Mechanics11/Seite14/Loesung5.1

Lösung
=5.1 a.=

$$w=\frac {\Delta \phi}{\Delta t}$$

$$w=\frac {26\pi}{300s}= \frac {13\pi}{150}*s^{-1}$$

Umlaufdauer: $$5min=300s $$ $$300s:13=23,1s$$

--Lenny 10:57, 18 April 2008 (UTC) --Benni.m 10:59, 18 April 2008 (UTC)

2.Lösung
a) Umlaufdauer ca. 23,1 s  $$\;w = frac{\Delta \phi }{\Delta t}$$  = $$\frac{26 \pi}{300s}$$ =  $$\frac{13 \pi}{150s^-1}$$

--Nikih 11:01, 18 April 2008 (UTC) --Kiprilo 11:03, 18 April 2008 (UTC)

5.1. Drehwurm:

a)

ges: Frequenz, Umlaufdauer, Winkelgeschwindigkeit

geg: $$\Delta t$$ = 5min = 5 * 60s = 300s

$$\Delta \phi$$ = 13 * $$2 \pi$$ = 26$$\pi$$

-> Umlaufdauer pro Runde: 300s : 13 = 23,1 s

-> Winkelgeschwindigkeit: $$\omega = \frac{\Delta \phi}{\Delta t} = \frac{26 \pi}{300 s } = \frac{13 \pi}{150 } s^{-1}$$

-> Frequenz: 13 : 300s = 0,043 $$\frac{1}{s}$$

b)

ges: $$\Delta \phi$$

geg: $$\Delta t$$ = 10s

$$\omega = \frac{13 \pi}{150 } s^{-1}$$

$$\omega = \frac{\Delta \phi}{\Delta t}$$

-> $$\Delta \phi$$ = $$\omega *  t$$

= $$\frac{13 \pi}{150 } s^{-1}$$ * 10s =  $$\frac{130 \pi}{150 }$$ = $$\frac{13 \pi}{15 }$$ = 2,7227

c)

ges: $$\Delta t$$

geg: $$\Delta \phi$$ = $$5 \pi$$

$$\omega $$ = $$\omega $$1

$$\frac{5 \pi}{\Delta t}$$= $$\frac{26 \pi}{300s}$$

-> $$\Delta t$$ = $$5 \pi$$ * 300s : $$26 \pi$$

= 57,69 s

d)

ges: b

geg: $$\Delta t$$ = 2min = 120s

r = 5,0m

$$\omega = \frac{13 \pi}{150 } s^{-1}$$

$$\frac{26 \pi}{300s}$$ = $$\frac{x2 \pi}{120s}$$

-> x = 5,2

b = x * U = x * $$2 \pi$$r

= 5,2 * $$2 \pi$$ * 5,0m

= $$52 \pi$$m = 163,36m

--Anki 12:39, 20 April 2008 (UTC)