Chapter 10, Problem 66P

The hour hand and the minute hand of Big Ben, the Parliament tower clock in London, are 2.70 m and 4.50 m long and have masses of 60.0 kg and 100 kg, respectively (see Fig. P10.17).

1. (a) Determine the total torque due to the weight of these hands about the axis of rotation when the time reads (i) 3:00, (ii) 5:15, (iii) 6:00, (iv) 8:20, and (v) 9:45. (You may model the hands as long, thin, uniform rods.)
2. (b) Determine all times when the total torque about the axis of rotation is zero. Determine the times to the nearest second, solving a transcendental equation numerically.

To determine

The torque due to the weight of the hands at time $3.00$, $5.15$, $6.00$, $8.20$, and $9.45$.

Answer to Problem 66P

1. (i) The torque due to the weight of the hands at time $3.00$ is $\underset{\_}{-794\text{N}\cdot \text{m}}$.
2. (ii) The torque due to the weight of the hands at time $5.15$ is $\underset{\_}{-2510\text{N}\cdot \text{m}}$.
3. (iii) The torque due to the weight of the hands at time $6.00$ is $\underset{\_}{0\text{N}\cdot \text{m}}$.
4. (iv) The torque due to the weight of the hands at time $8.20$ is $\underset{\_}{-1160\text{N}\cdot \text{m}}$.
5. (v) The torque due to the weight of the hands at time $9.45$ is $\underset{\_}{-2940\text{N}\cdot \text{m}}$.

Explanation of Solution

Consider that the weight of each hand is concentrating at the center of gravity

Write the expression for the total torque about the center of gravity.

  $\begin{array}{c}\tau =-m_{h}g\left(\frac{L_h}{2}\right)\sin {\theta }_h-m_{m}g\left(\frac{L_m}{2}\right)\sin {\theta }_m\\ =-\frac{g}{2}\left(m_h{L}_h\sin {\theta }_h+m_m{L}_m\sin {\theta }_m\right)\end{array}$        (I)

Here, $m_h$ is the mass of the hour hand, $m_m$ is the minute hand, $L_h$ is the length of the hour hand, $L_m$ is the length of the minute hand, ${\theta }_m$ is the angle inclined by minute hand, ${\theta }_h$ is the angle inclined by hour hand.

Consider $t=0$ as 12 o clock, hence the angular position of the hands are ${\theta }_h={\omega }_{h}t$, where ${\omega }_h=\frac{\pi }{6}\text{rad}/\text{h}$, and ${\theta }_m={\omega }_{m}t$, where ${\omega }_m=2\pi \text{rad}/\text{h}$.

Substitute, $\frac{\pi }{6}t$ for ${\theta }_h$, and $2\pi t$ for ${\theta }_m$ in equation (I).

  $\tau =-\frac{g}{2}\left(m_h{L}_h\sin \left(\frac{\pi }{6}t\right)+m_m{L}_m\sin \left(2\pi t\right)\right)$        (II)

Conclusion:

(i) Case 1: at $t=3.00\text{h}$.

Substitute, $60.0kg$ for $m_h$, $2.70\text{m}$ for $L_h$, $100\text{kg}$ for $m_m$, $4.50\text{m}$ for $L_m$, and $3.00\text{h}$ for $t$ in equation (II).

  $\begin{array}{c}\tau =-\frac{9.80\text{m}/{\text{s}}^2}{2}\left(\left(60.0kg\right)\left(2.70\text{m}\right)\sin \left(\frac{\pi }{6}\times 3.00\text{h}\right)+\left(100.0kg\right)\left(4.50\text{m}\right)\sin \left(2\pi \times 3.00\text{h}\right)\right)\\ =-794\text{N}\cdot \text{m}\left[\sin \left(\frac{\pi }{2}\right)+2.78\sin 6\pi \right]\\ =-794\text{N}\cdot \text{m}\end{array}$

(ii) Case 2: at $t=5.15\text{h}$.

Substitute, $60.0kg$ for $m_h$, $2.70\text{m}$ for $L_h$, $100\text{kg}$ for $m_m$, $4.50\text{m}$ for $L_m$, and $5\text{h+}\frac{15}{60}\text{h}$ for $t$ in equation (II).

