Lecture 16: Rotational Dynamics and Equilibrium

1. Lecture 16 • 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 Goals: Assignment: • HW7 due March 25th • After Spring Break Tuesday: Catch up

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

3.   Rotational Variables • Rotation about a fixed axis: • Consider a disk rotating aboutan axis through its center: • How do we describe the motion: (Analogous to the linear case )

4. v = w R x R    Rotational Variables... • Recall: At a point a distance R away from the axis of rotation, the tangential motion: • x =  R • v = R • a =  R

5. Comparison to 1-D kinematics Angular Linear And for a point at a distance R from the rotation axis: x = R v =  R aT =  R Here aT refers to tangential acceleration

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

7. 1 K= ½ m v2 = ½ m (w r)2 w 2 K= ½ m (2v)2 = ½ m (w 2r)2 System of Particles (Distributed Mass): • Twice the radius, four times the kinetic energy • The rotation axis matters too!

8. + + m2 m1 m1 m2 A special point for rotationSystem of Particles: Center of Mass (CM) • If an object is not held then it will rotate about the center of mass. • Center of mass: Where the system is balanced ! • Building a mobile is an exercise in finding centers of mass. mobile

9. System of Particles: Center of Mass • How do we describe the “position” of a system made up of many parts ? • Define the Center of Mass (average position): • For a collection of N individual point like particles whose masses and positions we know: RCM m2 m1 r2 r1 y x (In this case, N = 2)

10. RCM = (12,6) (12,12) 2m m m (0,0) (24,0) Sample calculation: • Consider the following mass distribution: XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters XCM= 12 meters YCM= 6 meters

11. System of Particles: Center of Mass • For a continuous solid, convert sums to an integral. dm r y where dmis an infinitesimal mass element. x

12. VCM  Connection with motion... • So for a rigid object which rotates about its center of mass and whose CM is moving: For a point p rotating:

13. Rotation & Kinetic Energy • Consider the simple rotating system shown below. (Assume the masses are attached to the rotation axis by massless rigid rods). • The kinetic energy of this system will be the sum of the kinetic energy of each piece: • K = ½m1v12 + ½m2v22 + ½m3v32 + ½m4v42 m4 m1 r1  r4 r2 m3 r3 m2

14. m4 m1 r1  r4 m3 r2 r3 m2 Rotation & Kinetic Energy • Notice that v1 = w r1 , v2 = w r2 , v3 = w r3 , v4 = w r4 • So we can rewrite the summation: • We recognize the quantity, moment of inertia orI, and write:

15. 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: m m L m m

16. Calculating Moment of Inertia... • For a single object, Idepends on the rotation axis! • Example:I1 = 4 m R2 = 4 m (21/2 L / 2)2 I1= 2mL2 I2= mL2 I= 2mL2 m m L m m

17. dr r R L dm r Moments of Inertia • For a continuous solid object we have to add up the mr2contribution for every infinitesimal mass element dm. • An integral is required to find I: Solid disk or cylinder of mass M and radius R, about perpendicular axis through its center. I = ½ M R2 • Some examples of I for solid objects: Use the table…

18. Ball 1 Ball 2 Exercise Rotational Kinetic Energy • ¼ • ½ • 1 • 2 • 4 • 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. • What is the ratio of the kinetic energy of Ball 2 to that of Ball 1 ?

19. Ball 1 Ball 2 Exercise Rotational Kinetic Energy • K2/K1 = ½ m wr22 / ½ m wr12 = 0.22 / 0.12 = 4 • 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

20. 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 r = 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 w2 (A)Disk 1 (B)Disk 2 (C)Same w2 w1 F F start d finish

21. 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 r = 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 I1w12 = ½ I2w22 w1 = (I2 / I1)½w2 and I2 < I1 (A)Disk 1 (B) Disk 2 (C)Same w2 w1 F F start d finish

22. Lecture 16 Assignment: • HW7 due March 25th • For the next Tuesday: Catch up

23. Lecture 16 Assignment: • HW7 due March 25th • After Spring Break Tuesday: Catch up