Answer:
I = 16.7kgm²
Explanation:
Since, Torque is given by,
\tau = F*r = I*\alpha
here, I = Moment of inertia = ??
\alpha = angular acceleration of wheel = a/r
F = tangential tension acting on the wheel = T
a = acceleration of bag of sand = 2.95 m/s^2
r = radius of wheel = d/2 = 120/2 = 60 cm = 0.60 m
from force balance on sand bag,
mg - T = m*a
T = m*(g-a)
m = mass of sand bag = 20 kg
So, I = T*r/\alpha = m*(g-a)*r/(a/r)
Using known values:
I = 20*(9.81 - 2.95)*0.60/(2.95/0.60) = 16.74
I = 16.7 kgm² = Moment of inertia of wheel experimentally
also, Moment of inertia of wheel theoretically(I') = M*r²
given, M = mass of wheel = 70 kg
I' = 70*0.60²= 25.2 kgm² = Moment of inertia of wheel theoretically