  $\begin{array}{c}\tau =-\frac{9.80\text{m}/{\text{s}}^2}{2}\left(\left(60.0kg\right)\left(2.70\text{m}\right)\sin \left(\frac{\pi }{6}\times \left(5\text{h+}\frac{15}{60}\text{h}\right)\right)+\left(100.0kg\right)\left(4.50\text{m}\right)\sin \left(2\pi \times \left(5\text{h+}\frac{15}{60}\text{h}\right)\right)\right)\\ =-2510\text{N}\cdot \text{m}\end{array}$

(iii) Case 3: at $t=6.00\text{h}$.

Substitute, $60.0kg$ for $m_h$, $2.70\text{m}$ for $L_h$, $100\text{kg}$ for $m_m$, $4.50\text{m}$ for $L_m$, and $6.00\text{h}$ for $t$ in equation (II).

  $\begin{array}{c}\tau =-\frac{9.80\text{m}/{\text{s}}^2}{2}\left(\left(60.0kg\right)\left(2.70\text{m}\right)\sin \left(\frac{\pi }{6}\times \left(6.00\text{h}\right)\right)+\left(100.0kg\right)\left(4.50\text{m}\right)\sin \left(2\pi \times \left(6.00\text{h}\right)\right)\right)\\ =0\text{N}\cdot \text{m}\end{array}$

(iv) Case 4: at $t=8.20\text{h}$.

Substitute, $60.0kg$ for $m_h$, $2.70\text{m}$ for $L_h$, $100\text{kg}$ for $m_m$, $4.50\text{m}$ for $L_m$, and $8\text{h+}\frac{20}{60}\text{h}$ for $t$ in equation (II).

  $\begin{array}{c}\tau =-\frac{9.80\text{m}/{\text{s}}^2}{2}\left(\left(60.0kg\right)\left(2.70\text{m}\right)\sin \left(\frac{\pi }{6}\times \left(8\text{h+}\frac{20}{60}\text{h}\right)\right)+\left(100.0kg\right)\left(4.50\text{m}\right)\sin \left(2\pi \times \left(8\text{h+}\frac{20}{60}\text{h}\right)\right)\right)\\ =-1160\text{N}\cdot \text{m}\end{array}$

(v) Case 5: at $9.45\text{h}$.

Substitute, $60.0kg$ for $m_h$, $2.70\text{m}$ for $L_h$, $100\text{kg}$ for $m_m$, $4.50\text{m}$ for $L_m$, and $9\text{h+}\frac{45}{60}\text{h}$ for $t$ in equation (II).

  $\begin{array}{c}\tau =-\frac{9.80\text{m}/{\text{s}}^2}{2}\left(\left(60.0kg\right)\left(2.70\text{m}\right)\sin \left(\frac{\pi }{6}\times \left(9\text{h+}\frac{45}{60}\text{h}\right)\right)+\left(100.0kg\right)\left(4.50\text{m}\right)\sin \left(2\pi \times \left(9\text{h+}\frac{45}{60}\text{h}\right)\right)\right)\\ =-2940\text{N}\cdot \text{m}\end{array}$

Therefore,

1. (i) The torque due to the weight of the hands at time $3.00$ is $\underset{\_}{-794\text{N}\cdot \text{m}}$.
2. (ii) The torque due to the weight of the hands at time $5.15$ is $\underset{\_}{-2510\text{N}\cdot \text{m}}$.
3. (iii) The torque due to the weight of the hands at time $6.00$ is $\underset{\_}{0\text{N}\cdot \text{m}}$.
4. (iv) The torque due to the weight of the hands at time $8.20$ is $\underset{\_}{-1160\text{N}\cdot \text{m}}$.
5. (v) The torque due to the weight of the hands at time $9.45$ is $\underset{\_}{-2940\text{N}\cdot \text{m}}$.

To determine

The time at which the total toque become zero.

Answer to Problem 66P

The time at which the total toque become zero is given below.

Explanation of Solution

The total torque become zero, when:

  $\sin \left(\frac{\pi t}{6}\right)+2.78\sin 2\pi t=0$        (III)

Use equation (III) to find the times at which torque become zero.

Conclusion:

The times at which torque become zero is given below

.