Answer:
[tex]4.73\cdot 10^{-6} kg m/s^2[/tex]
Explanation:
First of all, let's calculate the moment of inertia of the second hand. The moment of inertia of a slender rod rotating about one end is given by
[tex]I=\frac{1}{3}mL^2[/tex]
where m is the mass of the rod and L is its length. For the second hand, we have
m = 6.00 g = 0.006 kg
L = 15.0 cm = 0.15 m
So, the moment of inertia is
[tex]L=\frac{1}{3}(0.006 kg)(0.15 m)^2=4.5\cdot 10^{-5} kg m^2[/tex]
Then, we have to calculate the angular speed of the second hand, which is given by:
[tex]\omega = \frac{2 \pi}{T}[/tex]
where T is the period of the second hand, which is T=60 s. Substituting,
[tex]\omega = \frac{2 \pi}{60 s}=0.105 rad/s[/tex]
Now we can finally calculate the angular momentum of the second hand, which is equal to the product of the moment of inertia I and the angular speed [tex]\omega[/tex]:
[tex]L=I\omega =(4.5\cdot 10^{-5} kg m^2)(0.105 rad/s)=4.73\cdot 10^{-6} kg m/s^2[/tex]
