1. 4.1 Rotational kinematics 4.2 Moment of inertia 4.3 Parallel axis theorem 4.4 Angular momentum and rotational energy CHAPTER 4: ROTATIONAL MOTION

2. Part 1 Rotational kinematics

3. Riview of rotations 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. Klyde’s angular velocity is: (a) (a) the same as Bonnie’s (b) (b) twice Bonnie’s (c) (c) half Bonnie’s

4. REVIEW of ANGLE VELOCITY The angular velocity  of any point on a solid object rotating about a fixed axis is the same. –Both Bonnie & Klyde go around once (2pi radians) every two seconds.  Their “linear” speed v will be different since v =  r. How about their “linear” speed v ? The same or different? ?

5. Review: Rotational Variables. Rotation about a fixed axis: –Consider a disk rotating about an axis through its center: First, recall what we learned about Uniform Circular Motion: (Analogous to )  

6. Rotational Variables... Now suppose  can change as a function of time: We define the angular acceleration:    l Consider the case when  is constant. ç We can integrate this to find  and  as a function of time: constant

7. Rotational Variables... Recall also that for a point at a distance R away from the axis of rotation: –x =  R (distance in the circle) –v =  R And taking the derivative of this we find: –a =  R   R v x  constant

8. Example: Wheel And Rope 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

9. Solution Use a=  R to find  :  = a / R = (4 m/s 2 )/ 0.4 m = 10 rad/s 2 Now use this equations just as you would use the kinematic equations from the beginning of the semester. = 0 + 0(10) + (10)(10) 2 = 500 rad a R 

10. Part 2 Moment of Inertia

11. 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: rr1rr1 rr2rr2 rr3rr3 rr4rr4 m4m4 m1m1 m2m2 m3m3 

12. Compute: Kinetic Energy of Rotation system So: but v i =  r i which we write as: moment of inertia Define the moment of inertia about the rotation axis rr1rr1 rr2rr2 rr3rr3 rr4rr4 m4m4 m1m1 m2m2 m3m3  vv4vv4 vv1vv1 vv3vv3 vv2vv2 I has units of kg m 2.

13. Rotation & Kinetic Energy... Point Particle Rotating SystemThe 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. Moment of Inertia Notice that the moment of inertia I depends on the distribution of mass in the system. The further the mass is from the rotation axis, the bigger the moment of inertia. * For a given object, the moment of inertia will depend on where we choose the rotation axis (unlike the center of mass). * 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

15. Calculating Moment of Inertia We have shown that for N discrete point masses distributed about a fixed axis, the moment of inertia is: where r i is the distance from the mass i 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

16. Calculating Moment of Inertia... 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. Learning check 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. 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. 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. Check: 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. –Which 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

21. Solution a b c m m m L L l Masses m and lengths L as show: l Calculate moments of inerta: So (b) is correct: I a > I c > I b

22. Calculating Moment of Inertia... For a continuous solid object For a discrete collection of point masses we found: 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

23. Learn by heart 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

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.

25. Moment of Inertia 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. –Which 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

26. Hint 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. –The spherical shell (gold) will have a bigger moment of inertia. same mass & radius I SOLID < I SHELL solid hollow

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

28. Part 3 Parallel Axis Theorem

29. Suppose the moment of inertia of a solid object of mass M about an axis through the center of mass, I CM, is known. 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.

30. Parallel Axis Theorem: Example 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 We know So which agrees with the result on a previous slide. L D=L/2 M x CM I CM I END

31. Part 4 Angular momentum and rotational energy

32. Angular momentum Defination: Example: compute L for the system m=100g, a=10cm,  =(2  /5) rad/s a m L=2ma 2.  =2x0.1x 10 -4 x0.1  =2  10 -6.  r v L

33. Complete Motion by linear + rotation The total kinetic energy of a system of particles include 2 parts KRKR K CM l For a solid object rotating about its center of mass, we now see that the first term becomes: Substituting but

34. Connection with CM motion... 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.

35. Similarity Between Linear and Rotational Motions All physical quantities in linear and rotational motions show striking similarity. Quantities LinearRotational Mass Moment of Inertia Length of motionDistanceAngle (Radian) Speed Acceleration Force Torque Work Power Momentum Kinetic EnergyKineticRotational

36. Review of today’s lecture Rotational Kinematics –Analogy with one-dimensional kinematics Kinetic energy of a rotating system –Moment of inertia –Discrete particles Continuous solid objects Parallel axis theorem

37. Problem Use the law of energy conservation to compute the velocity of the ball when it hits to the ground. Hint : There are 2 motions (frictionless case) H=3m V 0 = 0,  0 =0 M=2kg, r=4cm A=45 Deg Change the ball by Cylinder have the same M and r