Question

A solid disk rotates in the horizontal plane at an angular velocity of 0.0639 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.149 kg·m2. From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of 0.409 m from the axis. The sand in the ring has a mass of 0.502 kg. After all the sand is in place, what is the angular velocity of the disk?

          A solid disk rotates in the horizontal plane at an angular
velocity of 0.0639 rad/s with respect to an axis perpendicular to
the disk at its center. The moment of inertia of the disk is 0.149
kg·m2. From above, sand is dropped straight down onto this rotating
disk, so that a thin uniform ring of sand is formed at a distance
of 0.409 m from the axis. The sand in the ring has a mass of 0.502
kg. After all the sand is in place, what is the angular velocity of
the disk?

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Added by Paul H.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Ryan Mcnamara

05/05/2022

Step 1

Given: Mass of sand, m = 0.502 kg Radius of sand from the center, r = 0.409 m Moment of inertia of the disk, I_disk = 0.149 kg·m^2 Using the formula for moment of inertia of a ring, I_ring = m * r^2, we can calculate the moment of inertia of the sand: I_sand = Show more…

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A solid disk rotates in the horizontal plane at an angular velocity of 0.0639 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.149 kg·m2. From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of 0.409 m from the axis. The sand in the ring has a mass of 0.502 kg. After all the sand is in place, what is the angular velocity of the disk?

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Transcript

00:01 Okay, so you've got a disk that's rotating, and it has an initial angular velocity of 0 .06, and initial rotational inertia of 0 .149.

00:11 And then sand is dropped on top of it, which changes the rotational inertia.

00:16 We're going to have to figure that out.

00:17 And it's also going to change the angular velocity, which we'll need to figure that out.

00:23 And that's going to be our final answer.

00:25 So the main concept here is that angular velocity initial equals angular.

00:31 Velocity final.

00:33 I said that wrong.

00:33 I'm sorry.

00:34 Angular momentum initial equals angular momentum final.

00:38 So conservation of angular momentum.

00:40 So the initial is pretty much given to us here.

00:43 It's going to be i times omega.

00:45 So 0 .0639 times 0 .19.

00:53 And that's going to be equal to i final times omega final.

00:58 So we don't know either of those.

01:00 We're ultimately trying to find the final angular velocity.

01:04 And in order to do that, we need to find the new rotational inertia.

01:07 So let me move to the other side here to make space, actually.

01:10 Let's just do a side calculation here.

01:13 Rotational inertia is equal to mr squared.

01:18 So after the sand is dropped, there are going to be two rotational inertia, though...

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