Question

A uniform cylinder of radius 40 cm and mass 28 kg is mounted so as to rotate freely about a horizontal axis that is parallel to and 6.0 cm from the central longitudinal axis of the cylinder. (a) What is the rotational inertia of the cylinder about the axis of rotation?

Answer 1

Answer:

I' = 2.34 kg.m².

Explanation:

Given ,R= 40 cm = 0.4 m

m = 28 kg

d= 6 cm = 0.06 m

The mass moment of inertia of the cylinder about center given as

$I=\dfrac{1}{2}mR^2$

By using Parallel axis theorem

$I'=\dfrac{1}{2}mR^2+md^2$

Now by putting the values

$I'=\dfrac{1}{2}mR^2+md^2$

$I'=\dfrac{1}{2}\times 28\times 0.4^2+28\times 0.06^2$

I'=2.34 kg.m²

There the answer will be 2.34 kg.m².