Chap. 11B - Rigid Body Rotation. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University

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1 Chap. 11B - Rigid Body Rotation A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University 007

2 Objectives: After completing this module, you should be able to: Define and calculate the moment of of inertia for simple systems. Define and apply the concepts of of Newton s second law, rotational kinetic energy, rotational work, rotational power,, and rotational momentum to to the solution of of physical problems. Apply principles of of conservation of of energy and momentum to to problems involving rotation of of rigid bodies.

3 Inertia of Rotation Consider Newton s s second law for the inertia of rotation to be patterned after the law for translation. F = 0 N a = 4 m/s F = 0 N R = 0.5 m = rad/s Linear Inertia, m 4 N m = 4 m/s = 5kg Rotational Inertia, I (0 N)(0.5 m) I = = =.5 kg m 4 m/s Force does for translation what torque does for rotation:

4 Rotational Kinetic Energy Consider tiny mass m: v = R K = ½mv K = ½m(R) K = ½(mR ) axis m 1 m m 4 m m 3 Sum to find K total: K = ½(mR ) (½ same for all m ) Object rotating at constant Rotational Inertia Defined: I = mr

5 Example 1: What is the rotational kinetic energy of the device shown if it rotates at a constant speed of 600 rpm? First: I = mr I = (3 kg)(1 m) + ( kg)(3 m) + (1 kg)( m) kg 3 m m 3 kg 1 m 1 kg I = 5 kg m = 600 rpm = 6.8 rad/s K = ½Iw = ½(5 kg m )(6.8 rad/s) K = 49,300 J

6 Common Rotational Inertias L L I 1 3 ml I 1 1 ml R R R I = mr I = ½mR I 5 mr Hoop Disk or cylinder Solid sphere

7 Example : A circular hoop and a disk each have a mass of 3 kg and a radius of 30 cm.. Compare their rotational inertias. ImR (3 kg)(0. m) R I = 0.10 kg m I = mr Hoop R I mr (3 kg)(0. m) 1 1 I = ½mR I = kg m Disk

8 Important Analogies For many problems involving rotation, there is an analogy to be drawn from linear motion. x f m A resultant force F produces negative acceleration a for a mass m. I R 4 kg 50 rad/s = 40 N m A resultant torque produces angular acceleration of disk with rotational inertia I. F ma I

9 Newton s s nd Law for Rotation How many revolutions required to stop? = I F R 4 kg 50 rad/s R = 0.0 m F = 40 N FR = (½mR ) F (40N) mr (4 kg)(0. m) = 100 rad/s 0 f - o 0 (50 rad/s) (100 rad/s ) = 1.5 rad = 1.99 rev

10 Example 3: What is the linear acceleration of the falling -kg mass? Apply Newton s nd law to rotating disk: I TR = (½MR ) T = ½MR a T = ½MR( ) ; R but a = R; = and T = ½Ma Apply Newton s nd law to falling mass: mg - T = ma mg - ½Ma T = ma ( kg)(9.8 m/s ) - ½(6 kg) a = ( kg) a 19.6 N - (3 kg) a = ( kg) a a = 3.9 m/s a R R = 50 cm M 6 kg a =? R = 50 cm 6 kg +a kg T T mg kg

11 Work and Power for Rotation Work = Fs = FR Work = FR s F Work Power = = t t = t s = R F Power = Power = Torque x average angular velocity

12 Example 4: The rotating disk has a radius of 40 cm and a mass of s 6 kg.. Find the work and power if the -kg mass is lifted 0 m in 4 s. s F kg 6 kg Work = = FR F=W s 0 m = = = 50 rad s = 0 m R 0.4 m F = mg = ( kg)(9.8 m/s ); F = 19.6 N Work = (19.6 N)(0.4 m)(50 rad) Work = 39 J Work Power = = t 39 J 4s Power = 98 W

13 The Work-Energy Theorem Recall for linear motion that the work done is equal to the change in linear kinetic energy: ½ f 0 Fx ½mv mv Using angular analogies, we find the rotational work is equal to the change in rotational kinetic energy: ½I f 0 ½I

14 Applying the Work-Energy Theorem: What work is needed to stop wheel rotating: F R 60 rad/s R = 0.30 m Work = r 4 kg F = 40 N First find I for wheel: I = mr = (4 kg)(0.3 m) = 0.36 kg m 0 ½I ½I Work = -½I f 0 Work = -½(0.36 kg m )(60 rad/s) Work = -648 J

15 Combined Rotation and Translation v cm v cm v cm Now consider a ball rolling without slipping. The angular velocity about the point P is same as for disk, so that we write: First consider a disk sliding without friction. The velocity of any part is equal to velocity v cm of the center of mass. R P v Or v R R v

16 Two Kinds of Kinetic Energy Kinetic Energy of Translation: K = ½mv Kinetic Energy of Rotation: K = ½I R P v Total Kinetic Energy of a Rolling Object: T 1 1 K mv I

17 Angular/Linear Conversions In many applications, you must solve an equation with both angular and linear parameters. It is necessary to remember the bridges: Displacement: Velocity: Acceleration: s R s R v R v R v R a R

18 Translation or Rotation? If you are to solve for a linear parameter, you must convert all angular terms to linear terms: s R v a I (?) mr R R If you are to solve for an angular parameter, you must convert all linear terms to angular terms: s R v R v R

