# Video: Determining the Angular Displacement of a Point on a Rotating Object

A piece of dust falls onto a spinning compact disc and sticks in place. The spin rate of the disk is 5.00 × 10² rpm and the piece of dust lands 4.30 cm from the disk’s centre. What is the total distance travelled by the piece of dust in 1.80 × 10² s 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.

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### 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.