1 vv3vv3 vv2vv2 which we write as: moment of inertia Define the moment of inertia about the rotation axis I has units of kg m 2. Recall text 9.6, systems of particles, CM

Physics 151: Lecture 20, Pg 14 Lecture 20, Act 1 Rotational Kinetic Energy l I have two basketballs. BB#1 is attached to a 0.1m long rope. I spin around with it at a rate of 2 revolutions per second. BB#2 is on a 0.2m long rope. I then spin around with it at a rate of 2 revolutions per second. What is the ratio of the kinetic energy of BB#2 to that of BB#1? A) 1/4 B) 1/2 C) 1 D) 2 E) 4 BB#1 BB#2

Physics 151: Lecture 20, Pg 15 Rotation & Kinetic Energy... Point Particle Rotating System 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 151: Lecture 20, Pg 16 Moment of Inertia Notice that the moment of inertia I depends on the distribution of mass in the system. çThe further the mass is from the rotation axis, the bigger the moment of inertia. l For a given object, the moment of inertia will depend on where we choose the rotation axis (unlike the center of mass). We will see that in rotational dynamics, the moment of inertia I appears in the same way that mass m does when we study linear dynamics ! See text: 10.4 l So where

Physics 151: Lecture 20, Pg 17 Calculating Moment of Inertia l We have shown that for N discrete point masses distributed about a fixed axis, the moment of inertia is: 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 See text: 10.5 See example 10.4 (similar)

Physics 151: Lecture 20, Pg 18 Calculating Moment of Inertia... l The squared distance from each point mass to the axis is: mm mm L r L/2 I = 2mL 2 See text: 10.5 See example 10.4 (similar)

Physics 151: Lecture 20, Pg 19 Calculating Moment of Inertia... Now calculate I for the same object about an axis through the center, parallel to the plane (as shown): mm mm L r I = mL 2 See text: 10.5 See example 10.4 (similar)

Physics 151: Lecture 20, Pg 20 Calculating Moment of Inertia... Finally, calculate I for the same object about an axis along one side (as shown): mm mm L r I = 2mL 2 See text: 10.5 See example 10.4 (similar)

Physics 151: Lecture 20, Pg 21 Calculating Moment of Inertia... For a single object, I clearly depends on the rotation axis !! L I = 2mL 2 I = mL 2 mm mm I = 2mL 2 See text: 10.5 See example 10.4 (similar)

Physics 151: Lecture 20, Pg 22 Lecture 20, Act 2 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. çWhich 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 151: Lecture 20, Pg 23 Lecture 20, Act 2 Moment of Inertia a b c l Label masses and lengths: m m m L L l Calculate moments of inerta: So (b) is correct: I a > I c > I b

Physics 151: Lecture 20, Pg 24 Calculating Moment of Inertia... l For a discrete collection of point masses we found: l For a continuous solid object we have to add up the mr 2 contribution for every infinitesimal mass element dm.  We have to do an integral to find I : r dm See text: 8-5

Physics 151: Lecture 20, Pg 25 Moments of Inertia Some examples of I for solid objects: Thin hoop (or cylinder) of mass M and radius R, about an axis through its center, perpendicular to the plane of the hoop. R See text: 10.5 Thin hoop of mass M and radius R, about an axis through a diameter. R see Example 10.5 in the text

Physics 151: Lecture 20, Pg 26 Moments of Inertia Solid disk or cylinder of mass M and radius R, about a perpendicular axis through its center. Some examples of I for solid objects: R L r dr

Physics 151: Lecture 20, Pg 27 Moments of Inertia... Some examples of I for solid objects: Solid sphere of mass M and radius R, about an axis through its center. R See text: 10.5 See Table 10.2, Moments of Inertia Thin spherical shell of mass M and radius R, about an axis through its center. R

Physics 151: Lecture 20, Pg 28 Recap of today’s lecture l Chapter 9, çCenter of Mass çElastic Collisions çImpulse