A solid disk of mass m1 = 9.5 kg and radius R = 0.18 m is rotating with a consta

Question

A solid disk of mass m1 = 9.5 kg and radius R = 0.18 m is rotating with a constant angular velocity of ? = 38.0 rad/s. A thin rectangular rod with mass m2 = 3.3 kg and length L = 2R = 0.36 m begins at rest above the disk and is dropped on the disk where it begins to spin with the disk.1) What is the initial angular momentum of the rod and disk system? 2) What is the initial rotational energy of the rod and disk system? 3) What is the final angular velocity of the disk? 4) What is the final angular momentum of the rod and disk system?5) What is the final rotational energy of the rod and disk system? 6) The rod took t = 6.3 s to accelerate to its final angular speed with the disk. What average torque was exerted on the rod by the disk?

Explanation / Answer

We could use the conservation of momentum, but I think that some momentum is lost (from the way that question 6 is phrased) Lets try the conservation of Energy (1/2) Id wi^2 = (1/2) (Id + Ir) wf^2 wf = final rotational speed wi = initial rotational speed Id = moment of Inertia of the disk Ir = moment of inertia of the rod. solve for wf^2 wf^2 = Id * wi^2 / (Id + Ir) and we know that Id = (1/2) M R^2 and we can look up Ir = (1/12) m L^2 (lets use big M for mass of the disk and little m for the mass of the rod, and L is the length of the rod.) wf^2 = wi^2 ((1/2) M R^2 / ((1/2) M R^2 + (1/12) m L^2) We can get rid of all the (1/2)'s ( and take (1/2) out of (1/12) which leaves (1/6) ) wf^2 = wi^2 M R^2 / (M R^2 + (1/6) m L^2) and we know that L = 2R wf^2 = wi^2 M R^2 / (M R^2 + (1/6) m (2R)^2) wf^2 = wi^2 M R^2 / (M R^2 + (1/6) m 4R^2) Now we can dump all the R^2 wf^2 = wi^2 M / (M + (4m/6)) wf = sqrt (wi^2 M / (M + (2m/3))) Now plug in your numbers wf = sqrt ((39 rad/sec)^2 10 kg / (10 kg + (2 * 3.6 kg / 3))) wf = 35.0 rad/sec Gosh, please check my math though. Ok the final momentum (Pf) is (Id + Ir) * wf Pf = (Id + Ir) * wf Pf = (((1/2) M R^2) + ((1/12) m L^2)) * wf and L = 2R Pf = (((1/2) M R^2) + ((1/12) m (2R)^2)) * wf Pf = (((1/2) M R^2) + ((1/12) m 4R^2)) * wf Pf = (((1/2) M R^2) + ((1/3) m R^2)) * wf Pull out the R^2 Pf = (((1/2) M) + ((1/3) m)) * wf *R^2 Pf = (((1/2) 10 kg) + ((1/3) 3.6 kg)) * (35 rad/s) *(0.26 m)^2 Pf = 14.669 Hmmm. that seems odd the final momentum is larger than the initial momentum? Well, the velocity got smaller but the moment of inertia got larger...... Well, I'm too tired to recheck it, check my math for me please, but at least the formulas and mthod may help you find a better answer. 5 . Ok, since we are using the conservation of energy to solve 4, I'm gonna say that final rotational energy = initial rotational energy = 257.049 You could check with Energy = (1/2) (Id + Ir) wf^2 6. Just like in translational motion, the change in momentum equals = force * time, here we use torque instead of force. T = change in momentum / time T = 14.669 - 13.182 / 5.7 sec T = 0.261 N-m

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