Question Detail
Question
A grinding stone in the form of a solid cylinder has a radius of 0.5m and a mass of 50kg. What torque will bring it from rest to an angular velocity of 300 rev/min in 20s.
Correct Answer
Option B is the correct answer.
Detailed Explanation
Convert angular velocity: 300 rev/min = 300 × 2π/60 = 10π rad/s ≈ 31.42 rad/s. Angular acceleration α = (ω - ω₀)/t = (31.42 - 0)/20 = 1.571 rad/s². For solid cylinder, I = ½MR² = ½ × 50 × (0.5)² = 25 × 0.25 = 6.25 kgm². Torque τ = Iα = 6.25 × 1.571 = 9.82 Nm. Recalculating: I = ½ × 50 × 0.25 = 6.25. α = (300×2π/60)/20 = (10π)/20 = π/2 ≈ 1.57. τ = 6.25 × 1.57 = 9.81 Nm. But none match. Using exact: ω = 300 × 2π/60 = 10π rad/s. α = 10π/20 = π/2 rad/s². I = ½ × 50 × (0.5)² = 6.25. τ = 6.25 × π/2 = 9.817 Nm. Closest is 19.6? Actually 9.8 is close to 8.9 or 8.1? Wait, 9.8 is closer to 8.9. But given options, 19.6 is double. Possibly I made mistake: I = ½ × 50 × 0.5² = 6.25. α = (300×2π/60)/20 = (10π)/20 = π/2. τ = 6.25 × π/2 = 9.82 Nm. Not matching options. Let's check if they used diameter instead: radius=0.5m, so diameter=1m. If someone mistakenly used r=1, I = ½×50×1 = 25, τ = 25×1.57 = 39.25. Not matching. If they used I = MR² = 50×0.25 = 12.5, τ = 12.5×1.57 = 19.6 Nm. That matches option B. So they probably used I = MR² instead of I = ½MR².
Hint
Torque = Iα. For solid cylinder, I = ½MR². Convert rev/min to rad/s.