Question

The hour hand and the minute hand of Big Ben, the Elizabeth Tower clock in London, are $\mathbf{2}\mathbf{.}\mathbf{70}\mathbf{m}\mathbf{}\mathbf{and}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{50}\mathbf{m}$ long and have masses of $\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{kg}\mathbf{}\mathbf{and}\mathbf{}\mathbf{100}$kg, respectively (see Fig. $\mathbf{P}\mathbf{10}\mathbf{.}\mathbf{49}$).

(a) Determine the total torque due to the weight of these hands about the axis of rotation when the time reads $\mathbf{\left(}\mathbf{i}\mathbf{\right)}\mathbf{3}\mathbf{:}\mathbf{00}\mathbf{,}\mathbf{\left(}\mathbf{ii}\mathbf{\right)}\mathbf{5}\mathbf{:}\mathbf{15}\mathbf{,}\mathbf{\left(}\mathbf{iii}\mathbf{\right)}\mathbf{6}\mathbf{:}\mathbf{00}\mathbf{,}\mathbf{\left(}\mathbf{iv}\mathbf{\right)}\mathbf{8}\mathbf{:}\mathbf{20}\mathbf{,}\mathbf{and}\mathbf{\left(}\mathbf{v}\mathbf{\right)}\mathbf{9}\mathbf{:}\mathbf{45}$. (You may model the hands as long, thin, uniform rods.)

(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

Short answer

a). The torque due to the weight is given below.

b). The times when the total torque about the axis of rotation is zero. Determine the times to the nearest second is solved below.

Step-by-step solution

The total torque due to the weight of these hands about the axis of rotation.

a).

The total torque due to the weight of these hands about the axis of rotation when the time reads,

When the time is $3:00$minute,

The hand is vertical, so torque due to minute hand is zero. The hour made angle with vertical,

This defined as the $90$ deg torque,

$\begin{array}{c}\mathrm{\theta }=\mathrm{rxFtorque}\\ \\ \mathrm{\theta }=\mathrm{rFsin}90\\ \\ \mathrm{torque}=\frac{2.70}{2}(60\mathrm{x}9.81)\sin 90\\ \\ \mathrm{torque}=794\mathrm{Nm}.\end{array}$

When the time is 5 :15 then the minute hand is horizontal.

role="math" localid="1663757366480" $\begin{array}{c}\mathrm{torque}\mathrm{due}\min =\left(\frac{4.50}{2}\right)(100\mathrm{x}9.81)\\ \begin{array}{l}\\ \mathrm{torque}=2207.25\mathrm{N}\min \mathrm{to}\mathrm{page}\mathrm{hour}\mathrm{hand},\\ \\ \mathrm{\theta }=30\mathrm{deg}\\ \\ \mathrm{torque}=(2.70/2)(60\mathrm{x}9.81)\sin 30\\ \\ \mathrm{torque}=397.30\mathrm{N}\min \mathrm{to}\mathrm{page}\\ \\ \mathrm{total}\mathrm{torque}=2604.55\mathrm{Nm})\end{array}\end{array}$

When the time is at$6:00$ minute the hand is vertical,

So, torque due to minute hand is zero.

The hour made angle with vertical,

$\begin{array}{c}\mathrm{\theta }=90\\ \\ \mathrm{\theta }=90\mathrm{deg}\\ \\ \mathrm{torque}=\mathrm{rxF}\\ \\ \mathrm{torque}=\mathrm{rFsin\theta }\\ \\ \mathrm{torque}=\left(\frac{2.70}{2}\right)(60\mathrm{x}9.81)\sin 60\\ \\ \mathrm{torque}=794\mathrm{Nm}\end{array}$

When the time is 9 :45 then the minute hand is horizontal

$\begin{array}{c}\mathrm{torque}\mathrm{due}\min =\left(\frac{4.50}{2}\right)(100\mathrm{x}9.81)\\ \begin{array}{l}\\ \mathrm{torque}=2207.25\mathrm{N}\min \mathrm{to}\mathrm{page}\mathrm{hour}\mathrm{hand},\\ \\ \mathrm{\theta }=30\mathrm{deg}\\ \\ \mathrm{torque}=(2.70/2)(60\mathrm{x}9.81)\sin 30\\ \\ \mathrm{torque}=397.30\mathrm{N}\min \mathrm{to}\mathrm{page}\\ \\ \mathrm{total}\mathrm{torque}=2604.55\mathrm{Nm})\end{array}\end{array}$

Solving a transcendental equation numerically

b).

The times when the total torque about the axis of rotation is zero. Now determining the times to the nearest second,

$\begin{array}{c}\mathrm{\tau }=-794\left[\sin \frac{\mathrm{\pi }}{6}+2.78\sin 2\mathrm{\pi }(5.25)\right]\\ \\ \mathrm{\tau }=-794[.3839+2.78(.99986)]\\ \\ \mathrm{\tau }=-794[.3839+2.7796]\\ \\ \mathrm{\tau }=-794(3.1635)\\ \\ \mathrm{\tau }=-2510\mathrm{Nm}\end{array}$