# Speed, diameter, and RPM/RPS

Discussion in 'Physics' started by iulian28ti, Mar 16, 2010.

1. ### iulian28ti Thread Starter Member

Dec 4, 2009
42
0
Here's my question.

A body moves along the path of a circle of diameter D at a speed V.

Whats the frequency ?

2. ### iulian28ti Thread Starter Member

Dec 4, 2009
42
0
I'm pretty sure there's something with Pi*D, and something

3. ### someonesdad Senior Member

Jul 7, 2009
1,585
142
I'll assume you mean what's the frequency of rotation, right?

If so, the way to solve the problem is to note that the distance traveled around the circumference of the circle constitutes one cycle. Then you find the time it takes to travel this distance and the frequency is the reciprocal of that time.

The frequency is in cycles per second (Hz or Hertz) -- so you can see the reciprocal has seconds per cycle (which is the period).

4. ### Wendy Moderator

Mar 24, 2008
21,416
2,948

$f=2{\pi}r={\pi}d$

where
d=diameter

5. ### davebee Well-Known Member

Oct 22, 2008
539
47
speed = distance/time

circumference = pi * diameter

period: (time per cycle) = circumference/speed = (pi * diameter)/speed

frequency = 1/period = speed/(pi * diameter)

So...

frequency = speed/(pi * diameter)

The placement of the variables does seem to make sense - a higher speed gives a higher frequency, and a larger diameter gives a lower frequency, as you'd expect, although this sort of analysis can't tell you whether pi is either needed, or is in the right place.

In looking at the units, the distances cancel, leaving the result having units of one over time, which is required of a frequency:

frequency = speed/(pi * diameter) = (distance/time) / (pi * diameter) = length/(time * pi * length) = 1/time

6. ### someonesdad Senior Member

Jul 7, 2009
1,585
142
Bill, your equation is dimensionally inconsistent, assuming you mean f to be frequency (you equated 1/time to distance). It's also nice to just give the OP a hint or a principal to solve the problem, not solve it for them -- it helps them learn, rather than just parrot back a result.

7. ### davebee Well-Known Member

Oct 22, 2008
539
47
What you say makes sense, someonesdad. Maybe I should have said less.

8. ### Wendy Moderator

Mar 24, 2008
21,416
2,948
This isn't the homework section, and the answer was correct (give that r and d both have dimensions). Just my 2¢

9. ### davebee Well-Known Member

Oct 22, 2008
539
47
Bill, the problem is not presented very rigorously, so any answer will be subject to one's interpretation of the original problem, but even with that said, how can the frequency (however that is interpreted) be independent of the speed, as you state?

I read the problem to mean that one circumnavigation of the circle by the body would be one cycle, with the frequency being how many cycles per second occurred.

With that interpretation, a speed of zero would lead to a frequency of zero cycles per second, a high body speed would correspond to a high frequency in cycles per second, etc.

How are you interpreting frequency so that it's independent of speed?

10. ### Wendy Moderator

Mar 24, 2008
21,416
2,948
OK, see what you're saying. I left the seconds off.

11. ### iulian28ti Thread Starter Member

Dec 4, 2009
42
0
Yup.... this is what i was looking for.

Thanks !

12. ### Duane P Wetick Senior Member

Apr 23, 2009
416
21
The time factor is missing from your equation. Frequency = 1/ time. You could say time is the Velocity / distance, therefore Frequency = 1/ velocity/ distance.

Cheers, DPW [ Everything has limitations...and I hate limitations.]