A torque of 12 N ∙ m is applied to a solid, uniform disk of radius 0.50 m. If the disk accelerates at 5.7 rad/s2, what is the mass of the disk?

Question

BlackTom AI · Accepted Answer

We start by identifying the rotational dynamics relation between torque, moment of inertia, and angular acceleration: τ = I α.
For a solid disk, the moment of inertia about its central axis is I = (1/2) M R^2. Here, R = 0.50 m, so I = (1/2) M (0.50)^2 = (1/2) M (0.25) = 0.125 M.
Substitute the given values into τ = I α: 12 N·m = (0.125 M)(5.7 rad/s^2).
Compute the product of the coefficient and α: 0.125 × 5.7 = 0.7125. So 12 = 0.7125 M.
Solve for M: M = 12 / 0.7125 ≈ 16.82 kg, which rounds to approximately 17 kg.
Now evaluate each option:
- Option 5 kg: If M were 5 kg, then I = 0.125 × 5 = 0.625 kg·m^2, and τ = I α = 0.625 × 5.7 ≈ 3.56 N·m, far from the given 12 N·m. This shows 5 kg cannot produce the required torque at the given acceleration.
- Option 17 kg: With M = 17 kg, I = 0.125 × 17 = 2.125 kg·m^2. Then τ = I α = 2.125 × 5.7 ≈ 12.11 N·m, which is very close to the given 12 N·m, consistent within rounding.
- Option 13 kg: I = 0.125 × 13 = 1.625 kg·m^2. τ = 1.625 × 5.7 ≈ 9.28 N·m, which is significantly less than 12 N·m, so this mass is not correct.
- Option 3 kg: I = 0.125 × 3 = 0.375 kg·m^2. τ = 0.375 × 5.7 ≈ 2.14 N·m, far too small to match the given torque and angular acceleration.
Overall, only the mass around 16.8–17 kg yields a torque consistent with the provided angular acceleration and radius, while the other masses lead to torques that are too small given τ = I α. The trend shows that increasing mass increases I and hence τ for the same α, so the incorrect options systematically underpredict the torque for the specified α.

A torque of 12 N ∙ m is applied to a solid, uniform disk of radius 0.50 m. If the disk accelerates at 5.7 rad/s2, what is the mass of the disk? 单项选择题

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