Mass of the rod M = 0.42 Kg
Time period T = 1.45 s
Angular displacement ? = 10.6 degree
a)
The rotational inertia of a uniform rod with pivot point at its
end is:
                Â
I = 1/3ML2
T = 2? Sqrt[(1/3ML^2)/Mg(L/2)]
   = 2? Sqrt[(2/3) (L/g)]
Squaring on both sides we get
T^2 = 4?^2 (2/3)(L/g)
  Â
       = (8?^2)(L/3g)
Length of the pendulum L = 3gT^2 / 8?^2
                                        =
3(9.8 m/s^2)(1.45s)^2 / 8?^2
                                      Â
= 0.784 m
b) Maximum Kinetic energy of  the rod =
Mg(L/2)(1-cos?)
                                                          Â
= (0.42 Kg)*(9.8 m/s^2)*(0.784 m/2)*(1-cos10.6)
                                                          Â
= 0.0275 J
