l Verizon 6:00 PM flipitphysics.com * 88%- A solid disk of mass m 9.5 kg and rad
ID: 2037669 • Letter: L
Question
l Verizon 6:00 PM flipitphysics.com * 88%- A solid disk of mass m 9.5 kg and radius R 0.2 m is rotating with a constant angular velocity of w-31 rad/s. A thin rectangular rod with mass m2 3.4 kg and length L- 2R 0.4 m begins at rest above the disk and is dropped on the disk where it begins to spin with the disk. What is the initial angular momentum of the rod and disk system? kg-m2/s Sam What is the initial rotational energy of the rod and disk system? What is the final angular velocity of the disk rad/s som "What is the final angular momentum of the rod and disk system kg-m2/s Somt 5What is the final rotational energy of the rod and disk system? The rod took t = 5.1 s to accelerate to its final angular speed with the disk. What average torque was exerted on the rod by the disk? Below is some space to write notes on this problemExplanation / Answer
(1) Initial angular momentum of the disk = Idiskwdisk
Idisk(moment of inertia of the disk) = (1/2)m1R2 = (1/2)*(9.5)*(0.2)2 = 0.19 kg-m2
hence initial angular momentum = 0.19*(31) = 5.89 kg-m2/s
(2) Initial rotational energy = (1/2)Iw2 = (1/2)*0.19*312 = 91.295 J
(3) Now when rod is dropped over disk then
final angular momentum of the disk and rod = (Idisk + Irod)W
where W is final angular velocity
Irod = m2L2 /12 = 3.4*0.42/12 = 0.0453 kg-m2
Now by conservation of momentum
Initial momentum = final momentum
(Idisk + Irod)W = 5.89
(0.19 + 0.0453)W = 5.89
W = 25.032 rad/s
(4) Since momentum remain conserved therefore
Final momentum = 5.89 kg-m2/s
(5) Final rotational energy of the system = (1/2)*(Idisk+Irod)W2
= (1/2)(0.19 + 0.0453)*(25.032)2 = 73.72 J
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