Question

A uniform solid disk of radius $\mathbf{R}\mathbf{}\mathbf{and}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{M}$ is free to rotate on a frictionless pivot through a point on its rim (Fig. P10.57). If the disk is released from rest in the position shown by the copper colored circle,

(a) What is the speed of its center of mass when the disk reaches the position indicated by the dashed circle?

(b) What is the speed of the lowest point on the disk in the dashed position?

(c) What If? Repeat part (a) using a uniform hoop.

Short answer

1. The speed of its center of mass when the disk reaches the position indicated by the dashed circle is $2\sqrt{\frac{\mathrm{Rg}}{3}}$.
2. The speed of the lowest point on the disk in the dashed position is $4\sqrt{\frac{\mathrm{Rg}}{3}}$.
3. A uniform hoop is $\sqrt{\mathrm{Rg}}$.

Step-by-step solution

Solution of “a” part.

a).

Let’s determine the moment of inertia of the disk about the axis of rotation which is going through a point on the rim.

About the center the Moment of inertia is 1 2MR2

Parallel axis theorem will give the moment of inertia about the axis in consideration:

$\begin{array}{l}\mathrm{I}=\frac{1}{2}{\mathrm{MR}}^2+{\mathrm{MR}}^2\\ \\ =\frac{3}{2}{\mathrm{MR}}^2\end{array}$

A frictionless rotation implies energy conservation which can be written as:

$\begin{array}{c}\mathrm{\Delta E}=0\\ \\ \mathrm{\Delta Ki}=-\mathrm{\Delta P}\\ \\ \frac{1}{2}{\mathrm{I\omega }}^2-0=-(-\mathrm{MgR})\\ \\ \frac{1}{2}\times \frac{3}{2}{\mathrm{MR}}^2\times {\mathrm{V}}^2{\mathrm{R}}^2=\mathrm{MgR}\end{array}$

$\begin{array}{l}{\mathrm{V}}^2=\frac{4}{3}\mathrm{gR}\\ \\ \mathrm{V}=2\sqrt{\frac{\mathrm{Rg}}{3}}\end{array}$

Hence, the speed of its center of mass when the disk reaches the position indicated by the dashed circle is $2\sqrt{\frac{\mathrm{Rg}}{3}}$.

Solution of “b” part.

b).

The lowest point on the disk and the center have the same angular speed but different speeds (located at different distances from the rotation axis).

$\begin{array}{l}\mathrm{\omega }\text{'}=\mathrm{\omega }\\ \\ \frac{\mathrm{V}\text{'}}{2\mathrm{R}}=\frac{\mathrm{V}}{\mathrm{R}}\\ \\ \mathrm{V}\text{'}=2\mathrm{V}\\ \\ \mathrm{V}\text{'}=4\sqrt{\frac{\mathrm{Rg}}{3}}\end{array}$

The speed of the lowest point on the disk in the dashed position is $4\sqrt{\frac{\mathrm{Rg}}{3}}$.

Uniform hoop

c).

If it was a hoop instead of a disk, the only change will be on the moment of inertia which is MR2 about the center and MR2 + MR2 = 2MR2 about the axis through the point on the rim:

$\begin{array}{c}\frac{1}{2}{\mathrm{I\omega }}^2-0=-(-\mathrm{MgR})\\ \\ \frac{1}{2}\times 2{\mathrm{MR}}^2\times {\mathrm{V}}^2{\mathrm{R}}^2=\mathrm{MgR}\\ \\ {\mathrm{V}}^2=\mathrm{gR}\\ \\ \mathrm{V}=\sqrt{\mathrm{Rg}}\end{array}$