Another Circular motion question

[laker_gurl3]

thanks so much for any help..work and formula used would be appreciated...

A record of diameter 30cm roates on a turntable at 33.3r/min.

a.) How fast is the outside edge of the record moving?

b.) how many times as fast would it move if the frequency were raised to 78 r/min.?

 

[Gale]

thanks so much for any help..work and formula used would be appreciated...

A record of diameter 30cm roates on a turntable at 33.3r/min.

a.) How fast is the outside edge of the record moving?

b.) how many times as fast would it move if the frequency were raised to 78 r/min.?

a) if the whole record moves at 33.3 revs per min, then a point on the outside edge does as well, which means that point has to go around the circumfrence (C) of the record within the minute, your speed= C/min.

b) same idea as part a, change the frequecny, and then compare.

 

[laker_gurl3]

so i did this out..for A.) i got 3138m/s
for B.) i got 7351m/s, therefore it's 2.34 times as fast...is that correct?

 

[Gale]

ok ya, i mean it had to go around the circumfrence 33.3 times per minute, but that's what you did, so good. Those are the numbers i got, except your units are wrong, its cm/min not m/s.

 

[dextercioby]

Gale,i get double for the first number...

$$v=\omega R \ [m \ s^{-1}]$$

R=0.3m

$$\omega=2\pi \nu=\left(2\pi \ \mbox{rad}\right) \left(\frac{33.3}{60} \ Hz \right) \simeq \frac{200\pi}{180} \mbox{rad} \ s^{-1}$$

Ergo

$$v\simeq \frac{\pi}{3} m \ s^{-1} = \frac{6000\pi}{3} cm \ (min)^{-1}$$

which is double that the # you referred to in post #4.

Daniel.


EDIT:As Gale pointed out,the radius is only 0.15m.So that explains the incorrect result i got.

 

[Gale]

you got double because you let R=.3 whereas .3 is the diameter.

 

[dextercioby]

****.

Daniel.