Motion of Rigid Body

40 rad

Step 1: Find angular acceleration: α = (ω_f − ω_i)/t = (20 − 0)/4 = 5 rad/s². Step 2: Use the rotational kinematic equation: θ = ω_i·t + (1/2)αt². Step 3: θ = 0 + (1/2)(5)(4²) = (1/2)(5)(16) = 40 rad. Alternatively, θ = (ω_i + ω_f)/2 × t = (0 + 20)/2 × 4 = 40 rad. Option (a) 80 rad uses ωt = 20×4 ignoring the uniform acceleration (treating it as constant velocity). Option (b) and (d) are arithmetic errors.

(1/2)md²

Step 1: Each atom is at distance d/2 from the axis of rotation (since the axis is midway between them). Step 2: Apply I = Σmᵢrᵢ²: each atom contributes m(d/2)² = md²/4. Step 3: Total I = md²/4 + md²/4 = md²/2 = (1/2)md². Option (a) md² incorrectly uses d as the distance instead of d/2. Option (c) 2md² uses 2m×d². Option (d) uses only one atom's contribution.

8 rad/s

Step 1: No external torque acts on the dancer, so angular momentum is conserved: L_i = L_f. Step 2: I_i × ω_i = I_f × ω_f → 4 × 2 = 1 × ω_f. Step 3: ω_f = 8/1 = 8 rad/s. When moment of inertia decreases, angular velocity must increase proportionally to conserve L = Iω. Option (b) 4 rad/s forgets to account for the factor-of-4 reduction in I. Option (d) 16 rad/s results from squaring rather than using direct proportion.

√(gh)

Step 1: Energy conservation: Mgh = (1/2)Mv² + (1/2)Iω². Step 2: For pure rolling, v = Rω, so ω = v/R. For a hoop, I = MR², so (1/2)Iω² = (1/2)MR²(v/R)² = (1/2)Mv². Step 3: Mgh = (1/2)Mv² + (1/2)Mv² = Mv², so v = √(gh). Option (a) √(2gh) is for a sliding object with no rotation. Options (c) and (d) correspond to a solid cylinder and solid sphere respectively.