Chapter 17, Problem 73P

(a)

To determine

The change in kinetic energy of the disk.

Answer to Problem 73P

The change in kinetic energy of the disk is $9.31\times {10}^{10}\text{J}$.

Explanation of Solution

Given info: The radius of copper disk is $28.0\text{m}$, thickness of copper disk is $1.20\text{m}$, the temperature at collision is $850°\text{C}$, angular speed of copper disk is $25.0\text{rad}/\text{s}$ and the temperature after radiation is $20°\text{C}$.

Write the equation for change in kinetic energy of the disk.

$\begin{array}{c}\Delta K.E.=\frac{1}{2}{I}_f{\omega }_f^2-\frac{1}{2}{I}_i{\omega }_i^2\\ =\frac{1}{2}\left(I_{\text{f}}{\omega }_{\text{f}}\right){\omega }_{\text{f}}-\frac{1}{2}{I}_{\text{i}}{\omega }_{\text{i}}^2\end{array}$

Here,

$I_{\text{i}}$ is the initial moment of inertia of the disk.

$I_{\text{f}}$ is the final moment of inertia of the disk.

${\omega }_i$ is the initial angular speed of the disk.

${\omega }_{\text{f}}$ is the final angular speed of the disk.

Write the equation of conservation of angular momentum.

$I_{\text{i}}{\omega }_{\text{i}}=I_{\text{f}}{\omega }_{\text{f}}$

Substitute $I_{\text{i}}{\omega }_{\text{i}}$ for $I_{\text{f}}{\omega }_{\text{f}}$ in the above equation to get the change in kinetic energy of the disk.

$\begin{array}{c}\Delta K.E.=\frac{1}{2}\left(I_{\text{f}}{\omega }_{\text{f}}\right){\omega }_{\text{f}}-\frac{1}{2}{I}_{\text{i}}{\omega }_{\text{i}}^2\\ =\frac{1}{2}\left(I_{\text{i}}{\omega }_{\text{i}}\right){\omega }_{\text{f}}-\frac{1}{2}{I}_{\text{i}}{\omega }_{\text{i}}^2\\ =\frac{1}{2}{I}_{\text{i}}{\omega }_{\text{i}}\left({\omega }_{\text{f}}-{\omega }_{\text{i}}\right)\end{array}$ (1)

Write the formula for initial moment of inertia $I_{\text{i}}$.

$\begin{array}{c}{I}_{\text{i}}=\frac{1}{2}m{r}^2\\ =\frac{1}{2}\left(\rho V\right)r^2\\ =\frac{1}{2}\left(\rho \left(\pi r^{2}t\right)\right)r^2\\ =\frac{1}{2}\rho \pi r^{4}t\end{array}$

Here,

m is the mass of the disk.

r is the radius of disk.

$\rho$ is the density of copper disk.

t is the thickness of copper disk.

The density of copper is $8920\text{kg}/{\text{m}}^3$.

Substitute $8920\text{kg}/{\text{m}}^3$ for $\rho$, $28.0\text{m}$ for r and $1.20\text{m}$ for t in the above equation to get the initial moment of inertia.

$\begin{array}{c}{I}_{\text{i}}=\frac{1}{2}\left(8920\text{kg}/{\text{m}}^3\right)\pi {\left(28.0\text{m}\right)}^4\left(1.20\text{m}\right)\\ =1.033\times {10}^{10}\text{kg}\cdot {\text{m}}^2\end{array}$

Write the equation of conservation of angular momentum to calculate the final angular speed of the disk.

$\begin{array}{c}{I}_{\text{i}}{\omega }_{\text{i}}=I_{\text{f}}{\omega }_{\text{f}}\\ \frac{1}{2}m{r}_{\text{i}}^2{\omega }_{\text{i}}=\frac{1}{2}m{r}_{\text{f}}^2{\omega }_{\text{f}}\\ \frac{1}{2}m{r}_{\text{i}}^2{\omega }_{\text{i}}=\frac{1}{2}m{r}_{\text{i}}^2{\left(1-\alpha \left|\Delta T\right|\right)}^2{\omega }_{\text{f}}\\ {\omega }_{\text{f}}=\frac{{\omega }_{\text{i}}}{{\left(1-\alpha \left|\Delta T\right|\right)}^2}\end{array}$

Further solve the above equation to calculate the final angular speed of the disk.

$\begin{array}{c}{\omega }_{\text{f}}=\frac{{\omega }_{\text{i}}}{{\left(1-\alpha \left|\Delta T\right|\right)}^2}\\ =\frac{{\omega }_{\text{i}}}{{\left(1-\alpha \left(T_{\text{c}}-T_{\text{r}}\right)\right)}^2}\end{array}$

Here,

$\alpha$ is the coefficient of linear expansion for copper.

$\Delta T$ is the change in temperature.

$T_{\text{c}}$ is the temperature at collision.

$T_{\text{r}}$ is the temperature after radiation.

The value of coefficient of linear expansion $\alpha$ for copper is $17\times {10}^{-6}°{\text{C}}^{-1}$.

