Find out the moment of inertia of a semi circular disc of mass 10 kg and radius 3m about an axis passing through its centre of mass and perpendicular to the plane? (Take π2 = 10)

The moment of inertia (I) of a full circular disc about an axis passing through its centre of mass and perpendicular to the plane is given by the formula:

I = 1/2 * m * r^2

where m is the mass of the disc and r is the radius.

However, in this case, we have a semi-circular disc, which is half of a full disc. Therefore, the moment of inertia of the semi-circular disc is half of that of a full disc. So, we have:

I = 1/2 * (1/2 * m * r^2)

Substituting the given values, m = 10 kg and r = 3 m, we get:

I = 1/2 * (1/2 * 10 kg * (3 m)^2)

I = 1/2 * (1/2 * 10 kg * 9 m^2)

I = 1/2 * (45 kg*m^2)

I = 22.5 kg*m^2

However, the question also mentions to take π^2 = 10. The moment of inertia of a semi-circular disc is actually 1/2 * m * r^2 * π/2. So, we need to multiply our answer by π/2.

I = 22.5 kg*m^2 * π/2

Substituting π^2 = 10, we get π = sqrt(10). So, π/2 = sqrt(10)/2.

I = 22.5 kg*m^2 * sqrt(10)/2

I = 22.5 kg*m^2 * 1.58

I = 35.55 kg*m^2

So, the moment of inertia of the semi-circular disc about an axis passing through its centre of mass and perpendicular to the plane is 35.55 kg*m^2.