Physics 1501: Lecture 19, Pg 1 Physics 1501: Lecture 19 Today’s Agenda l Announcements çHW#7: due Oct. 21 l Midterm 1: average = 45 % … l Topics çRotational Kinematics çRotational Energy çMoments of Inertia

Physics 1501: Lecture 19, Pg 2 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 1501: Lecture 19, Pg 3 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) a R

Physics 1501: Lecture 19, Pg 4 Wheel And Rope... Use a =  R to find  :  = a / R = 4m/s 2 / 0.4m = 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 1501: Lecture 19, Pg 5 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 1501: Lecture 19, Pg 6 Rotation & Kinetic Energy... 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. So: but v i =  r i

Physics 1501: Lecture 19, Pg 7 Lecture 19, 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) 1/4B) 1/2 C) 1D) 2E) 4 BB#1 BB#2

Physics 1501: Lecture 19, Pg 8 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 1501: Lecture 19, Pg 9 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 ! l So where

Physics 1501: Lecture 19, Pg 10 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 1501: Lecture 19, Pg 11 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 1501: Lecture 19, Pg 12 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 1501: Lecture 19, Pg 13 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 1501: Lecture 19, Pg 14 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 1501: Lecture 19, Pg 15 Lecture 19, 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 1501: Lecture 19, Pg 16 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 1501: Lecture 19, Pg 17 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 1501: Lecture 19, Pg 18 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 Thin spherical shell of mass M and radius R, about an axis through its center. R

Physics 1501: Lecture 19, Pg 19 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

Physics 1501: Lecture 19, Pg 20 Parallel Axis Theorem Suppose the moment of inertia of a solid object of mass M about an axis through the center of mass is known, = I CM l The moment of inertia about an axis parallel to this axis but a distance R away is given by: I PARALLEL = I CM + MR 2 So if we know I CM, it is easy to calculate the moment of inertia about a parallel axis.