Question

A solid disk with a mass of 0.8 kg and a radius of 12cm is rotating at a rate of 2.6 rev/s. Another disk with half the mass and half the radius is dropped on top of the first disk. What is the new angular speed of both disks when they are rotating together?

Answer 1

Answer:

The new angular speed of both disks when they are rotating together is 2.311 revolutions per second.

Explanation:

This situation can be analyzed by the Principle of Conservation of Angular Momentum since there are no external forces and moments being exerted on the system, whose definition is presented below:

$I_{o}\cdot \omega_{o} = I_{f}\cdot \omega_{f}$ (1)

Where:

$\omega_{o}$, $\omega_{f}$ - Initial and final angular speeds, measured in revolutions per second.

$I_{o}, I_{f}$ - Initial and final moments of inertia, measured in kilograms per square meter.

The initial and final moments of inertia of the system are, respectively:

$I_{o} = \frac{1}{2}\cdot m_{o}\cdot r_{o}^{2}$ (2)

$I_{f} = \frac{1}{2}\cdot \left[m_{o}\cdot r_{o}^{2}+\frac{1}{2}\cdot m_{o}\cdot \left(\frac{1}{2}\cdot r_{o} \right)^{2}\right]$

$I_{f} = \frac{1}{2}\cdot m_{o}\cdot r_{o}^{2}+\frac{1}{16}\cdot m_{o}\cdot r_{o}^{2}$

$I_{f} = \frac{9}{16}\cdot m_{o}\cdot r_{o}^{2}$ (3)

By applying (2) and (3) in (1), we obtain the following expression:

$\frac{1}{2}\cdot m_{o}\cdot r_{o}^{2}\cdot \omega_{o} = \frac{9}{16}\cdot m_{o}\cdot r_{o}^{2}\cdot \omega_{f}$

$\frac{1}{2}\cdot \omega_{o} = \frac{9}{16}\cdot \omega_{f}$

$\omega_{f} = \frac{8}{9}\cdot \omega_{o}$ (4)

If we know that $\omega_{o} = 2.6\,\frac{rev}{s}$, then the new angular speed of both disks when they are rotating together is:

$\omega_{f} = \frac{8}{9}\cdot \left(2.6\,\frac{rev}{s} \right)$

$\omega_{f} = 2.311\,\frac{rev}{s}$

The new angular speed of both disks when they are rotating together is 2.311 revolutions per second.