Chapter 13, Problem 22P

To determine

To calculate: The change in the angular frequency.

Answer to Problem 22P

The change in the angular frequency is $-2.8\times {10}^{-3}\text{rad}/\sec$ .

Explanation of Solution

Given:

Change in temperature $\Delta T=55°\text{C}$

Coefficient of thermal heat expansion $25\times {10}^{-6}\text{/}°\text{C}$

Formula used:

Moment of inertia

  $I=\frac{1}{2}m{r}^2$

Here,

  $m$ is the mass,

  $r$ is the radius.

Conservation of angular momentum:

  $\begin{array}{l}{L}_0=L_{final}\\ I_0{\omega }_0=I_{final}{\omega }_{final}\\ {\omega }_{final}=\frac{I_0{\omega }_0}{I_{final}}\end{array}$

Here,

  $L$ is the angular momentum,

  $I$ is moment of inertia,

  $\omega$ is angular velocity.

Calculation:

  $\begin{array}{}$ I=\frac{1}{2}m{r}^2$$ L_0=L_{final}$$ I_0{\omega }_0=I_{final}{\omega }_{final}$$ {\omega }_{final}=\frac{I_0{\omega }_0}{I_{final}}$$ =\frac{\frac{1}{2}m{r}_0{}^2{\omega }_0}{\frac{1}{2}m{r}^2}$$ ={\frac{r_0}{r^2}}^2{\omega }_0$$ \frac{\Delta \omega }{\omega }=\frac{{\omega }_{final}-{\omega }_0}{{\omega }_0}$$ =\frac{\frac{r_0{}^2{\omega }_0}{r^2}-{\omega }_0}{{\omega }_0}$$ =\frac{r_0{}^2}{r^2}-1$$ =\frac{r_0{}^2}{{(r_0+\Delta r)}^2}-1$$ =\frac{r_0{}^2}{{(r_0+\alpha r_0\Delta T)}^2}-1$$ =\frac{1}{{(1+\alpha \Delta T)}^2}-1$$ =\frac{1-(1+2\alpha \Delta T+{(\alpha \Delta T)}^2)}{{(1+\alpha \Delta T)}^2}$$ =\frac{-2\alpha \Delta T-{(\alpha \Delta T)}^2}{{(1+\alpha \Delta T)}^2}$$ =-\alpha \Delta T\frac{2+\alpha \Delta T}{{(1+\alpha \Delta T)}^2}$\end{array}$

Now considered that the value of $\alpha \Delta T<<1$

  $\begin{array}{l}\frac{\Delta \omega }{\omega }=-\alpha \Delta T\frac{2+\alpha \Delta T}{{(1+\alpha \Delta T)}^2}\approx -2\alpha \Delta T\\ =-2\alpha \Delta T=-2(25\times {10}^{-6}{\text{/ }}^{\text{0}}\text{C})(55{}^{\text{0}}\text{C})\\ =-2.8\times {10}^{-3}\text{rad}/\sec \end{array}$

Conclusion:

Because of the heating effect, there is no change in the torque with the change in the radius. And the change in the angular frequency is $-2.8\times {10}^{-3}\text{rad}/\sec$ .