A tangential for of 8.0 N acts on a wheel which com plates its first revolution from rest in 20 s s the force acts 0.30 m from the center of the wheel. Calculate a. The angular acceleration b. The moment of inertia of the wheel
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To find the angular acceleration, we first need to use the formula for torque (τ), which is τ = force (F) × distance (r). Here, F = 8.0 N and r = 0.30 m, so τ = 8.0 N × 0.30 m = 2.4 N·m. The angular displacement (θ) for one complete revolution is 2π radians. The average angular acceleration (α) can be determined from the formula θ = 0.5αt². Plugging in θ = 2π and t = 20 seconds gives us: 2π = 0.5α(20)² 2π = 0.5α(400) α = (4π) / 400 = 0.0314 rad/s² (approximately). For part b, we know that τ = Iα, where I is the moment of inertia. Rearranging gives us I = τ / α. From part a, τ = 2.4 N·m and α = 0.0314 rad/s², thus: I = 2.4 / 0.0314 ≈ 76.4 kg·m². So the moment of inertia of the wheel is approximately 76.4 kg·m².
