Question

Figure 10 - 27shows three flat disks (of the same radius) that can rotate about their centers like merry-go-rounds. Each disk consists of the same two materials, one denser than the other (density is mass per unit volume). In disks 1and 3, the denser material forms the outer half of the disk area. In disk 2, it forms the inner half of the disk area. Forces with identical magnitudes are applied tangentially to the disk, either at the outer edge or at the interface of the two materials, as shown. Rank the disks according to (a) the torque about the disk center, (b) the rotational inertia about the disk center, and (c) the angular acceleration of the disk, greatest first.

Short answer

(a) The rank of the discs according to their torque${\mathrm{\tau }}_1={\mathrm{\tau }}_2>{\mathrm{\tau }}_3$

(b) Rotational inertia${\mathrm{I}}_1={\mathrm{I}}_3>{\mathrm{I}}_2$

(c) Angular acceleration is${\mathrm{\alpha }}_2>{\mathrm{\alpha }}_1>{\mathrm{\alpha }}_3$

Step-by-step solution

Step 1: Given

Figure 10-27 with disk made up of two different materials and force acting on them at different points.

Determining the concept

The torque on the disc is equal to the moment of force. Torque is also equal to the product of the moment of inertia and angular acceleration. By using the geometry of all three discs and the location of the point of application of the force, that disc can be easily ranked.

Formulae are as follows:

$$\begin{gathered} \mathrm{\tau }=\mathrm{r}\times \mathrm{F} \\ =\mathrm{rF}\sin \mathrm{\varphi } \\ \mathrm{\tau }=\mathrm{Ia} \end{gathered}$$

Where, r is radius, F is force and is torque, I is moment of inertia and αis angular acceleration

(a) Determining the ranks of the discs according to torque

It’s known that,

$\mathrm{\tau }=\mathrm{r}\times \mathrm{F}$

In all cases, the applied force is the same. So the magnitude of the torque can be compared using the distance betweenthe pointof application of forceand the axisof rotation.

Disk 1 and disk 2 have the same distance between the point of application of force and axis of rotation whereas disk 3 has a smaller distance between the point of application of force and axis of rotation.

Therefore,${\mathrm{\tau }}_1={\mathrm{\tau }}_2>{\mathrm{\tau }}_3$

(b) Determining the ranks of the discs according to rotational inertia

As it’s knownthat rotationalinertia depends on the mass and distribution of mass from the axis of rotation. The wider the distribution, the larger would be the rotational inertia. In the case of disk 1 and disk 3, denser material is distributed in the same way and is more widely distributed as compared to disk 1. So the rotational inertiaof disk1 and disk 3 is the same but greater than disk 2.

Hence,${\mathrm{I}}_1={\mathrm{I}}_3>{\mathrm{I}}_2$

(c) Determining the ranks of the discs according to angular acceleration

Write the torque in terms of rotational inertia and angular acceleration as,

$$\begin{gathered} \mathrm{\tau }=\mathrm{I\alpha } \\ \mathrm{\alpha }=\frac{\mathrm{\tau }}{\mathrm{I}} \end{gathered}$$

From the above equation, it can be seen that angular acceleration depends directly on the torque and inversely on rotational inertia.

Now disk 2 has a larger torque and smaller rotational inertia. So it must have the largest angular acceleration. Disk 3 has the smallest torque and largest rotational inertia, so it must have a smaller angular momentum.

Therefore,${\mathrm{\alpha }}_2>{\mathrm{\alpha }}_1>{\mathrm{\alpha }}_3$