Question

Moment of Inertia (M.I.) of four bodies having same mass ‘M‘ and radius ‘2R‘ are as follows:

I₁ = M.I. of solid sphere about its diameter

I₂ = M.I. of solid cylinder about its axis

I₃ = M.I. of solid circular disc about its diameter.

I₄ = M.I. of thin circular ring about its diameter

If 2(I₂ + I₃) + I₄ = x⋅ I₁ then the value of x will be ______.

Answer 1

Correct Answer: 5

To solve the problem, we must calculate the moment of inertia (M.I.) for each given body and use them in the provided equation: 2(I₂ + I₃) + I₄ = x ⋅ I₁.

Step 1: Calculate I₁
The moment of inertia of a solid sphere about its diameter is given by: I₁ = (2/5)MR², where M is the mass and R is the radius. Since the radius here is 2R, then I₁ = (2/5)M(2R)² = (8/5)MR².

Step 2: Calculate I₂
The moment of inertia of a solid cylinder about its axis is: I₂ = (1/2)MR². For radius 2R, I₂ = (1/2)M(2R)² = 2MR².

Step 3: Calculate I₃
The moment of inertia of a solid circular disc about its diameter is: I₃ = (1/4)MR² + (1/12)Mh². Assuming negligible height, I₃ ≤ (1/4)M(2R)² = MR².

Step 4: Calculate I₄
The moment of inertia of a thin circular ring about its diameter is: I₄ = (1/2)MR². Hence, for radius 2R, I₄ = (1/2)M(2R)² = 2MR².

Step 5: Formulate equation
Plug the calculated values into the given equation:

2(I₂ + I₃) + I₄ = x ⋅ I₁

Substitute the values:

2(2MR² + MR²) + 2MR² = x ⋅ (8/5)MR²

Simplify the left side:

2(3MR²) + 2MR² = x ⋅ (8/5)MR²

6MR² + 2MR² = x ⋅ (8/5)MR²

8MR² = x ⋅ (8/5)MR²

Cancel MR² from both sides:

8 = x ⋅ (8/5)

Multiply both sides by 5/8 to isolate x:

x = 8 ⋅ (5/8) = 5

Conclusion:
The value of x is 5, falling within the expected range of 5 to 5.