# MOMENT OF INERTIA | Moment of inertia of thin spherical solid sphere | thick spherical shell | right circular cone | truncated cone | Linear acceleration of a body rolling down in an inclined plane

## Moment of inertia of thin spherical shell about it’s diameter

A thin uniform spherical shell has a radius of a and mass M. We have to calculate its moment of inertia about any axis through its centre.

Let, radius of Spherical shell = a

mass/area = σ

Total mass M = 4πa2σ

We take on elementary ring perpendicular to diameter AB.

radius of the ring – asinθ

Mass of the ring = 2πasinθ.adθ.σ

Moment of inertia of the elementary ring about AB axis = I = ma2

Moment of inertia of the whole spherical Shell about its diameter (AB) ## Moment of inertia of a solid sphere-about its diameter

According to the figure a sphere of mass M and radius a is shown, whose density is ρ. We have to calculate the moment of inertia of the sphere about the diameter AB. We can assume the sphere to be made up of many discs whose surfaces are parallel to AB and the center is on this axis. This discs has a center at O the width of this disc is dx.

Let the radius of the solid sphere = a.

mass per unit volume = ρ

Total mass M = We can divide a solid sphere into a no. of concentric spherical shell (thin). Let us consider such a shell of radius x and thickness dx.

Mass of the elementary shell = Moment of inertia of the spherical shell about diameter  (AB)  = So,Moment of inertia of the whole sphere about the diameter (AB) ## Moment of inertia of a thick spherical shell about its diameter

Let, Let us consider a thin spherical shell of radius x a ( a2 > x >a1) and thickness dx –

Mass of the elementary shell =  ## Moment of inertia of a right circular cone about an axis passing through vertex perpendicular to axis

Here we will look at the derivation as well as the calculation for finding the moment of inertia of a uniform right circular cone about an axis.

Let us consider a right circular cone. Its have –

height =h

mass per unit volume = ρ Total mass M = We divide the solid cone into a number of coaxial discs having different radii

Let us take such a disc having radius x , thickness dy at a distance Y from the vertex of the-cone.

Moment of inertia of the elementary disc about-its own diameter = According to parallel axis theorem, Moment of inertia about the x axis passing through vertex, as from figure Moment of inertia of the whole solid cone about the required axis (x) ## Moment of inertia of a right circular solid cone about its own axis Let us consider a right circular cone having

height =h

mass per unit volume= ρ

Total mass M = We divide the solid cone into a no. of co-axial discs having different radii

Let us take such an elementary thick disc having radius x  and thickness at a distance y from vertex

Moment of inertia of the disc about its own axis (Y) = and from figure, Moment of inertia of the solid cone about its own axis (Y) Moment of inertia of a hollow cone

about an axis passing through vertex and perpendicular  to its own axis.

## Moment of inertia of a truncated cone about its own axis Let the truncated solid cone has,

the radii of the ends – a, b

height of base – h

height of upper portion – h1

mass per unit volume – ρ

Total mass M =  We divide the cone into a number of coaxial circular discs has different radii. Let us take such an elementary disc of radius x , Thickness dy at a distance y from vertex,

Moment of inertia of the thick disc about OY From figure Moment of inertia of the truncated cone about Y axis, ## Linear acceleration of a body rolling down in an inclined plane

Let, f be the linear acceleration of centre of mass and α be the angular acceleration of the body. For rotation around the axis through centre of mass, net torque = Iα. As no slipping is there, so the point of contact of the body with the plane is instantaneously at rest. Let a body of mass m is at A and it rolls down in an inclined plane AC , at an angle of inclination θ.

Let at  Q , linear velocity =v

angular velocity = w

So, gain in  kinetic energy of the body = v = wr  , r is the radius of circular body

I = mk² ,  where k is the radius of Gyration

Loss in potential energy = If f be the linear acceleration, From energy conservation,

gain in kinetic energy = loss in potential energy 1.For a solid sphere : 2.For a hollow sphere : 3.For a ring 4.For a circular disc : 5.For a solid cylinder : 6.For a hollow cylinder  