Question

A thin uniform rod (mass = 0.420 kg) swings about an axis that passes through one end of the rod and is perpendicular to the plane of the swing. The rod swings with a period of 1.45 s and an angular amplitude of 10.6

Answer 1

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:

                  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)

   

       = (8?^2)(L/3g)

Length of the pendulum L = 3gT^2 / 8?^2

                                        = 3(9.8 m/s^2)(1.45s)^2 / 8?^2

                                        = 0.784 m

b) Maximum Kinetic energy of  the rod = Mg(L/2)(1-cos?)

                                                            = (0.42 Kg)*(9.8 m/s^2)*(0.784 m/2)*(1-cos10.6)

                                                            = 0.0275 J