Chapter 12: Rotation of Rigid Bodies. Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics

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1 Chapter 12: Rotation of Rigid Bodies Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics

2 Translational vs Rotational 2 / / 1/ 2 m x v dx dt a dv dt F ma p mv KE mv Work Fd P Fv 2 / / 1/ 2 I d dt d dt I L I KE I Work P 2 c t s r v r a r a r Fr L pr Connection

3 Center of Mass The geometric center or average location of the mass.

4 Rotational & Translational Motion Objects rotate about their Center of Mass. The Center of Mass Translates as if it were a point particle. v d r CM dt CM

5 Center of Mass The Center of Mass Translates as if it were a point particle and, if no external forces act on the system, momentum is then conserved. This means: EVEN if the bat EXPLODED into a thousand pieces, all the pieces would move so that the momentum of the CM is conserved that is, the CM continues in the parabolic trajectory!!!! THIS IS VERY VERY IMPORTANT!

System of Particles Center of Mass A projectile is fired into the air and suddenly explodes With no explosion, the projectile would follow the dotted line After the explosion, the center of mass of

6 System of Particles Center of Mass A projectile is fired into the air and suddenly explodes With no explosion, the projectile would follow the dotted line After the explosion, the center of mass of the fragments still follows the dotted line, the same parabolic path the projectile would have followed with no explosion! If no external forces act on the system, then the velocity of the CM doesn t change!!

7 Center of Mass: Stability If the Center of Mass is above the base of support the object will be stable. If not, it topples over.

8 Balance and Stability This dancer balances en pointe by having her center of mass directly over her toes, her base of support. Slide 12-88

9 Center of Mass The geometric center or average location of the mass. Extended Body: System of Particles: r CM 1 M r dm x CM i i mx M i

10 CM Lecture Problem 1. Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the meter stick. A) What is the CM of the system? B) What is the torque acting on the system? C) What is the rotational inertia of the system about the end? D) If this body is released from rest in a horizontal position, what is the angular acceleration at the release? E) What is the angular speed of the meter stick as it swings through its lowest position?

11 Prelab A meter stick has a mass of 75.0 grams and has two masses attached to it 50.0 grams at the 20.0cm mark and grams at the 75.0 cm mark. (a) Find the center of mass of the system - that is, at what mark on the meter stick should the fulcrum be placed so that the system balances? (b) A fulcrum is then placed at the CM. Sketch and label the system. (c) Show that the net torque about the cm is zero. (d) Calculate the rotational inertia of the system about the CM axis. Box answers, make it neat.

12 Center of Mass The geometric center or average location of the mass. Extended Body: r CM 1 M r dm

As m 0 and is replaced by dm, the sums become Before these can be

13 Center of Mass of a Solid Object Divide a solid object into many small cells of mass m. As m 0 and is replaced by dm, the sums become Before these can be integrated: dm must be replaced by expressions using dx and dy. Integration limits must be established. Slide 12-41

14 Example 12.2 The Center of Mass of a Rod Slide 12-42

15 Example 12.2 The Center of Mass of a Rod Slide 12-43

16 Example Extended Body: 1 rcm rdm M You must generate an expression for the density and the mass differential, dm, from geometry and by analyzing a strip of the sign. We assume the sign has uniform density. If M is the total mass then the total volume and density is given by: x CM V 1 M abt, 2 1 abt 2 M 2My dm dv ( ytdx) ytdx dx 1 abt ab 2 1 M x dm Where a, b and t are the width, height and thickness of the sign, respectively. Then the mass element for the strip shown is: a 3 a 1 2My 2 b 2 x 2 x dx x ( x) dx a M ab ab a a

17 Newton s 2 nd Law for Rotation The net external torques acting on an object around an axis is equal to the rotational inertia times the angular acceleration. I Acceleration thing Force thing Inertia thing The rotational equation is limited to rotation about a single principal axis, which in simple cases is an axis of symmetry.

18 Torque: Causes Rotations Fr sin Fd lever arm: d rsin The moment arm, d, is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force The horizontal component of F (F cos ) has no tendency to produce a rotation

19 Torque: Causes Rotations Fr sin Fd The direction convention is: Counterclockwise rotations are positive. Clockwise rotations are negative.

20 Newton s 1 st Law for Rotation If the sum of the torques is zero, the system is in rotational equilibrium. boy 500N 1.5m 750Nm 0 girl 250N 3m 750Nm

21 Torque Is there a difference in torque? (Ignore the mass of the rope) NO! In either case, the lever arm is the same! What is it? 3m

22 CM Lecture Problem 1. Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the meter stick. A) What is the CM of the system? B) What is the torque acting on the system? C) What is the rotational inertia of the system about the end? D) If this body is released from rest in a horizontal position, what is the angular acceleration at the release? E) What is the angular speed of the meter stick as it swings through its lowest position?

23 Newton s 2 nd Law for Rotation The net external torques acting on an object around an axis is equal to the rotational inertia times the angular acceleration. I Acceleration thing Force thing Inertia thing The rotational equation is limited to rotation about a single principal axis, which in simple cases is an axis of symmetry.

24 Torque is a Vector! = r x F The direction is given by the right hand rule where the fingers extend along r and fold into F. The Thumb gives the direction of.

plane formed by A and B The best way to determine this direction is to use the

25 The Vector Product The magnitude of C is AB sin and is equal to the area of the parallelogram formed by A and B The direction of C is perpendicular to the plane formed by A and B The best way to determine this direction is to use the right-hand rule A B ˆ i ˆ j k ˆ A B A B A B A B A B A B y z z y x z z x x y y x

26 COMPARE! rf Fr sin Fd CROSS PRODUCT F and d must be mutually perpendicular! L r p L mvr sin CROSS PRODUCT L and p must be mutually perpendicular! DOT PRODUCT W F d Fd cos F and d must be mutually PARALLEL!

27 A B ˆ i ˆ j k ˆ A B A B A B A B A B A B y z z y x z z x x y y x Two vectors lying in the xy plane are given by the equations A = 5i + 2j and B = 2i 3j. The value of AxB is a. 19k b. 11k c. 19k d. 11k e. 10i j

28 CM Lecture Problem 1. Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the meter stick. A) What is the CM of the system? B) What is the torque acting on the system? C) What is the rotational inertia of the system about the end? D) If this body is released from rest in a horizontal position, what is the angular acceleration at the release? E) What is the angular speed of the meter stick as it swings through its lowest position?

29 Net external torques

30 Find the Net Torque Fr sin Fd Fd F d ( 20 N)(.5 m) (35 N)(1.10 msin60) OR ( 20 N)(.5 m) (35N cos30)(1.10 m) F and d must be mutually perpendicular! 23.3 Nm CCW

Chapter 12: Rotation of Rigid Bodies. Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics

Chapter 12: Rotation of Rigid Bodies. Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics Chapter 1: Rotation of Rigid Bodies Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics Translational vs Rotational / / 1/ m x v dx dt a dv dt F ma p mv KE mv Work Fd P Fv / / 1/ I