Physics total angular momentum clock question?

Question

The Parliament Building 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. Treat the hands as long, thin uniform rods.

Answer 1

"But at my back, I always hear, Time's winged chariot hurrying near..."

The "Parliament Building tower clock in London" is called "Big Ben."

Treating the hands of the clock as long, thin uniform rods, their moments of inertia about the central pivot point are as follows:

H hand: (4/3) (60.0) (2.70/2)^2kg m^2 = 145.8 kg m^2.

M hand: (4/3) (100) (4.50/2)^2kg m^2 = 675.0kg m^2.

The hour hand goes around once in 12 hours or 12x3600s = 43200s. So its angular speed is 2 π / 43200 rad/s = 0.000145444... rad/s. Therefore its angular momentum is:

(154.8)(0.000145444..) kg m^2/s = 0.02251... kg m^2/s.

The minute hand goes around once in 1 hour or 3600s. So its angular speed is 2 π / 3600 rad/s = 0.00174533... rad/s. Therfore its angular momentum is:

(675.0)(0.00174533...) kg m^2/s = 1.178... kg m^2/s.

Clearly the minute hand, with its greater mass, length, and "dizzying angular speed," completely dominates the total angular momentum. However, for what it's worth, that total angular momentum is:

≈ 1.200(6) kg m^2/s.

Live long and prosper.