Chapter 8, Problem 56P

To determine

To Find: Final frequency of the wheel.

Answer to Problem 56P

  $8.57\text{rad/s}$

Explanation of Solution

Given Information:

Mass of uniform disc, $\text{m =}\text{5}\text{.0 kg}$

Diameter of disc = $\text{0}\text{.40}\text{m}$

Mass of clay, ${\mathrm{m}}_{\mathrm{c}\mathrm{l}\mathrm{a}\mathrm{y}}=3.1\text{kg}$

Radius of flat disc, $\mathrm{r}=8.0\text{cm}\text{or}0.08\text{m}$

Formula Used:

According to law of conservation of angular momentum

  $\begin{array}{l}\mathrm{L}=\mathrm{I}\mathrm{\omega }\\ {\mathrm{I}}_{\mathrm{I}}{\mathrm{\omega }}_{\mathrm{I}}={\mathrm{I}}_{\mathrm{F}}{\mathrm{\omega }}_{\mathrm{F}}\\ {\text{I}}_{\mathrm{I}}=\frac{1}{2}{\mathrm{m}}_{\mathrm{d}}{\mathrm{r}}_{\mathrm{d}}{}^2\\ {\mathrm{I}}_{\mathrm{F}}=\frac{1}{2}{\mathrm{m}}_{\mathrm{d}}{\mathrm{r}}_{\mathrm{d}}{}^2+\frac{1}{2}{\mathrm{m}}_{\mathrm{c}}{\mathrm{r}}_{\mathrm{c}}{}^2\end{array}$

Here,

  $\mathrm{L}$ is the angular momentum,

  $\mathrm{I}$ is the moment of inertia,

  $\mathrm{\omega }$ is the angular velocity.

  ${\mathrm{m}}_{\mathrm{d}}$ is the mass of disc,

  ${\mathrm{m}}_{\mathrm{c}}$ is the mass of clay,

  $\mathrm{\omega }$ is the angular velocity.

  ${\mathrm{r}}_{\mathrm{d}}$ is the radius of disc.

  ${\mathrm{r}}_{\mathrm{c}}$ is the radius of flat disc.

Calculation:

From the above equation

  ${\mathrm{\omega }}_{\mathrm{F}}=\frac{{\mathrm{I}}_{\mathrm{I}}{\mathrm{\omega }}_{\mathrm{I}}}{{\mathrm{I}}_{\mathrm{F}}}$

Here, ${\mathrm{\omega }}_{\mathrm{I}}=2\mathrm{\pi }{\mathrm{f}}_1$

  $\begin{array}{c}=2\times \mathrm{\pi }\times 1.5\\ =3\mathrm{\pi }\end{array}$

Substitute the values in the expression of law of conservation.

  $\begin{array}{l}{\mathrm{\omega }}_{\mathrm{F}}=\frac{3\mathrm{\pi }\times \left(\frac{1}{2}\right)(5){(0.2)}^2}{\frac{1}{2}(5){(0.2)}^2+\frac{1}{2}(3.1){(0.08)}^2}\\ {\mathrm{\omega }}_{\mathrm{F}}=8.57\text{rad/s}\end{array}$

Conclusion: Final frequency of the wheel is

  $8.57\text{rad/s}$ .