Question

Big Ben Fig. the Parliament tower clock in London, has hour and minute hands with lengths of 2.70 m and 4.50 m and masses of 60.0 kg and 100 kg, respectively. Calculate the total angular momentum of these hands about the centre point.(You may model the hands as long, thin rods rotating about one end. Assume the hour and minute hands are rotating at a constant rate of one revolution per12 hours and 60 minutes, respectively.)

Short answer

$L=1.\text{2}kg\cdot m^2/s$

Step-by-step solution

Concept

The angular momentum is the amount of rotation of the body, which is the product of its inertia time and angular velocity.

Angular momentum of hour hand.

Consider the given data as below.

The length of hour hand,$L_h=2.70m$,

The mass of hour hand,$m_h=60.0kg$

The momentum of inertia of a thin rod of mass and length rotating about one end is given as follows:

$I=\frac{1}{3}m{L}^2$

The angular momentum L is defined as the product of moment of inertia I and angular speed$\omega$.

$L=I\omega$

The moment of inertial I_h of hour hand rotating about its one end is given as follows:

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

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

Substitute 2.70m for L_h, and 60.0kg for m_h in the above equation.

$\begin{array}{c}{I}_h=\frac{1}{3}(60.0\text{Invalid <msup> element}kg)\\ =145.8kg\cdot m^2\end{array}$

The hour hand makes one revolution per 12hours. Thus, the angular speed of hour hand is given as follows:

$\begin{array}{c}{\omega }_h=\frac{1rev}{12hour}\left(\frac{2\text{π}rad}{1rev}\right)\left(\frac{1h}{3600s}\right)\\ =1.454\times {10}^{-4}rad/s\end{array}$

The angular momentum L_h of hour hand is given as follows:

$L_h=I_h{\omega }_h$

Substitutefor 145.8 Kg.m² , and $1.454\times {10}^{-4}rad/s$for ${\omega }_h$in the above equation.

$\begin{array}{c}{L}_h=(145.8kg\cdot m^2)(1.454\times {10}^{-4}rad/s)\\ =0.0211kg\cdot m^2/s\end{array}$

Angular momentum of minute hand.

Consider the given data as below.

The mass of minute hand, $m_m=100kg$

The length of minute hand, $L_m=4.50m$

The moment of inertial of minute hand rotating about its one end is given as follows:

$I_m=\frac{1}{3}{m}_m{L}_m^2$

Here, m_m is mass of minute hand, and L_m is length of minute hand.

Substitute 4.50m for L_m, and 100 Kg for m_m in the above equation.

$\begin{array}{c}{I}_m=\frac{1}{3}(100kg){(4.50m)}^2\\ =675kg\cdot m^2\end{array}$

The minute hand makes one revolution per. Thus, the angular speed of minute hand is given as follows:

$\begin{array}{c}{\omega }_m=\frac{1rev}{60min}\left(\frac{2\text{π}rad}{1rev}\right)\left(\frac{1min}{60s}\right)\\ =1.745\times {10}^{-3}rad/s\end{array}$

The angular momentum L_m of minute hand is given as follows:

$L_m=I_m{\omega }_m$

Substitute known values in the above equation, you get

$\begin{array}{c}{L}_m=(675kg\cdot m^2)(1.745\times {10}^{-3}rad/s)\\ =1.177kg\cdot m^2/s\end{array}$

 Total angular momentum

The total angular momentum L is equal to the sum of angular momentum L_h of hour hand and angular momentum L_m of minute hand.

$\begin{array}{c}L=L_h+L_m\\ L=(0.0211kg\cdot m^2/s)+(1.177kg\cdot m^2/s)\\ =1.\text{2}kg\cdot m^2/s\end{array}$

Thus, the total angular momentum (rounding off to significant figures) is $1.\text{2}kg\cdot m^2/s$

Result

$L=1.\text{2}kg\cdot m^2/s$