1. Lecture 34, Page 1 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Physics 2211: Lecture 34 l Rotational Kinematics çAnalogy with one-dimensional kinematics l Kinetic energy of a rotating system çMoment of inertia

2. Lecture 34, Page 2 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm 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 analogs of ones we have already learned when studying linear kinematics and dynamics.

3. Lecture 34, Page 3 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Example 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: (1) (1) the same as Bonnie’s (2) (2) twice Bonnie’s (3) (3) half Bonnie’s

4. Lecture 34, Page 4 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Example 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  ).

5. Lecture 34, Page 5 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm 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 ) angular velocity, rad/s

6. Lecture 34, Page 6 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm 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

7. Lecture 34, Page 7 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Rotational Variables l Recall also that for a point at a distance R away from the axis of rotation:  s = R  (  in radians)  v = R  And taking the derivative of this we find:  a = R  constant    R v s

8. Lecture 34, Page 8 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Summary (with comparison to 1-D kinematics) And for a point at a distance R from the rotation axis: s = R  v = R  a = R  Angular Linear

9. Lecture 34, Page 9 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm 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

10. Lecture 34, Page 10 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm 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 

11. Lecture 34, Page 11 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Physics 2211: Lecture 35 l Moments of Inertia çDiscrete particles çContinuous solid objects l Parallel axis theorem

12. Lecture 34, Page 12 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm 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: vv4vv4 vv1vv1 vv3vv3 vv2vv2 l So But v i = r i  or Moment of Inertia (rotational inertial) about the rotation axis ( I has units of kg m 2.) rr1rr1 rr2rr2 rr3rr3 rr4rr4 m4m4 m1m1 m2m2 m3m3 

13. Lecture 34, Page 13 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm 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.

14. Lecture 34, Page 14 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm l So where Moment of Inertia (Rotational 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!

15. Lecture 34, Page 15 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Calculating Moment of Inertia l We have shown that for N discrete point masses distributed about a fixed axis, the moment of inertia is: 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 where r i is the distance from the mass to the axis of rotation.

16. Lecture 34, Page 16 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm 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

17. Lecture 34, Page 17 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm 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

18. Lecture 34, Page 18 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm 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

19. Lecture 34, Page 19 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm 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

20. Lecture 34, Page 20 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Calculating Moment of Inertia l For a discrete collection of point masses we found: r dm 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 :

21. Lecture 34, Page 21 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm 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

22. Lecture 34, Page 22 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm 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. Moments of Inertia Some examples of I for solid objects:

23. Lecture 34, Page 23 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Example 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 (1) solid aluminum(2) hollow gold(3) same

24. Lecture 34, Page 24 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Example 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. I SOLID < I SHELL same mass & radius solid hollow

25. Lecture 34, Page 25 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm More Moments of Inertia Some examples of I for solid objects 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

26. Lecture 34, Page 26 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm Moment of Inertia Calculation Example Thin rod of mass M and length L, about a perpendicular axis through its end. L x y

27. Lecture 34, Page 27 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm 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.

28. Lecture 34, Page 28 Physics 2211 Spring 2005 © 2005 Dr. Bill Holm 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. We know So which agrees with the result on a previous slide. L D=L/2 M x CM I CM I END I PARALLEL = I CM + MD 2