Question

The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12 kg/m2 . a) Determine the location ӯ of the center of mass G of the pendulum, then calculate the mass moment of inertia of the pendulum about z axis passing through G. b) Calculate the mass moment of inertia about z axis passing the rotation center O.

Answer 1

Answer:

The answer is below

Explanation:

a) The location ӯ of the center of mass G of the pendulum is given as:

$y=\frac{0+(\pi*(0.3\ m) ^2*12kg/m^2*1.8\ m-\pi*(0.1\ m) ^2*12kg/m^2*1.8\ m)+0.75\ m*1.5\ m *3\ kg/m}{(\pi*(0.3\ m) ^2*12kg/m^2-\pi*(0.1\ m) ^2*12kg/m^2)+3\ kg/m^2*0.8\ m+3\ kg/m^2*1.5\ m} \\\\y=0.88\ m$

b)  the mass moment of inertia about z axis passing the rotation center O is:

$I_G=\frac{1}{12}*3(0.8)(0.8)^2+ 3(0.8)(0.888)^2-\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8-\\0.888)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8-0.888)^2+\frac{1}{12}*3(1.5)(1.5)^2+\\3(1.5)(0.888-0.75)^2\\\\I_G=13.4\ kgm^2$

c) The mass moment of inertia about z axis passing the rotation center O is:

$I_o=\frac{1}{12}*3(0.8)(0.8)^2+ \frac{1}{3}* 3(1.5)(1.5)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8)^2-\\\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8)^2\\\\I_o=13.4\ kgm^2$