please help....easy problem?

Question

1)If the hanging mass is 100. g and drops a distance 25.5 cm in a time of 6.0 s, what is the moment of inertia of the support rod and shaft? The radius of the shaft is 0.50 cm.

2)Now two masses each of 180 g are placed on the rod at a distance of 15.0 cm from the point of rotation. What is the TOTAL moment of inertia of the masses plus rod and shaft?

3)How long does it take for the hanging mass to fall the same distance 25.5 cm?

please show work so I can understand the problem

Answer 1

1. I'm assuming that this mass is attached to a string wrapped around a shaft of radius 0.50 cm and forces the shaft to rotate frictionlessly as the mass descends.

The governing equation is

ma1 + T/r=mg where

m- mass of the object

T= IA
I- the moment of inertia of the rod
a1 - tangential acceleration we have
T - torque on the shaft
r - radius of the shaft
g- acceleration due to gravity

A- angular acceleration
A=a1/r we have

ma1 + I a1 /r^2=mg

I= m(g-a1)r^2/a1
now since h=0.5a1t^2
a1= 2h/(t)^2
Finally we can find moment of inertia
I= m(g-a1)r^2/a1

2) Use parallel axis theorem
I=Icm + mR^2
if asymmetrical or if symmetrical simply add
I=I1+ I2+I3


3) If the mass hangs how can it fall. However time in free fall is t=sqrt(2h/g)