Circular motion: a. Fest - ma; Weight: F;= mg,; Pur =- At g= 9.8 m/s: R Kinetic energy: K-m v: Potential energy: Ug = mgy E - Er E- U+K Rotational motion: 1 rev = 2n rad; v = o r; a-a r: a, - e - o, t+ at?; 28a = o - o K=-lo ? o= + at: T-rx F; T=rFsing; 19. An Atwood machine (similar to an elevator with a counter-weight) is initially at rest. On one side is a mass of 2.00 kg and on the other side is a mass of 1.00 kg. They are connected by a massless wire that passes over a pulley. The pulley has a mass of 4.00 kg. a radius of 20.0 cm and no friction. The heavier mass then falls for 50.0 cm. What is the linear speed at that point? ET = la; Laa mR? Ipaint ma= mr Isoep = mR Iadicenter) Irediend)= mR² mR? Ibali = 14 a. b. 4 K-? w- , dw C. 3.14 1= leam + MD Par = At work: W-t 0; P= dt d. 6.28 Rolling: Veom- Ra K= Fumas = H.Fn Incline: Fe- mgsino Fe- mgcos0 Angular momentum: Lpoiat mass m rxv L=mrvsin 0; L=m (r.vy - r,V)k L- lo m,X1 +m,X2 X com m1y1+m2y2 Y com = m1 +m2 I o1 =1 202 m, +m2

Circular motion: a. Fest - ma; Weight: F;= mg,; Pur =- At g= 9.8 m/s: R Kinetic energy: K-m v: Potential energy: Ug = mgy E - Er E- U+K Rotational motion: 1 rev = 2n rad; v = o r; a-a r: a, - e - o, t+ at?; 28a = o - o K=-lo ? o= + at: T-rx F; T=rFsing; 19. An Atwood machine (similar to an elevator with a counter-weight) is initially at rest. On one side is a mass of 2.00 kg and on the other side is a mass of 1.00 kg. They are connected by a massless wire that passes over a pulley. The pulley has a mass of 4.00 kg. a radius of 20.0 cm and no friction. The heavier mass then falls for 50.0 cm. What is the linear speed at that point? ET = la; Laa mR? Ipaint ma= mr Isoep = mR Iadicenter) Irediend)= mR² mR? Ibali = 14 a. b. 4 K-? w- , dw C. 3.14 1= leam + MD Par = At work: W-t 0; P= dt d. 6.28 Rolling: Veom- Ra K= Fumas = H.Fn Incline: Fe- mgsino Fe- mgcos0 Angular momentum: Lpoiat mass m rxv L=mrvsin 0; L=m (r.vy - r,V)k L- lo m,X1 +m,X2 X com m1y1+m2y2 Y com = m1 +m2 I o1 =1 202 m, +m2