Question Detail
Question
A grindstone is suspended by a wire. When a torque of 0.15 Nm is applied, it turns through 10°. If released, it oscillates with a period of 1.0 s. What is the moment of inertia of the grindstone? (1° = 0.01745 rad)
Correct Answer
Option B is the correct answer.
Detailed Explanation
From Worked Example 11 on page 30: First, convert the angle to radians: θ = 10° × 0.01745 rad/° = 0.1745 rad. The torsion constant K = τ/θ = 0.15 Nm / 0.1745 rad = 0.86 Nm/rad. For a torsional pendulum, the period T = 2π√(I/K). Rearranging for I: I = K * T² / (4π²) = (0.86 Nm/rad) * (1.0 s)² / (4π²) = 0.0217 kg m², which rounds to 0.022 kg m².
Hint
First, find the torsion constant (K) using the applied torque and angle. Then, use the period formula for a torsional pendulum to find the moment of inertia (I).