Chapter 13, Problem 23PQ

To determine

The total rotational kinetic energy of the clock hands around their axis of rotation.

Answer to Problem 23PQ

The total rotational kinetic energy of the clock hands around their axis of rotation is $6.89\times {10}^{-5}\text{J}$ .

Explanation of Solution

The hour hand and minute hand are modelled as thin rods.

Write the expression for the rotational inertia of a thin rod about an axis through one end.

  $I=\frac{1}{3}M{L}^2$                                                                                                                  (I)

Here, $I$ is the rotational inertia of a thin rod about an axis through one end, $M$ is the mass of rod and $L$ is the length of the rod.

Write the expression for the rotational kinetic energy of hour hand.

  $K_h=\frac{1}{2}{I}_h{\omega }_h^2$                                                                                                            (II)

Here, $K_h$ is the rotational kinetic energy of hour hand, $I_h$ is the rotational inertia of hour hand and ${\omega }_h$ is the angular speed of hour hand.

Write the expression for the rotational inertia of hour hand.

  $I_h=\frac{m_h{L}_h^2}{3}$                                                                                                              (III)

Here, $m_h$ is the mass of hour hand and $L_h$ is the length of hour hand.

Write the expression for the rotational kinetic energy of minute hand.

  $K_m=\frac{1}{2}{I}_m{\omega }_m^2$                                                                                                          (IV)

Here, $K_m$ is the rotational kinetic energy of minute hand, $I_m$ is the rotational inertia of minute hand and ${\omega }_h$ is the angular speed of minute hand.

Write the expression for the rotational inertia of minute hand.

  $I_m=\frac{m_m{L}_m^2}{3}$                                                                                                              (V)

Here, $m_m$ is the mass of minute hand, $L_m$ is the length of minute hand.

The total rotational kinetic energy is the sum of the clock hands around their axis is the sum of rotational kinetic energy of each hand.

Write the expression for the total rotational kinetic energy.

  $K_R=K_h+K_m$                                                                                                                   (VI)

Here, $K_R$ is the total rotational kinetic energy.

Conclusion:

Substitute $25.0\text{kg}$ for $m_h$ and $2.00\text{m}$ for $L_h$ in equation (III) to get $I_h$ .

  $\begin{array}{c}{I}_h=\frac{\left(25.0\text{kg}\right){\left(2.00\text{m}\right)}^2}{3}\\ =33.3\text{kg}\cdot {\text{m}}^2\end{array}$

Hour hand rotates $2\pi \text{rad}$ in 12 hour. Thus, angular velocity of each hour hand is $\frac{2\pi }{12}\text{rad}/\text{h}$ .

Convert $\frac{2\pi }{\text{12}}\text{rad}/\text{h}$ into the unit of $\text{rad}/\text{s}$ .

  $\begin{array}{c}\frac{2\pi }{12}\text{rad}/\text{h}=\frac{2\pi }{\text{12}}\frac{\text{rad}}{\text{h}}\left(\frac{1\text{h}}{3600\text{s}}\right)\\ =1.45\times {10}^{-4}\text{rad}/\text{s}\end{array}$

Substitute $33.3\text{kg}\cdot {\text{m}}^2$ for $I_h$ and $1.45\times {10}^{-4}\text{rad}/\text{s}$ for ${\omega }_h$ in equation (II) to get $K_h$ .

  $\begin{array}{c}{K}_h=\frac{1}{2}\left(33.3\text{kg}\cdot {\text{m}}^2\right){\left(1.45\times {10}^{-4}\text{rad}/\text{s}\right)}^2\\ =35.00\times {10}^{-8}\text{J}\end{array}$

Substitute $15.0\text{kg}$ for $m_m$ and $3.00\text{m}$ for $L_m$ in equation (V) to get $I_m$ .

  $\begin{array}{c}{I}_m=\frac{\left(15.0\text{kg}\right){\left(3.00\text{m}\right)}^2}{3}\\ =45.0\text{kg}\cdot {\text{m}}^2\end{array}$

Minute hand rotates $2\pi \text{rad}$ in one hour. Thus, angular velocity of each clock hand is $2\pi \text{rad}/\text{h}$ .

Convert $2\pi \text{rad}/\text{h}$ into the unit of $\text{rad}/\text{s}$ .

  $\begin{array}{c}2\pi \text{rad}/\text{h}=2\pi \frac{\text{rad}}{\text{h}}\left(\frac{1\text{h}}{3600\text{s}}\right)\\ =1.75\times {10}^{-3}\text{rad}/\text{s}\end{array}$

Substitute $45.0\text{kg}\cdot {\text{m}}^2$ for $I_m$ and $1.75\times {10}^{-3}\text{rad}/\text{s}$ for ${\omega }_m$ in equation (IV) to get $K_m$ .

  $\begin{array}{c}{K}_m=\frac{1}{2}\left(45.0\text{kg}\cdot {\text{m}}^2\right){\left(1.75\times {10}^{-3}\text{rad}/\text{s}\right)}^2\\ =6.89\times {10}^{-5}\text{J}\end{array}$

Substitute $35.00\times {10}^{-8}\text{J}$ for $K_h$ and $6.89\times {10}^{-5}\text{J}$ for $K_m$ in equation (VI) to get $K_R$ .

  $\begin{array}{c}{K}_R=35.00\times {10}^{-8}\text{J}+6.89\times {10}^{-5}\text{J}\\ =\text{6}\text{.89}\times {\text{10}}^{-5}\text{J}\end{array}$

Therefore, the total rotational kinetic energy of the clock hands around their axis of rotation is $6.89\times {10}^{-5}\text{J}$ .