Question

The figure at the left shows a uniform disk of radius R = 0.8 m and 6-kg mass with as mall hole a distance d from the disk’s center that can serve as a pivot point. What should be the distance d so that the period of this physical pendulum is 2.7 s?

Answer 1

I around center
Assume you do not have a moment of inertia table so
integral rho (2 pi r t) r^2 dr = rho 2 pi t integral r^3 dr
= rho 2 pi t R^4/4 = rho pi t R^4/2
mass = rho pi t R^2
so integral= mass * R^2/2 = I around center
I around point distance d from center =m d^2 + m R^2/2
at angle T radians from straight down
gravity perpendicular to line to center = m g d sinT
for small T that is m g d T
torque about center = -m g d T
torque = -I d^2T/dt^2
-m g d T = [ m d^2 + m R^2/2 ] d^2 T/ dt^2
if T = a sin w t
d^2 T/dt^2 = - a w^2sin wt = - w^2 T
so
-m g d T = - [ m d^2 + m R^2/2 ] w^2 T
w^2 = g d /[ d^2 + R^2/2]
w = 2 pi f = 2 pi /period
4 pi^2 / period = g d /[ d^2 + R^2/2]

Answer 2

I didn't get any of that but thank you. I'll do my best to review it.