Answer:
Explanation:
STEP 1
Given
Radius of cylinder = r = 25cm, 2.5m
mass = 27kg
cylinder is mounted so as to rotate freely about a horizontal axis that is parallel to and 60cm to the central logitudinal axis of the cylinder
height = 0.6m
part 1
The cylinder is mounted so as to rotate freely about a horizontal axis tha is paralle to 60cm from the central longitudinal axis of then cylinder. The rotational inertia of the cylinder about the axis of rotation is given by
I = Icm + mh²
∴ I = 1/2mr² + mh² = 1/2x27x (0.5)² + 20 x (0.6)²
I=13.09kg.m²
where
Icm is the rotational inertia of the cylinder about its central axis
m is the mass of the cylinder
h is the distance between the axis of the rotation and the central axis of the cylinder
r is the radius of the cylinder
I=13.09kg.m²
part2
from the conservation of the total mechanical energy of the meter stick, the change in gravitational potential energyof the meter stick plus the change in kinetic energy must be zero
Δk + Δu = 0
1/2 I(w²-w²) = Ui-Uf
1/2 x 13.09w² = mgh
∴w=√20 x 9.8 x 0.6/(1/2 x 13.09) =117.6/6.5
w=18.09rad/s