A disc of radius R and thickness R÷6 has moment of inertia I about an axis passing through its centre and perpendicular to its plane . Disc is melted and recast into a solid sphere. The moment of inertia of a sphere about its diameter is
volumedisk=volumesphere
PI*R^3/6=4/3 PI r^3
r=R/2
Isphere=.40 MR^2
Idisk=1/2 Mr^2= .50 MR^2/4=.125 MR^2
so the sphere has a moment about .40/.125 greater
To find the moment of inertia of the sphere about its diameter, we can use the concept of conservation of angular momentum.
The moment of inertia of a disc about its axis passing through its center and perpendicular to its plane is given by the formula:
I_disc = (1/4) * M * R^2
Where M represents the mass of the disc, and R represents the radius of the disc.
Let's denote the moment of inertia of the sphere about its diameter as I_sphere.
Now, according to the law of conservation of angular momentum, the moment of inertia of the disc should be equal to the moment of inertia of the sphere. Since both the disc and the sphere have the same mass, we can equate their moments of inertia:
I_disc = I_sphere
(1/4) * M * R^2 = I_sphere
Now, we need to express the moment of inertia of the sphere in terms of its mass and radius. The moment of inertia of a solid sphere about its diameter is given by the formula:
I_sphere = (2/5) * M * R^2
Substituting this into the equation, we get:
(1/4) * M * R^2 = (2/5) * M * R^2
Simplifying the equation, we find:
(1/4) = (2/5)
To solve for (2/5) * M * R^2 = I_sphere, we can multiply both sides of the equation by (5/2) to get:
(5/2) * (2/5) * M * R^2 = (5/2) * I_sphere
Canceling out the factors of (5/2), we find:
M * R^2 = (5/2) * I_sphere
Therefore, the moment of inertia of a sphere about its diameter is:
I_sphere = M * R^2 / (5/2)
I_sphere = (2/5) * M * R^2
Hence, the moment of inertia of the sphere about its diameter is (2/5) times the mass of the sphere times the square of its radius.