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A sanding disk with rotational inertia 1.2 10-3 kg·m2 is attached to an electric drill whose motor delivers a torque of 10 N·m about the central axis of the disk. What are the following values about the central axis at the instant the torque has been applied for 70 ms?

Respuesta :

Answer:

Angular momentum = 0.7 kg.m²/s

Angular velocity = 583.3 rad/s

Explanation:

1. The torque τ  is related to the angular momentum L by the relation

τ = ΔL/Δt

ΔL = τΔt

τ = 10 N. m

Δt = 70 ms = 70 × 10⁻³s

ΔL = (10 N. m) × (70 × 10⁻³s) = 700 × 10⁻³ kg.m²/s = 0.7 kg.m²/s

2. The rotational inertia I relates the angular momentum L to the angular velocity w

L = Iw

w = L/I

L =  0.7 kg.m²/s

I = 1.2 × 10⁻³ kg.m²

w = (0.7 kg.m²/s)/(1.2 × 10⁻³ kg.m²) = 583.3 rad/s

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