A uniform cylinder of radius 25 cm and mass 27 kg is mounted so as to rotate freely about a horizontal axis that is parallel to and 60.0 cm from the central longitudinal axis of the cylinder. (a) What is the rotational inertia of the cylinder about the axis of rotation? kg·m2 (b) If the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates, what is the angular speed of the cylinder as it passes through its lowest position? rad/s

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

the answer is 100. i just did this lesson and got a 100% on it. : )

-fishtank