Question

Question 9 10 pts A uniform thin straight rod of mass m = 2.96 kg and length L = 1.28 m can oscillate freely in a verticle plane about one of its end O, as a compound pendulum. Using parallel axis theorem, calculate the moment of inertia of the rod I in kg m2, about an axis through the end O and perpendicular to the rod. Give your answer to 3 decimal places Note: The moment of inertia of a uniform rod about a perpendicular axis passing through its centre of mass is given by l = (1/12m L2.)

Hi, need some help with full workings for this question, thanks.

Answer 1

from the question m = 2.96 kg L = 1.28 m To calculate moment of Inertia of the rod in kgm: - we have center of mass I_c = mL^2/12 By using parallel-axis theorem I = I_c + md^2 = mL^2/12 + mL^2/4 = mL^2 + 8 mL^2/12 = 4 mL^2/12 = mL^2/3 = 2.96 times (1.28)^2/3 = 2.96 times 1.6384/3 I = 4.849664/3 I = 1.616 kgm^2