1.a. The moment of inertia here is
"I=\\dfrac 1{12}mL^2=0.00853\\text{ kg}\u00b7\\text m^2."
b. In this case, we will need a parallel axis theorem:
"I=\\dfrac 1{12}mL^2+m(L-x)^2=0.0134\\text{ kg}\u00b7\\text m^2."
2. The angular moments of the clock hands are:
"L_h=I_h\\omega_h=\\bigg[\\dfrac130.032\u00b70.15^2\\bigg]\\bigg[\\dfrac{2\\pi 0.15}{3600}\\bigg],\\\\\\space\\\\\nI_h=6.28\u00b710^{-8}\\text{ kg}\u00b7\\text{m}^2\\text{\/s}.\\\\\\space\\\\\nL_m=I_m\\omega_m=\\bigg[\\dfrac130.046\u00b70.25^2\\bigg]\\bigg[\\dfrac{2\\pi 0.25}{60}\\bigg]^2,\\\\\\space\\\\\nI_m=2.51\u00b710^{-5}\\text{ kg}\u00b7\\text{m}^2\\text{\/s}."
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