17. A disk of mass m1 and radius R spins in the horizontal...

17. A disk of mass $m_1$ and radius R spins in the horizontal plane at angular speed $\omega_1$ about a vertical axis through its center. A second disk of mass $m_2$ and radius R, which is initially not rotating, is then dropped onto it from above, and they rotate together about the same axis after the collision. The final rotational kinetic energy of the system is 1/3 of the initial rotational kinetic energy. What is the value of $m_2$ ? Formula: rotational inertia for a disk about its CM: $I_{CM} = \frac{1}{2} mR^2$ (a) $m_1$ (b) $2m_1$ (c) $3m_1$ (d) $m_1 / 2$ (e) $m_1 / 3$

Transcript

00:01 All right, so we have this disc that has a diameter of 10 centimeters, which means that its radius is 0 .05 meters or 5 centimeters.

00:12 The mass of the disc is 1 .5 kilograms and the point on the outer rim is has a translational speed of 2 meters per second and then eventually we're going to drop a 0 .5 kilogram mass on the outer rim as well.

00:26 But the first two parts are before we do that.

00:29 So let's look at part a where we need to find the angular momentum of the disc before we drop on the new weight.

00:36 So the angular momentum can be given by this equation i times omega where omega is the rotational speed and i is the rotational inertia.

00:44 Now for a solid disc like this one the rotational inertia is one half times the mass of the disc times the radius of the disc squared and the rotational inertia can be found by taking the translational speed divided by, or sorry, the rotational velocity can be found by taking the translational speed divided by the radius.

01:05 And when we plug those in to this equation, we get one half mr times v because that r in the denominator of omega cancels one of the r's from the rotational inertia.

01:20 So then we just need to plug in the values for the mass 1 .5 kilograms, the radius 0 .05 meters, and the rotation or the translational speed of 2 meters per second and we get a rotational momentum of 0 .075 kilogram meters squared per second.

01:41 Now in part b we are going to be looking for the kinetic energy, the rotational kinetic energy, and that can be found by using this equation one half times the rotational inertia times the rotational speed squared.

01:55 And now we could do the same thing we did before where we plug everything in using these equations there.

02:04 Or we can recognize that there's an i omega in there and so this becomes one half times the rotational momentum times omega.

02:13 Then we just have to do it once for omega since we already know what l is.

02:17 So plugging in for l, it's 0 .075 and then omega is v over r, so 2 divided by 0 .05.

02:26 We get a rotational kinetic energy of 1 .5 joules...

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