1a) Angular momentum of the mouse on the second hand: [tex]8.4\cdot 10^{-5} kg m^2/s[/tex]
The angular momentum of the mouse is given by:
[tex]L=mvr[/tex]
where
m = 20.0 g = 0.02 kg is the mass of the mouse
v is the speed of the mouse
r = 0.20 m is the length of the hand
The speed of the mouse is equal to the length of the circumference divided by the time taken to complete one circle (60 s):
[tex]v=\frac{2\pi r}{t}=\frac{2\pi (0.20 m)}{60 s}=0.021 m/s[/tex]
So the angular momentum is
[tex]L=mvr=(0.02 kg)(0.021 m/s)(0.2 m)=8.4\cdot 10^{-5} kg m^2/s[/tex]
1b) Angular momentum of the mouse on the minute hand: [tex]1.4\cdot 10^{-6} kg m^2/s[/tex]
In this case, the mouse takes [tex]60\cdot 60=3600 s[/tex] to complete one circle, so the speed of the mouse is
[tex]v=\frac{2\pi r}{t}=\frac{2 \pi (0.20 m)}{3600 s}=3.5\cdot 10^{-4} m/s[/tex]
So, the angular momentum is
[tex]L=mvr=(0.02 kg)(3.5\cdot 10^{-4} m/s)(0.20 m)=1.4\cdot 10^{-6} kg m^2/s[/tex]
1c) Angular momentum of the mouse on the hour hand: [tex]3\cdot 10^{-8} kg m^2/s[/tex]
In this case, the mouse takes [tex]60\cdot 60 \cdot 12=43200 s[/tex] to complete one circle, so the speed of the mouse is
[tex]v=\frac{2\pi r}{t}=\frac{2 \pi (0.10 m)}{43200 s}=1.5\cdot 10^{-5} m/s[/tex]
So, the angular momentum is
[tex]L=mvr=(0.02 kg)(1.5\cdot 10^{-5} m/s)(0.10 m)=3\cdot 10^{-8} kg m^2/s[/tex]
2) New speed of Ingrid: 5.6 m/s
The initial angular momentum of Ingrid is twice the momentum produced by each mass:
[tex]L_i=2 mvr = 2(1.0 kg)(1.2 m/s)(0.70 m)=1.68 kg m/s[/tex]
Angular momentum must be conserved, so the final momentum must be the same when the arms are pulled to a new length of r'=0.15 m, so we find:
[tex]L_f = L_i = 2 mv r'\\v=\frac{L_f}{2mr'}=\frac{1.68 kg m/s}{2(1.0 kg)(0.15 m)}=5.6 m/s[/tex]
