Question

moment of interia of hollow cylinder write a given data​

Answer 1

Answer:Generally, the derivation involves 3 primary steps. It includes;

Splitting the cylinder into infinitesimally thin disks and stating the moment of inertia.

Using both parallel and perpendicular axis theorems to determine the expression.

Integrating over the length of the cylinder.

1. Cutting the cylinder into infinitesimally thin disks

We will consider the cylinder having mass M, radius R, length L and the z-axis which passes through the central axis.

Here,

Density ρ = M / V

Next, we will consider the moment of inertia of the infinitesimally thin disks with thickness dz.

First, we assume that dm is the mass of each disk, We get;

dm = ρ X Volume of disk

dm = M / V X (πr2.dz)

We take V = area of circular face X length which is = πr2L. Now we obtain;

dm = M / πr2L X (πr2.dz)

dm = M / L X dz

The moment of inertia about the central axis is given as;

dlz = ½ dmR2

2. Use of Perpendicular Axis Theorem

We now apply the perpendicular axis theorem which gives us;

dlz = dlx + dly

Here, if we need to consider that both x and y moments of inertia are equal by symmetry.

dlx = dly

We need to combine the equations for the perpendicular axis theorem and symmetry. We get;

dlx = dlz / 2

Now we substitute lz from the equation above.

dlx = ½ x ½ dmR2

dlx = ¼ dmR2

Alternatively, for the x-axis, we use the parallel axis theorem to find the moment of inertia. We get;

dlx = ¼ dmR2 + dmz2

3. Integration

Now we conduct integration over the length of the cylinder to express the mass element dm in terms of z. We take the integral from z=0 to z=L.

Ix = o∫L dlx

Ix = o∫L ¼ dzR2 + o∫L ¼ z2 m / l dz

Ix = ¼ M / L R2z + M / L Z3 / 3]0L

Since it is a definite form of integral we ignore the constant. We will now have;

Ix = ¼ M / L R2 L + M / 3L 2L3 / 23

Ix = ¼ MR2 + 1/12 ML2

Explanation: