Physics, asked by sandeeppallanti, 1 month ago

Derive the expression for the moment of inertia of a cylinder length T.radius r and density *w* about longitudinal centroidal axis and about the centroidal transverse axis​

Answers

Answered by amitbobbypathak
0

Answer:

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) 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  (M / L) R2 dz+ o∫L 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 + ML3 / 3

Ix = MR2 /4+ ML2/3

⇒ Check Other Object’s Moment of Inertia:

Moment Of Inertia Of A Solid Cylinder

Moment Of Inertia Of A Hollow Cylinder

Moment Of Inertia Of A Rectangular Plate

Moment Of Inertia Of Rectangle

Moment Of Inertia Of Rod

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