19 Example (a): Find velocity v of a disk if given its total kinetic energy E. Total energy: E = ½mv + ½I E mv I ; I mr ; v E mv mr E mv mv R ; 4 E 3mv 4 E or v 4 3m v R

20 Example (b) Find angular velocity of a disk given its total kinetic energy E. Total energy: E = ½mv + ½I E mv I ; I mr ; vr E m( R) mr ; E mr mr E 3mR 4 E or 4 3mR

21 Strategy for Problems Draw and label a sketch of the problem. List givens and state what is to be found. Write formulas for finding the moments of inertia for each body that is in rotation. Recall concepts involved (power, energy, work, conservation, etc.) and write an equation involving the unknown quantity. Solve for the unknown quantity.

22 Example 5: A circular hoop and a circular disk, each of the same mass and radius, roll at a linear speed v.. Compare the kinetic energies. Two kinds of energy: K T = ½mv K r = ½I Total energy: E = ½mv + ½I = v R Disk: ½ ½ ½ v E mv mr R E = ¾mv Hoop: ½ ½ v E mv mr R E = mv v v

23 Conservation of Energy The total energy is still conserved for systems in rotation and translation. However, rotation must now be considered. Begin: (U + K t + K R ) o = End: (U + K t + K R ) f Height? Rotation? velocity? mgh o ½ ½mv o = mgh f ½ f ½mv f Height? Rotation? velocity?

24 Example 6: Find the velocity of the -kg mass just before it strikes the floor. mgh o ½ ½mv o = mgh f ½ f ½mv f R = 50 cm 6 kg kg h = 10 m mgh mv I v mgh0 mv ( MR ) R ()(9.8)(10) () v (6) v I MR.5v = 196 m /s v = 8.85 m/s

25 Example 7: A hoop and a disk roll from the top of an incline. What are their speeds at the bottom if the initial height is 0 m? m mgh o = ½mv + ½I Hoop: I = mr v mgh0 ½mv ½( mr ) R mgh o = ½mv + ½mv ; mgh o = mv 0 m Hoop: v = 14 m/s v gh 0 (9.8 m/s )(0 m) Disk: I = ½mR ; mgh o = ½mv + ½I v mgh0 ½mv ½(½ mr ) R v gh v = 16. m/s

26 Angular Momentum Defined Consider a particle m moving with velocity v in a circle of radius r. Define angular momentum L: L = mvr Substituting v= r, gives: L = m(r) r = mr For extended rotating body: L = (mr ) v = r axis m 1 m m 4 m m 3 Object rotating at constant Since I = mr, we have: L = I Angular Momentum

27 Example 8: Find the angular L = m momentum of a thin 4-kg rod of length m if it rotates about its midpoint at a speed of 300 rpm. m = 4 kg For rod: I = ml = (4 kg)( m) I = 1.33 kg m rev rad 1 min rad/s min 1 rev 60 s L = I(1.33 kg m )(31.4 rad/s) L = 1315 kg m /s

28 Impulse and Momentum Recall for linear motion the linear impulse is equal to the change in linear momentum: F t mv mv Using angular analogies, we find angular impulse to be equal to the change in angular momentum: f 0 t I I f 0

29 Example 9: A sharp force of 00 N is applied to the edge of a wheel free to rotate. The force acts for 0.00 s. What is the final angular velocity? I = mr = ( kg)(0.4 m) I = 0.3 kg m Applied torque FR t = 0.00 s R F kg 0 rad/s R = 0.40 m F = 00 N Impulse = change in angular momentum t = I f o 0 FR t = I f f FRt (00 N)(0.4 m)(0.00 s) I 0.3 m f = 0.5 rad/s

30 Conservation of Momentum In the absence of external torque the rotational momentum of a system is conserved (constant). 0 I f f o = t I f f o I o = kg m ; = 600 rpm I f = 6 kg m ; =? f I00 I f ( kg m )(600 rpm) 6 kg m f = 00 rpm

31 Summary Rotational Analogies Quantity Linear Rotational Displacement Displacement x Radians Inertia Mass (kg) I (kgm ) Force Newtons N Torque N m Velocity v m/s Rad/s Acceleration a m/s Rad/s Momentum mv (kg m/s) I (kgm rad/s)

32 Analogous Formulas Linear Motion F = ma Rotational Motion = I K = ½mv K = ½I Work = Fx Power = Fv Fx = ½mv f -½mv o Work = Power = I = ½I f - ½I o

33 Summary of Formulas: I = mr K 1 I Work I I o o f f ½I ½I f 0 Power t Height? Rotation? velocity? mgh o ½ ½mv o = mgh f ½ f ½mv f Height? Rotation? velocity?

34 CONCLUSION: Chapter 11B Rigid Body Rotation

Lecture 17. Last time we saw that the rotational analog of Newton s 2nd Law is

Lecture 17. Last time we saw that the rotational analog of Newton s 2nd Law is Lecture 17 Rotational Dynamics Rotational Kinetic Energy Stress and Strain and Springs Cutnell+Johnson: 9.4-9.6, 10.1-10.2 Rotational Dynamics (some more) Last time we saw that the rotational analog of