Question

Suppose a piece of dust has fallen on a CD. If the spin rate of the CD is 500 rpm, and the piece of dust is 4.3 cm from the center, what is the total distance traveled by the dust in 3 minutes? (Ignore accelerations due to getting the CD rotating.)

Answer 1

Answer:

s = 405.27 m

Explanation:

given,

spin rate of the CD = 500 rpm

distance of dust, r = 4.3 cm

time = 3 minute = 3 x 60 = 180 s

total distance traveled by the dust, d = ?

$\omega = 500\times \dfrac{2\pi}{60}$

$\omega = 52.36\ rad/s$

we know,

$\Delta \theta = \omega\ t$

$\Delta \theta =52.36\times 180$

$\Delta \theta = 9424.8\ rad$

distance traveled by the dust particle

s = Δ θ x r

s = 9424.8  x 0.043

s = 405.27 m

Hence, distance traveled by the dust particle is 405.27 m