Big ben, 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 center 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 per 12 hours and 60 minutes, respectively.)

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aristocles aristocles · Answer 1 · 2019-01-14T06:26:55+01:00

Mass of Hour hand = 60 kg

length of hour hand = 2.70 m

Now moment of inertia of the hour hand is given by

[tex]I = \frac{mL^2}{3}[/tex]

[tex]I = \frac{60(2.70)^2}{3}[/tex]

[tex]I = 145.8 kg m^2[/tex]

now the angular speed of hour hand is

[tex]\omega = \frac{2\pi}{T}[/tex]

[tex]\omega = \frac{2\pi}{12\times3600}[/tex]

[tex]\omega = 1.45\times 10^{-4} rad/s[/tex]

now angular momentum is given as

[tex]L_1 = I\omega = 145.8 \times 1.45 \times 10^{-4} = 0.021 kg m^2/s[/tex]

Mass of minute hand = 100 kg

length of minute hand = 4.50 m

Now moment of inertia of the minute hand is given by

[tex]I = \frac{mL^2}{3}[/tex]

[tex]I = \frac{100(4.50)^2}{3}[/tex]

[tex]I = 675 kg m^2[/tex]

now the angular speed of minute hand is

[tex]\omega = \frac{2\pi}{T}[/tex]

[tex]\omega = \frac{2\pi}{60\times60}[/tex]

[tex]\omega = 1.74\times 10^{-3} rad/s[/tex]

now angular momentum is given as

[tex]L_2 = I\omega = 675 \times 1.74 \times 10^{-3} = 1.178 kg m^2/s[/tex]

Total angular momentum is given as

[tex]L = L_1 + L_2[/tex]

[tex]L = 0.021 + 1.178 = 1.2 kg m^2/s[/tex]