Physics 111: Lecture 17, Pg 1 Physics 111: Lecture 17 Today’s Agenda l Rotational Kinematics çAnalogy with one-dimensional kinematics l Kinetic energy of a rotating system çMoment of inertia çDiscrete particles çContinuous solid objects l Parallel axis theorem

Physics 111: Lecture 17, 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 pulleys are without mass. l Rotation is extremely important, however, and we need to understand it! l Most of the equations we will develop are simply rotational analogues of ones we have already learned when studying linear kinematics and dynamics.

Physics 111: Lecture 17, Pg 3 Lecture 17, Act 1 Rotations l Bonnie sits on the outer rim of a merry-go-round, and Klyde sits midway between the center and the rim. The merry-go- round makes one complete revolution every two seconds. çKlyde’s angular velocity is: (a) (a) the same as Bonnie’s (b) (b) twice Bonnie’s (c) (c) half Bonnie’s

Physics 111: Lecture 17, Pg 4 Lecture 17, Act 1 Rotations The angular velocity  of any point on a solid object rotating about a fixed axis is the same.  Both Bonnie & Klyde go around once (2  radians) every two seconds.  (Their “linear” speed v will be different since v =  r).

Physics 111: Lecture 17, 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 )   Spin round blackboard

Physics 111: Lecture 17, Pg 6 Rotational Variables... Now suppose  can change as a function of time: l We define the angular acceleration:    Consider the case when  is constant.  We can integrate this to find  and  as a function of time: constant

Physics 111: Lecture 17, Pg 7 Rotational Variables... l Recall also that for a point at 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  constant

Physics 111: Lecture 17, Pg 8 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

Physics 111: Lecture 17, Pg 9 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 = 4 m/s 2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2  radians)a R

Physics 111: Lecture 17, Pg 10 Wheel And Rope... Use a =  R to find  :  = a / R = 4 m/s 2 / 0.4 m = 10 rad/s 2 l Now use the equations we derived above just as you would use the kinematic equations from the beginning of the semester. = 0 + 0(10) + (10)(10) 2 = 500 rad a R 

Physics 111: Lecture 17, Pg 11 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 

Physics 111: Lecture 17, Pg 12 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.

Physics 111: Lecture 17, Pg 13 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 111: Lecture 17, Pg 14 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! Inertia Rods l So where

Physics 111: Lecture 17, Pg 15 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

Physics 111: Lecture 17, Pg 16 Calculating Moment of Inertia... l The squared distance from each point mass to the axis is: mm mm L r L/2 so I = 2mL 2 Using the Pythagorean Theorem

Physics 111: Lecture 17, Pg 17 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

Physics 111: Lecture 17, Pg 18 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

Physics 111: Lecture 17, Pg 19 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

Physics 111: Lecture 17, Pg 20 Lecture 17, 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 111: Lecture 17, Pg 21 Lecture 17, 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 111: Lecture 17, Pg 22 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

Physics 111: Lecture 17, 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 its center, perpendicular to the plane of the hoop. R Thin hoop of mass M and radius R, about an axis through a diameter. R Hoop

Physics 111: Lecture 17, Pg 24 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 R Solid disk or cylinder of mass M and radius R, about a perpendicular axis through its center. Sphere and disk

Physics 111: Lecture 17, Pg 25 Lecture 17, Act 3 Moment of Inertia l Two spheres have the same radius and equal masses. One is made of solid aluminum, and the other is made from a hollow shell of gold. çWhich one has the biggest moment of inertia about an axis through its center? same mass & radius solid hollow (a) solid aluminum(b) hollow gold(c) same

Physics 111: Lecture 17, Pg 26 Lecture 17, Act 3 Moment of Inertia l Moment of inertia depends on mass (same for both) and distance from axis squared, which is bigger for the shell since its mass is located farther from the center. çThe spherical shell (gold) will have a bigger moment of inertia. same mass & radius I SOLID < I SHELL solid hollow

Physics 111: Lecture 17, Pg 27 Moments of Inertia... Some examples of I for solid objects (see also Tipler, Table 9-1): Thin rod of mass M and length L, about a perpendicular axis through its center. L Thin rod of mass M and length L, about a perpendicular axis through its end. L Rod

Physics 111: Lecture 17, Pg 28 Parallel Axis Theorem Suppose the moment of inertia of a solid object of mass M about an axis through the center of mass, I CM, is known. l The moment of inertia about an axis parallel to this axis but a distance D away is given by: I PARALLEL = I CM + MD 2 So if we know I CM, it is easy to calculate the moment of inertia about a parallel axis.

Physics 111: Lecture 17, Pg 29 Parallel Axis Theorem: Example l Consider a thin uniform rod of mass M and length D. Figure out the moment of inertia about an axis through the end of the rod. I PARALLEL = I CM + MD 2 L D=L/2 M x CM We know So which agrees with the result on a previous slide. I CM I END

Physics 111: Lecture 17, Pg 30 Connection with CM motion l Recall what we found out about the kinetic energy of a system of particles in Lecture 15: K REL K CM l For a solid object rotating about its center of mass, we now see that the first term becomes: Substituting but

Physics 111: Lecture 17, Pg 31 Connection with CM motion... l So for a solid object which rotates about its center or mass and whose CM is moving:  V CM We will use this formula more in coming lectures.

Physics 111: Lecture 17, Pg 32 Recap of today’s lecture l Rotational Kinematics (Text: 9-1) çAnalogy with one-dimensional kinematics l Kinetic energy of a rotating system çMoment of inertia (Text: 9-2, 9-3, Table 9-1) çDiscrete particles (Text: 9-3) çContinuous solid objects(Text: 9-3) l Parallel axis theorem (Text: 9-3) l Look at textbook problems l Look at textbook problems Chapter 9: # 7, 11, 27, 31, 33, 37