Recent content by Ang09

The original ques did not state the mass of the t-shape pendulum nor the mass of each cardboard. Only inform that the two cardboard are identical/similar. So I assume that each cardboard's mass is m. Need that when I relate the MoI of the pendulum to its period.

Sorry.. i am not clear in my working.. the m in the 0.02 m and 0.18 m are the unit (metre). the m in the moment of inertia is the mass of each strip.

The computation of the centre of mass is based on something that i search online. [Link deleted by the Mentors] My computation was:- COM of the horizontal part y1: 0.02 m from the base of the horizontal part COM of the vertical part y2: (0.14 + 0.04=) 0.18 m from the base of the horizontal...

d1 is the distance between pivot and centre of mass of the vertical portion d2 is the distance between pivot and centre of mass of the horizontal portion (Refer to picture) The 0.08 m is derived based on the diagram to relate the centre of mass of the vertical and horizontal portion to the pivot.

h = d1 + 0.08 d1 = h - 0.08 d2 = h + 0.08 I of the vertical portion = 1/12 m (l^2 + b^2) + md1^2 = 1/12 m (0.28^2 + 0.04^2) + m(h - 0.08)^2 I of the horizontal portion = 1/12 m (l^2 + b^2) + md2^2 = 1/12 m (0.28^2 + 0.04^2) + m(h + 0.08)^2 The moment of inertia for the whole T-shape about...