MHB Angular and Linear Speed of a Point

[karush]

View attachment 1429

my ? on (i) is since it is asking for an angle and not a arc length
then the angle generated would be just

$$\displaystyle 8\frac{\pi}{12}=\frac{2\pi}{3}$$

or not?

 

[MarkFL]

Yes, that's right, although you might want to get in the habit of carrying the units as a means of making sure your result is dimensionally correct:

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

 

[karush]

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

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

 

[MarkFL]

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

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

Yes, those are correct.(Sun)