Substitute $25.0\text{rad}/\text{s}$ for ${\omega }_{\text{i}}$, $17\times {10}^{-6}°{\text{C}}^{-1}$ for $\alpha$, $850°\text{C}$ for $T_{\text{c}}$ and $20°\text{C}$ for $T_{\text{r}}$ in the above equation to get the final angular speed of the disk.

$\begin{array}{c}{\omega }_{\text{f}}=\frac{\left(25.0\text{rad}/\text{s}\right)}{{\left(1-\left(17\times {10}^{-6}°{\text{C}}^{-1}\right)\left(850°\text{C}-20°\text{C}\right)\right)}^2}\\ =25.7207\text{rad}/\text{s}\end{array}$

Substitute $\left(1.033\times {10}^{10}\text{kg}\cdot {\text{m}}^2\right)$ for $I_{\text{i}}$, $25.0\text{rad}/\text{s}$ for ${\omega }_{\text{i}}$ and $25.7207\text{rad}/\text{s}$ for ${\omega }_{\text{f}}$ in equation (1) to calculate the change in kinetic energy of the disk.

$\begin{array}{c}\Delta K.E.=\frac{1}{2}\left(1.033\times {10}^{10}\text{kg}\cdot {\text{m}}^2\right)\left(25.0\text{rad}/\text{s}\right)\left(\left(25.7207\text{rad}/\text{s}\right)-\left(25.0\text{rad}/\text{s}\right)\right)\\ =9.31\times {10}^{10}\text{J}\end{array}$

Conclusion:

Therefore, the change in kinetic energy of the disk is $9.31\times {10}^{10}\text{J}$.

(b)

To determine

The change in internal energy of the disk.

Answer to Problem 73P

The change in internal energy of the disk is $-8.47\times {10}^{12}\text{J}$.

Explanation of Solution

Given info: The radius of copper disk is $28.0\text{m}$, thickness of copper disk is $1.20\text{m}$, the temperature at collision is $850°\text{C}$, angular speed of copper disk is $25.0\text{rad}/\text{s}$ and the temperature after radiation is $20°\text{C}$.

Write the equation to calculate the change in internal energy of the disk.

$\begin{array}{c}\Delta E_{\mathrm{int}}=Q\\ =mc\Delta T\\ =\left(\rho V\right)c\left(T_{\text{r}}-T_{\text{c}}\right)\\ =\rho \pi r^{2}tc\left(T_{\text{r}}-T_{\text{c}}\right)\end{array}$

Here,

$\Delta E_{\mathrm{int}}$ is the change in internal energy of the disk.

Q is the energy required to change the temperature of substance.

c is the specific heat of the copper disk.

Specific heat of copper disk is $387\text{J}/\text{kg}\cdot °\text{C}$.

Substitute $8920\text{kg}/{\text{m}}^3$ for $\rho$, $28.0\text{m}$ for r, $1.20\text{m}$ for t, $387\text{J}/\text{kg}\cdot °\text{C}$ for c, $20°\text{C}$ for $T_{\text{r}}$ and $850°\text{C}$ for $T_{\text{c}}$ in the above equation to get the change in internal energy of the disk.

$\begin{array}{c}\Delta E_{\mathrm{int}}=\left(8920\text{kg}/{\text{m}}^3\right)\pi {\left(28.0\text{m}\right)}^2\left(1.20\text{m}\right)\left(387\text{J}/\text{kg}\cdot °\text{C}\right)\left(20°\text{C}-850°\text{C}\right)\\ =-8.4683\times {10}^{12}\text{J}\\ \cong -\text{8}\text{.47}\times {10}^{12}\text{J}\end{array}$

Conclusion:

Therefore, the change in internal energy of the disk is $-8.47\times {10}^{12}\text{J}$.

(c)

To determine

The amount of radiated energy.

Answer to Problem 73P

The amount of radiated energy is $-8.38\times {10}^{12}\text{J}$.

Explanation of Solution

Given info: The radius of copper disk is $28.0\text{m}$, thickness of copper disk is $1.20\text{m}$, the temperature at collision is $850°\text{C}$, angular speed of copper disk is $25.0\text{rad}/\text{s}$ and the temperature after radiation is $20°\text{C}$.

Write the equation for change in kinetic energy of the disk to calculate the amount of radiated energy.

$\begin{array}{c}\Delta K.E.=T_{\text{ER}}-\Delta E_{\mathrm{int}}\\ T_{\text{ER}}=\Delta K.E.+\Delta E_{\mathrm{int}}\end{array}$

Here,

$T_{\text{ER}}$ is the amount of radiated energy.

Substitute $9.31\times {10}^{10}\text{J}$ for $\Delta K.E.$ and $-8.47\times {10}^{12}\text{J}$ for $\Delta E_{\mathrm{int}}$ in the above equation to get the amount of radiated energy.

$\begin{array}{c}{T}_{\text{ER}}=\left(9.31\times {10}^{10}\text{J}\right)+\left(-8.47\times {10}^{12}\text{J}\right)\\ =\left(9.31\times {10}^{10}\text{J}\right)-\left(8.47\times {10}^{12}\text{J}\right)\\ =-8.38\times {10}^{12}\text{J}\end{array}$

Conclusion:

Therefore, the amount of radiated energy is $-8.38\times {10}^{12}\text{J}$.