### Video Transcript

A piece of dust falls onto a
spinning compact disc and sticks in place. The spin rate of the disk is 5.00
times 10 to the two rpm and the piece of dust lands 4.30 centimetres from the disk’s
centre. What is the total distance
travelled by the piece of dust in 1.80 times 10 to the two seconds of rotation? Assume that the disk is negligibly
slowed by the dust falling on it and that the time taken for the dust to accelerate
to the same speed as the part of the disk that it lands on is negligible.

We’ll call this total distance
travelled by the piece of dust lowercase 𝑠. In this scenario, we’re told that a
piece of dust falls at distance—we’ve called 𝑟—of 4.30 centimetres from the centre
of a rotating disc. The disc is spinning at an angular
speed—we’ve called capital Ω—5.00 times 10 to the two revolutions every minute. After the dust has been on the
rotating disk for a time of 1.80 times 10 to the two seconds, we want to know the
linear distance 𝑠 it has travelled. The linear distance 𝑠 will equal
the number of times the disc rotates in this amount of time multiplied by two times
𝜋 times 𝑟.

We’re given the distance 𝑟 from
the centre of rotation at which the dust settles. But we don’t yet know how many
times the disc rotates in our amount of time 𝑡. To figure that out, we can use our
angular rotation rate capital Ω. To make capital Ω as useful as
possible to us, we’ll convert it from its given units of revolutions per minute to
revolutions per second. To do that, we can multiply capital
Ω by one minute per 60 seconds, which is effectively multiplying it by one. But what it does do is it cancels
out the units of minutes and leaves us with time units of seconds. The total number of rotations our
disk will make in the time 𝑡 is equal to that time multiplied by the rotational
rate.

When we plug in our given value for
𝑡, we see that the units of seconds cancel out and the units of our number of
rotations will indeed be in revolutions. When we plug these numbers in for
𝑁 rotation in our equation for 𝑠 and then insert the value of our radius 𝑟
converting it from units of centimetres to units of metres, we see looking at this
overall expression that the units of revolutions cancel out. And we’ll be left with an answer in
units of metres. When we calculate 𝑠, we find that
to three significant figures, it’s 405 metres. That’s the distance this piece of
dust would travel in the amount of time 𝑡.