Question

Four solid sphere each of diameter √5 cm and mass 0.5 kg are placed with their centres at the corners of a square of side 4 cm.

The moment of inertia of the system about the diagonal of the square is $N×{10}^{-4}kg{m}^{2}$.

Find the value of N.

✔️✔️ Proper solution needed ✔️✔️​

Answer 1

Given information :

Side of square ( s ) = 4 cm = 0.04 m

Mass of sphere ( m ) = 0.5 kg .

Radius of the sphere ( r ) = diameter / 2

⇒ √5/2 cm ⇒ √5/200 m

NOTE :

Remember to convert cm to m .

Also the diameter and side both should be converted .

If the question was given in g , then we would have converted to kg .

Moment of inertia of sphere through centre = 2/5 m r²

Moment of inertia at a distance of s/√2 from centre = 2/5 m r² + m ( s/√2 )²

⇒ m ( 2/5 r² + s²/2 )

Moment of inertia of the whole system = N × 10⁻⁴ kg m².

Moment of inertia of the whole system = 2 ( m ( 2/5 r² + s²/2 ) + 2/5 m r² )

⇒ 2 ( 2/5 m r² + 2/5 m r² + m s² /2 )

⇒ 2 ( 4/5 m r² + m s² / 2)

⇒ 2 ( 4/5 × 0.5 × (√5/200 )² + 0.5 ×(0.04)²/2 )

⇒ 2 ( 4/5 × 1/2 × 5/(40000) + 0.5 × 16/20000 )

⇒ 2 ( 2/5 × 5/40000 + 16/40000 )

⇒ 2 ( 2/40000 + 16/40000)

⇒ 2 × 18/40000

⇒ 36/40000

⇒ 9/10000

⇒ 9 × 10⁻⁴ kg m²

N × 10⁻⁴ kg m² = 9 × 10⁻⁴ kg m²

Comparing both the sides we get :

⇒ N = 9

The value of n is 9 .

Answer 2

Hello Friend..❤️❤️

The answer of u r question..✌️✌️

Value of n = 9

For more information refer to the attachment!!

Thank you..☺️☺️