Question

Big Ben, the nickname for the clock in Elizabeth Tower (named after the Queen in 2012) in London, has an hour hand long with a mass of 60.0 kgand a minute hand 4.50 mlong with a mass of 100 kg(Fig. P10.49). Calculate the total rotational kinetic energy of the two hands about the axis of rotation. (You may model the hands as long, thin rods rotated about one end. Assume the hour and minute hands are rotating at a constant rate of one revolution per 12 hours and 60 minutes, respectively.)

Short answer

Hence, total rotational kinetic energy is, $K=1.04\times {10}^{-3}\mathrm{J}$.

Step-by-step solution

Define rotational kinetic energy

Rotational energy or angular kinetic energy is kinetic energy due to the rotation of an object and is part of its total kinetic energy.

Given values

$\begin{array}{c}{\omega }_h=\frac{2\pi }{12 \mathrm{hrs}}\\ =1.45\times {10}^{-4}\mathrm{rad}/\sec \end{array}$

$\begin{array}{c}{m}_h=60 \mathrm{kg},\\ L_h=2.7\mathrm{m}\\ {\omega }_m=\frac{2\pi }{60 \min }\\ =1.75\times {10}^{-3} \mathrm{rod}/\sec \end{array}$

$\begin{array}{c}{m}_m=100 \mathrm{kg},\\ L_{\mathrm{m}}=4.5 \mathrm{m}\end{array}$

Find the rotational kinetic energy

Now moment of inertia of hour hand,

$I_h=\frac{1}{3}{m}_h{h}^2$

Substitute the values,

$\begin{array}{c}{I}_h=\frac{1}{3}\times 60\times 2.7^2\\ =146{\mathrm{kgm}}^2\end{array}$

M. I of the minute hand

$\begin{array}{c}{I}_m=\frac{1}{3}\left(100\right)(4.5{)}^2\\ =675 {\mathrm{kgm}}^2\end{array}$

Hence, total rotational kinetic energy is,

$\begin{array}{c}k=\frac{1}{2}\times 146\times {\left(1.45\times {10}^{-4}\right)}^2+\frac{1}{2}\times 675\times {\left(1.75\times {10}^{-3}\right)}^2\\ K=1.04\times {10}^{-3}\mathrm{J}\end{array}$

So, the energy is$1.04\times {10}^{-3}\mathrm{J}$.