Question Detail
Question
A grindstone is suspended by a wire from its center. A torque of 0.15 Nm applied to it turns it through 10°. When released, it oscillates with a period of 1.0 s. Find the moment of inertia of the grindstone.
Correct Answer
Option A is the correct answer.
Detailed Explanation
From Worked Example 11 (page 30), first, convert the angle to radians: 10° = 10 * (π/180) rad ≈ 0.1745 rad. The torsion constant K = τ/θ = 0.15 Nm / 0.1745 rad ≈ 0.86 Nm/rad. The period of a torsional pendulum is T = 2π√(I/K). Squaring both sides: T² = 4π²I/K. So, I = KT² / (4π²) = (0.86 Nm/rad) * (1.0 s)² / (4π²) ≈ 0.022 kgm².
Hint
First, calculate the torsion constant K from the given torque and angular displacement (remember to convert degrees to radians). Then use the formula for the period of a torsional pendulum, T = 2π√(I/K), to solve for the moment of inertia I.