Starting from rest, a 25 cm diameter disk takes 5.0 s to reach its final angular velocity of 2500 rpm (rotations per minute). Assume that the angular acceleration is constant. The disk has a moment of inertia of
2.3
×
10
−
5
kg m
2
How much torque is applied to the disk in these 5.0 s?

Question

Starting from rest, a 25 cm diameter disk takes 5.0 s to reach its final angular velocity of 2500 rpm (rotations per minute). Assume that the angular acceleration is constant. The disk has a moment of inertia of 
2.3
×
10
−
5
kg m
2
 How much torque is applied to the disk in these 5.0 s?

BlackTom AI · Accepted Answer

Question restatement: Starting from rest, a 25 cm diameter disk takes 5.0 s to reach its final angular velocity of 2500 rpm. With constant angular acceleration, and given the disk's moment of inertia I = 2.3 × 10^−5 kg·m^2, what torque is applied to the disk during these 5.0 s?

Option 1: 1.2 × 10^−3 N·m
This is the correct scale. To verify, convert the final angular speed: 2500 rpm = 2500 × 2π rad per minute = 5000π rad/min. Converting to rad/s gives ω_f = (5000π)/60 ≈ 261.8 rad/s. With ω_i = 0 and Δt = 5.0 s, the constant angular acceleration is α = Δω/Δt ≈ 261.8/5.0 ≈ 52.36 rad/s^2. The torque is τ = Iα = (2.3 × 10^−5) × 52.36 ≈ 1.20 × 10^−3 N·m. This matches Option 1.

Option 2: 2.1 × 10^−3 N·m
If you compute torque as τ = Iα with the same α ≈ 52.36 rad/s^2, you’d need I ≈ τ/α ≈ (2.1 × 10^−3)/(52.36) ≈ 4.0 × 10^−5 kg·m^2, which is more than ten times the given I. This shows a mismatch between the stated I and the implied torque, so this option is inconsistent with the given data.

Option 3: 8.0 × 10^−5 N·m
This would correspond to a very small α or a much smaller I than stated. If τ were 8.0 × 10^−5 N·m and I = 2.3 × 10^−5 kg·m^2, then α = τ/I ≈ (8.0 × 10^−5)/(2.3 × 10^−5) ≈ 3.5 rad/s^2. Over 5 seconds, Δω ≈ αΔt ≈ 17.5 rad/s, which is far from the final ω_f ≈ 261.8 rad/s. Hence this option is not compatible with the given final speed.

Option 4: 5.5 × 10^−3 N·m
Using I = 2.3 × 10^−5 kg·m^2, the required α would be α = τ/I ≈ (5.5 × 10^−3)/(2.3 × 10^−5) ≈ 239 rad/s^2. Over 5 s, Δω ≈ 1195 rad/s, which would lead to a final ω far larger than the stated 261.8 rad/s. This shows an inconsistency with the provided ω_f and Δt.

Option 5: 3.9 × 10^−4 N·m
Similarly, τ ≈ 3.9 × 10^−4 N·m implies α ≈ τ/I ≈ (3.9 × 10^−4)/(2.3 × 10^−5) ≈ 17 rad/s^2. Over 5 s, Δω ≈ 85 rad/s, far from the required 261.8 rad/s, so this option does not align with the given final speed.

Summary across options: The only choice that coherently uses the given I, the time interval, and the final angular speed to produce the required angular acceleration and resulting torque is Option 1, which yields τ ≈ 1.2 × 10^−3 N·m. Each of the other options either uses an inconsistent inertia value, an incorrect acceleration, or leads to a final angular speed that does not match the provided data.

Starting from rest, a 25 cm diameter disk takes 5.0 s to reach its final angular velocity of 2500 rpm (rotations per minute). Assume that the angular acceleration is constant. The disk has a moment of inertia of 2.3 × 10 − 5 kg m 2 How much torque is applied to the disk in these 5.0 s?

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