Question

Rotational Motion Physics Problem

The problem is in the image posted bellow. Thank you for your help.

Answer 1

First the moments of ineria of the two discs about their own axes:
Ia = (1/2)(.791)(.168^2) = .0112 kg m^2
Ib = (1/2)(.439)(.062^2) = 8.44 *10^-4 kg m^2
now axis transformation to point p
for disk A, distance from center to p = .168
so add .791 * .168^2 = .0223
for disk B, distance requires trig , law of cosines
angle PAB = 180 - 64 = 116
cos 116 = -.438
AB = dist center A to center B = .168+.062= .230
PB^2 = .230^2 + .168^2 + .23*.168*.438
= .098
so add .429*.098 = .042
final I = .0112 + .000844 + .0223 + .042
= .0764 Kg m^2

Answer 2

I had tried something similar to this in my work and got an answer close to yours. It was incorrect. I don't know where we went wrong.

Answer 3

Well, our method is correct. There might be an arithmetic error.