System of Particles & Rotational Motion Test - 3

For a ring moment of inertia ,I=M*R*R   M=Mass and R=Radius

In terms of radius of gyration 

I=M*K*K

M*R*R=M*K*K

Therefore ,K=R

We know that MI of a ring is mr²
Where m is mass of the ring and r is its radius
When we have ratio of I = 2:1
And ratio of r = 2:1
We get ratio of r² = 4:1
Thus to make this ratio 2:1 , that ratio of masses must be 1:2

If K is radius of gyration, then moment of inertia for the rod is given as -

I = MK²                        (1)

But moment of inertia of uniform rod of length L and Mass M about an axis passing through its center the perpendicular to its length  is given as - 

comparing equation (1) and (2), we get

We know that for a solid sphere of mass S and radius r, moment of inertia I = ⅖ Sr2 
Similarly for a hollow sphere of mass H and radius r, moment of inertia I = ⅔ Hr² 
Thus if there radii and I are equal the ratio of there mass can be find by equating both as
I =  ⅖ Sr²  = ⅔ Hr²
Hence we het H/S = ⅗
So if S is 5kg we get H = 3kg

Moment of inertia of solid cylinder
MR²/2=0.2*10*10^-2*10*10^-2  Kg m²
=2*10^-3/2
=10^-3 Kg m²

Given,
Mass of A=1,
Mass of B=2.
diameter if A=2,
diameter if B=1.
radius (r) of A=d/2=2/2=1.
radius (r) of B=d/2=1/2.
we know ,
moment of inertia of disc=MR²/2.
moment of inertia (I)of A/moment of inertia (I)of B=MR2/2/MR2/2.
(I) of A/(I) of B=1×1²/2/2×(1/2)²/2.
=1×1/2/2×(1/4)/2.
=1/2/(1/2)/2.
=1/2/1/4.
=4/2.
=2/1.

Moment of inertia of solid sphere I_s= 2/5MR²
moment of inertia of hollow sphere I_h =2/3MR²
given mass of solid sphere =√45 kg.
I_s=I_h
2MR²/5=2MR²/3
given their masses are equal 2 (√45)²/5= 2 R²/3
45/5=R²/3
9=R²/3
9×3=R²
27=R²
√27=R
√3×9=R
3√3 m=R.

The moment of inertia of a body is independent of. The moment of inertia of a body is independent of its angular velocity.It depends upon the position and orientation of the axis of rotation,shape and size of the body and distribution of mass of the body about the axis of rotation.

The moment of inertia (M.I.) of a sphere about its diameter=2MR²/5
According to the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.
The M.I. about a tangent of the sphere =2MR²/5+MR²=7MR²/5

A ring has a larger moment of inertia because its entire mass is concentrated at the rim at maximum distance from the axis.