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# Physics 2053C – Fall 2001

Physics 2053C – Fall 2001. Chapter 8 Rotational Motion. Rotational Motion. Uses all the above concepts and, Rolling without slipping. v =  x /  t a =  v /  t When a = constant: v = v o + a t x = x o + v o t + ½ a t 2.  =   /  t  =   /  t

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## Physics 2053C – Fall 2001

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1. Physics 2053C – Fall 2001 Chapter 8 Rotational Motion Dr. Larry Dennis, FSU Department of Physics

2. Rotational Motion • Uses all the above concepts and, • Rolling without slipping. • v = x/t • a = v/t • When a = constant: • v = vo + at • x = xo + vot + ½at2 •  = /t •  = /t • When  = constant: •  = o + t •  = o + ot + ½t2

3. F1 F2 R1 R2 Rotational Forces  Torque •  = I = RFsin or RF

4. Rotational Energy & Momentum • Kinetic Energy: K = ½I2 • Rotational energy is often conserved. • Rotational energy must be included in total energy. • Angular Momentum: L = I • Angular momentum is conserved.

5. R M M2 R2 R1 M1 Moments of Inertia I = MR2 I = M1R12 + M2R22

6. Moments of Inertia Hollow Disk I = MR2 Solid Disk I = ½MR2 Solid Sphere I = 2/5MR2

7. Conservation of Angular Momentum Change in angular momentum = Torque If then the angular momentum remains constant. (I)before=(I)after

8. Conservation of Angular Momentum • Examples: • Figure Skater • Diver • Gymnast

9. CAPA 9 & 10 • A day-care worker pushes tangentially on a hand-driven merry-go-round and is able to accelerate it from rest to a spinning rate of 18.0 rpm in 9 seconds. Assume the merry-go-round is a disk of radius 2.7 m and has a mass of 750 kg, and two children (each with a mass of 25 kg) sit opposite each other on the edge. • Calculate the torque required to produce the acceleration, neglecting any frictional torque. • What is the force required?  = I

10. CAPA 9 & 10 • A day-care worker pushes tangentially on a hand-driven merry-go-round and is able to accelerate it from rest to a spinning rate of 18.0 rpm in 9 seconds. Assume the merry-go-round is a disk of radius 2.7 m and has a mass of 750 kg, and two children (each with a mass of 25 kg) sit opposite each other on the edge. • Calculate the torque required to produce the acceleration, neglecting any frictional torque. • What is the force required? I = 2*McRc2 + ½MdRd2 I = (2*Mc +½Md )R2

11. CAPA 9 & 10 • A day-care worker pushes tangentially on a hand-driven merry-go-round and is able to accelerate it from rest to a spinning rate of 18.0 rpm in 9 seconds. Assume the merry-go-round is a disk of radius 2.7 m and has a mass of 750 kg, and two children (each with a mass of 25 kg) sit opposite each other on the edge. I = (2*Mc +½Md )Rc2 I = (2*25+½*750)2.72 I = 3098 kg*m2

12. CAPA 9 & 10 A day-care worker pushes tangentially on a hand-driven merry-go-round and is able to accelerate it from rest to a spinning rate of 18.0 rpm in 9 seconds. Assume the merry-go-round is a disk of radius 2.7 m and has a mass of 750 kg, and two children (each with a mass of 25 kg) sit opposite each other on the edge. 8. Calculate the torque required to produce the acceleration, neglecting any frictional torque.  = I  = /t = 18 rpm*1/60 min/s * 2 rad/s /9 s = 0.209 rad/s2

13. CAPA 9 & 10 A day-care worker pushes tangentially on a hand-driven merry-go-round and is able to accelerate it from rest to a spinning rate of 18.0 rpm in 9 seconds. Assume the merry-go-round is a disk of radius 2.7 m and has a mass of 750 kg, and two children (each with a mass of 25 kg) sit opposite each other on the edge. 8. Calculate the torque required to produce the acceleration, neglecting any frictional torque.  = I  = 3098 kg*m2 *0.209 rad/s2  = 649 kg*m2/s2 = 649 N-m

14. F R CAPA 9 & 10 A day-care worker pushes tangentially on a hand-driven merry-go-round and is able to accelerate it from rest to a spinning rate of 18.0 rpm in 9 seconds. Assume the merry-go-round is a disk of radius 2.7 m and has a mass of 750 kg, and two children (each with a mass of 25 kg) sit opposite each other on the edge. 9. What is the force required?  = FR sin F =  /Rsin = 649 N-m/( 2.7 m * 1 ) F = 240 N

15. CAPA 11 & 12 (I)initial=(I)final Iii=Iff f=Iii /If

16. CAPA 12 (½I2)initialand(½I2)final are not the same. ½Iii2<½Iff2

17. Quiz #5 – Chapter 8 Sample Questions: Chap. 8: 2, 8, 12 Sample Problems: Chap. 8: 17, 43, 66 Answers to 66: A) 1.19 rad/s B) Before = 2000 J After = 1190 J

18. Next Time • Quiz on Chapter 8 • Begin Chapter 9 • Please see me with any questions or comments. See you on Wednesday.

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