# Mechanics11/Seite14/Loesung5.1

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# 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)