- Thread starter
- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,928

my ? on (i) is since it is asking for an angle and not a arc length

then the angle generated would be just

\(\displaystyle \displaystyle 8\frac{\pi}{12}=\frac{2\pi}{3}\)

or not?

- Thread starter karush
- Start date

- Thread starter
- #1

- Jan 31, 2012

- 2,928

my ? on (i) is since it is asking for an angle and not a arc length

then the angle generated would be just

\(\displaystyle \displaystyle 8\frac{\pi}{12}=\frac{2\pi}{3}\)

or not?

- Admin
- #2

\(\displaystyle \theta=\omega t=\frac{\pi}{12}\frac{\text{rad}}{\text{s}} \cdot8\text{s}=\frac{2\pi}{3}\text{ rad}\)

- Thread starter
- #3

- Jan 31, 2012

- 2,928

so (ii) would be just $\text {s}=\theta\text{ r}$

$\displaystyle\frac{2\pi}{3}\text{rad}\cdot60\text { cm}=40\ \pi\text { cm}$

and (iii) would be

$\displaystyle \frac{5\pi\text { cm}}{s}$

$\displaystyle\frac{2\pi}{3}\text{rad}\cdot60\text { cm}=40\ \pi\text { cm}$

and (iii) would be

$\displaystyle \frac{5\pi\text { cm}}{s}$

Last edited:

- Admin
- #4

Yes, those are correct.so (ii) would be just $\text {s}=\theta\text{ r}$

$\displaystyle\frac{2\pi}{3}\text{rad}\cdot60\text { cm}=40\ \pi\text { cm}$

and (iii) would be

$\displaystyle \frac{5\pi\text { cm}}{s}$