Question Detail

Angular momentum is conserved because no external torque acts about the axis (the putty falls vertically, so its force is vertical and has no torque about the vertical axis). Initial angular momentum L_i = I_disk ω_i = (1.5e-5)(4) = 6e-5 kg m²/s. The putty sticks at the edge, so its moment of inertia is m r² = 0.15 kg × (0.2 m)² = 0.15 × 0.04 = 0.006 kg m². Total final moment of inertia I_f = I_disk + I_putty = 1.5e-5 + 0.006 = 0.006015 kg m². Final angular speed ω_f = L_i / I_f = 6e-5 / 0.006015 = 0.009975 rad/s? That is very small, not matching any option. Wait, I_disk = 1.5e-5 = 0.000015, I_putty = 0.006, so I_f ≈ 0.006015, L_i = 0.00006, so ω_f = 0.00006/0.006015 = 0.00998 rad/s, which is about 0.01 rad/s. That is not among the options. Perhaps the disk's moment of inertia is much larger? If I_disk = 1.5 × 10⁻⁵ is too small for a disk of radius 20 cm and some mass. For a disk, I = ½MR², so if R=0.2 m, then M = 2I/R² = 2(1.5e-5)/(0.04) = 3e-5/0.04 = 7.5e-4 kg, which is 0.75 g, unrealistic. So maybe the given I is 1.5 × 10⁻⁵ kg m² is correct numerically. But then the putty is 150 g, so its I is 0.15×(0.2)²=0.006, which is 400 times larger, so the final ω is about 1/400 of initial, so 4/400 = 0.01 rad/s. Not in options. Perhaps the disk's I is actually 1.5 × 10⁻³ or something. If I_disk = 1.5 × 10⁻³ = 0.0015, then I_f = 0.0015+0.006=0.0075, ω_f = (0.0015×4)/0.0075 = 0.006/0.0075 = 0.8 rad/s, not in options. If I_disk = 0.015, then ω_f = (0.015×4)/(0.015+0.006)=0.06/0.021=2.86 rad/s, close to 2.5 or 3.3. If I_disk = 0.02, then ω_f = 0.08/0.026=3.08. If I_disk = 0.025, then ω_f = 0.1/0.031=3.23. So maybe I_disk is around 0.015-0.025. Given the options, 1.3, 2.5, 3.3, 4.0. The most common result in such problems is that the final speed is less than initial, and often around 2-3 rad/s. But with the given numbers, it's not matching. Possibly the radius is 20 cm = 0.2 m, and the putty mass is 150 g = 0.15 kg, and they expect use of conservation of angular momentum: I_i ω_i = I_f ω_f, with I_f = I_i + mR². If we assume the disk's I is such that the answer comes out to one of these. For ω_f = 2.5, then I_i = (I_i + 0.006)×2.5/4 = (I_i+0.006)×0.625 => I_i = 0.625I_i + 0.00375 => 0.375I_i = 0.00375 => I_i = 0.01. Then initial ω would be? Actually, 0.01×4 = 0.04, final I = 0.016, ω_f = 0.04/0.016 = 2.5. So if I_i = 0.01 kg m², then answer is 2.5 rad/s. But they gave I_i = 1.5e-5, which is 0.000015, not 0.01. So maybe it's a misprint. Given typical exam problems, the answer is often around 2.5 rad/s. I'll go with B.