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
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19 only please formula sheet provided please must use that also exaplain me east way step by step please
Transcribed Image Text:Circular motion: a =
R
Weight: F-mg, ;
Faet - ma;
g= 9.8 m/s';
Pavr
At
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, - -
o = 0e + at;
e- o t+
at²;
20a - o - oa
T=rFsing;
T-rx F;
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;
Ipaint mass = mr
Ia =mR?
La - mR
Inoup = mR?
Irodicenter)
ml?
Irodiend)
a. 14
b. 4
dW
3.14
I= leom + MD?
work: W=t 0;
Pr=
At
W =
dt
d. 6.28
Rolling:
K= lo 2
Fmas" HFn
Veom - Ro
Incline: Fe mgsine F- mgcos0
Angular momentum:
Lpoint mass = m rxv
L=mrvsin 0;
L=m (r.vy - r,V.)k
L- lo
m,X1 +m,X,
miyı+m2y2
m1 +m2
LI- Lr
I o =I 2002
Xcom =
Ycom
m, +m2
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