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Answer 1

Question:

Four particles, each of mass 'm' are situated at the vertices of a square ABCD of side 'a' as shown in the figure. The moment of inertia of the system about line BD will be

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Answer:

I = ma²

Explanation:

The moment of inertia of the system is given by,

  I = Σmr²

where

m denotes mass of the particle

r denotes the perpendicular distance of the particle from the axis.

  In the given system, we have to find the moment of inertia of the system about line BD.

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ABCD is a square of side a.

Hence, diagonal of the square = a√2

Half of the diagonal = a√2/2 = a/√2

Mass of the four particles is same.

m₁ = m₂ = m₃ = m₄ = m

Let

r₁ be the perpendicular distance of the particle situated at A from the line BD

r₂ be the perpendicular distance of the particle situated at B from the line BD

r₃ be the perpendicular distance of the particle situated at C from the line BD

r₄ be the perpendicular distance of the particle situated at D from the line BD

So,

r₁ = AO = half of the diagonal of the square.

r₁ = a/√2

r₂ = 0

As the particle is situated on the line BD.

r₃ = OC = half of the diagonal of the square

r₃ = a/√2

r₄ = 0

As the particle is situated on the line BD.

 The moment of inertia of the system :

I = m₁r₁² + m₂r₂² + m₃r₃² + m₄r₄²

I = m(a/√2)² + m(0)² + m(a/√2)² + m(0)²

I = ma²/2 + 0 + ma²/2 + 0

I = ma²/2 + ma²/2

I = ma²

Therefore, the moment of inertia of the system about the line BD will be ma²

Answer 2

Question:

Four particles, each of mass 'm' are situated at the vertices of a square ABCD of side 'a' as shown in the figure. The moment of inertia of the system about line BD will be

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Answer:

I = ma²

Explanation:

The moment of inertia of the system is given by,

I = Σmr²

where

m denotes mass of the particle

r denotes the perpendicular distance of the particle from the axis.

In the given system, we have to find the moment of inertia of the system about line BD.

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ABCD is a square of side a.

Hence, diagonal of the square = a√2

Half of the diagonal = a√2/2 = a/√2

Mass of the four particles is same.

m₁ = m₂ = m₃ = m₄ = m

Let

r₁ be the perpendicular distance of the particle situated at A from the line BD

r₂ be the perpendicular distance of the particle situated at B from the line BD

r₃ be the perpendicular distance of the particle situated at C from the line BD

r₄ be the perpendicular distance of the particle situated at D from the line BD

So,

r₁ = AO = half of the diagonal of the square.

r₁ = a/√2

r₂ = 0

As the particle is situated on the line BD.

r₃ = OC = half of the diagonal of the square

r₃ = a/√2

r₄ = 0

As the particle is situated on the line BD.

The moment of inertia of the system :

I = m₁r₁² + m₂r₂² + m₃r₃² + m₄r₄²

I = m(a/√2)² + m(0)² + m(a/√2)² + m(0)²

I = ma²/2 + 0 + ma²/2 + 0

I = ma²/2 + ma²/2

I = ma²

Therefore, the moment of inertia of the system about the line BD will be ma²