Question

suppose a piece of dust finds itself on a cd. if the spin rate of the cd is 395 rpm, and the piece of dust is 4.0 cm from the center, what is the total distance (in m) traveled by the dust in 8 minutes? (ignore accelerations due to getting the cd rotating.)

Explanation:

Step1: Convert radius to meters

The radius \(r = 4.0\ cm=0.04\ m\) (since \(1\ m = 100\ cm\)).

Step2: Find the number of revolutions \(N\)

The spin rate is \(n = 395\ rpm\) (revolutions per minute). In \(t = 8\) minutes, the number of revolutions \(N=nt\).
So \(N=395\times8 = 3160\) revolutions.

Step3: Find the circumference \(C\) of the circular path

The circumference of a circle is \(C = 2\pi r\). Substituting \(r = 0.04\ m\), we get \(C=2\pi\times0.04=\frac{2\times3.14\times0.04}{1}=0.2512\ m\) (using \(\pi\approx3.14\)).

Step4: Calculate the total distance \(d\)

The total distance \(d = N\times C\). Substituting \(N = 3160\) and \(C=0.2512\ m\), we have \(d=3160\times0.2512\).

$$
LATEXBLOCK0
$$

Answer:

\(794\ m\) (rounded to the nearest whole number)