Chapter 15, Problem 27SP

A solid sphere of mass m and radius b is spinning freely on its axis with angular velocity $\omega$. When heated by an amount $\Delta T$, its angular velocity changes to $\omega$. Find ${\omega }_0\text{/}\omega$ if the linear expansion coefficient for the material of the sphere is $\alpha$.

To determine

The ratio $\frac{{\omega }_0}{\omega }$ when a solid sphere, spinning freely on its axis, is heated by an amount $\Delta T$.

Answer to Problem 27SP

Solution:

$1+2\alpha \Delta T+{\left(\alpha \Delta T\right)}^2$

Explanation of Solution

Given data:

A solid sphere of mass $m$ and radius $b$ is spinning freely on its axis with angular velocity ${\omega }_0$.

The angular velocity of the sphere changes to $\omega$ when heated by an amount $\Delta T$.

The linear expansion coefficient for the material of the sphere is $\alpha$.

Formula used:

Write the expression for angular momentum:

$L=I\omega$

Here, $L$ is the angular momentum, $I$ is the moment of inertia, and $\omega$ is the angular velocity.

Write the expression for conservation of angular momentum:

$L_i=L_f$

Here, $L_i$ is the initial angular momentum and $L_f$ is the final angular momentum.

Write the expression for moment of inertia of solid sphere:

$I=\frac{2}{5}m{r}^2$

Here, $I$ is the moment of inertia of solid sphere, $m$ is the mass of sphere, and $r$ is the radius of sphere.

Write the expression for linear expansion of solids:

$\begin{array}{c}L-L_0=\alpha L_0\Delta T\\ L=L_0+\alpha L_0\Delta T\\ =L_0\left(1+\alpha \Delta T\right)\end{array}$

Here, $L$ is the length of the solid at temperature $T$, $\alpha$ is the coefficient of linear expansion, and $L_0$ is the length of solid at temperature $T_0$.

Write the expression for binomial expansion of the term ${\left(1+x\right)}^2$:

${\left(1+x\right)}^2=1+2x+x^2$

Explanation:

Recall the expression for the initial moment of inertia of solid sphere:

$I_i=\frac{2}{5}m{r}^2$

Substitute $b$ for $r$

$I_i=\frac{2}{5}m{b}^2$

Recall the expression for the initial angular momentum of solid sphere:

$L_i=I_i{\omega }_i$

Here, $L_i$ is the initial angular momentum, $I_i$ is the initial moment of inertia of solid sphere, and ${\omega }_i$ is the initial angular velocity.

Substitute $\frac{2}{5}m{b}^2$ for $I_i$ and ${\omega }_0$ for ${\omega }_i$

$\begin{array}{c}{L}_i=\left(\frac{2}{5}m{b}^2\right)\left({\omega }_0\right)\\ =\frac{2}{5}m{b}^2{\omega }_0\end{array}$

When the sphere is heated by an amount $\Delta T$, the radius of the sphere changes due to thermal expansion of the material of sphere.

Recall the expression for the final radius of solid sphere:

$r=r_0\left(1+\alpha \Delta T\right)$

Here, $r$ is the final radius of sphere and $r_0$ is the initial radius of sphere.

Substitute $b$ for $r_0$

$r=b\left(1+\alpha \Delta T\right)$

Recall the expression for the final moment of inertia of solid sphere:

$I_f=\frac{2}{5}m{r}^2$

Substitute $b\left(1+\alpha \Delta T\right)$ for $r$

$I_f=\frac{2}{5}m{\left(b\left(1+\alpha \Delta T\right)\right)}^2$

Recall the expression for the final angular momentum of solid sphere when it is heated by an amount $\Delta T$:

$L_f=I_f{\omega }_f$

Here, $L_f$ is the final angular momentum, $I_f$ is the final moment of inertia of solid sphere, and ${\omega }_{{}_f}$ is the final angular velocity.

Substitute $\frac{2}{5}m{\left(b\left(1+\alpha \Delta T\right)\right)}^2$ for $I_f$ and $\omega$ for ${\omega }_f$

$\begin{array}{c}{L}_f=\left(\frac{2}{5}m{\left(b\left(1+\alpha \Delta T\right)\right)}^2\right)\left(\omega \right)\\ =\frac{2}{5}m{b}^2{\left(1+\alpha \Delta T\right)}^2\omega \end{array}$

Apply the conservation of angular momentum on the solid sphere:

$L_i=L_f$

Substitute $\frac{2}{5}m{b}^2{\omega }_0$ for $L_i$ and $\frac{2}{5}m{b}^2{\left(1+\alpha \Delta T\right)}^2\omega$ for $L_f$

$\frac{2}{5}m{b}^2{\omega }_0=\frac{2}{5}m{b}^2{\left(1+\alpha \Delta T\right)}^2\omega$

Solve the expression

$\begin{array}{l}{\omega }_0={\left(1+\alpha \Delta T\right)}^2\omega \\ \frac{{\omega }_0}{\omega }={\left(1+\alpha \Delta T\right)}^2\end{array}$ ……(1)

The expression for binomial expansion of the term ${\left(1+\alpha \Delta T\right)}^2$ is as follows:

${\left(1+\alpha \Delta T\right)}^2=1+2\left(\alpha \Delta T\right)+{\left(\alpha \Delta T\right)}^2$

Substitute $1+2\left(\alpha \Delta T\right)+{\left(\alpha \Delta T\right)}^2$ for ${\left(1+\alpha \Delta T\right)}^2$ in equation (1)

$\frac{{\omega }_0}{\omega }=1+2\left(\alpha \Delta T\right)+{\left(\alpha \Delta T\right)}^2$

Conclusion:

The ratio $\frac{{\omega }_0}{\omega }$ when a solid sphere, spinning freely on its axis, is heated by an amount $\Delta T$ is $1+2\left(\alpha \Delta T\right)+{\left(\alpha \Delta T\right)}^2$.