A wooden disk with a moment of inertia of 5 kg m2 rotates with an angular speed of 50 rad/s. A metal cylinder with rotational inertia 20 kg m2 is dropped from rest onto the wooden cylinder so that their centers coincide.(a) When the cylinders no longer slip against each other, what is the final angular speed of the rotating system?
Question
A wooden disk with a moment of inertia of 5 kg m2 rotates with an angular speed of 50 rad/s. A metal cylinder with rotational inertia 20 kg m2 is dropped from rest onto the wooden cylinder so that their centers coincide.(a) When the cylinders no longer slip against each other, what is the final angular speed of the rotating system?
Solution
To solve this problem, we need to use the principle of conservation of angular momentum. The total initial angular momentum of the system is the sum of the angular momentum of the wooden disk and the metal cylinder. Since the metal cylinder is initially at rest, its angular momentum is zero. The initial angular momentum of the wooden disk is given by the product of its moment of inertia and its angular speed.
Initial angular momentum (L_initial) = Moment of inertia of the wooden disk * Angular speed of the wooden disk L_initial = 5 kg m^2 * 50 rad/s = 250 kg m^2 rad/s
After the metal cylinder is dropped onto the wooden disk, the two objects rotate together at the same angular speed. The total final angular momentum of the system is the sum of the angular momentum of the wooden disk and the metal cylinder.
Final angular momentum (L_final) = Moment of inertia of the wooden disk * Final angular speed + Moment of inertia of the metal cylinder * Final angular speed
Since angular momentum is conserved, the initial and final angular momentum are equal.
L_initial = L_final 250 kg m^2 rad/s = (5 kg m^2 + 20 kg m^2) * Final angular speed
Solving for the final angular speed gives:
Final angular speed = 250 kg m^2 rad/s / 25 kg m^2 = 10 rad/s
So, the final angular speed of the rotating system is 10 rad/s.
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