1 / 18

180 likes | 279 Vues

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

Télécharger la présentation
## Physics 2053C – Fall 2001

**An Image/Link below is provided (as is) to download presentation**
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.
Content is provided to you AS IS for your information and personal use only.
Download presentation by click this link.
While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

**Physics 2053C – Fall 2001**Chapter 8 Rotational Motion Dr. Larry Dennis, FSU Department of Physics**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 + ½t2**F1**F2 R1 R2 Rotational Forces Torque • = I = RFsin or RF**Rotational Energy & Momentum**• Kinetic Energy: K = ½I2 • Rotational energy is often conserved. • Rotational energy must be included in total energy. • Angular Momentum: L = I • Angular momentum is conserved.**R**M M2 R2 R1 M1 Moments of Inertia I = MR2 I = M1R12 + M2R22**Moments of Inertia**Hollow Disk I = MR2 Solid Disk I = ½MR2 Solid Sphere I = 2/5MR2**Conservation of Angular Momentum**Change in angular momentum = Torque If then the angular momentum remains constant. (I)before=(I)after**Conservation of Angular Momentum**• Examples: • Figure Skater • Diver • Gymnast**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**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**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**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**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**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**CAPA 11 & 12**(I)initial=(I)final Iii=Iff f=Iii /If**CAPA 12**(½I2)initialand(½I2)final are not the same. ½Iii2<½Iff2**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**Next Time**• Quiz on Chapter 8 • Begin Chapter 9 • Please see me with any questions or comments. See you on Wednesday.

More Related