**already found the answer to part A...just need part B*** A uniform rod of mass

Question

**already found the answer to part A...just need part B***

A uniform rod of mass 3.00 x 10^-2 kg and length 0.350m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.180kg, are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.30 x 10^-2 m on each side from the center of the rod, and the system is rotating at an angular velocity 31.0rev/min. Without otherwise changing the system, the catches are released, and the rings slide outward along the rod and fly off at the ends.

a) What is the angular speed of the system at the instant when the rings reach the ends of the rod?

3.60 (rev/min)

b) What is the angular speed of the rod after the rings leave it?

? (rev/min)

Explanation / Answer

This is solved by conservation of angular momentum, which is I*w, where I=moment of inertia, w = angular velocity. Calculate I1, with the rings in the first postion (that is the sum of the I of the rod Ir, plus the I from the rings, which for each ring is m_2*r^2; so the initial momentum is (Ir + 2*m_2*r^2)*w1. At the time the rings are about to leave the rod, they are at L/2 from the center of rotation, so the angular momentum then is [Ir + 2*m_2*(L/2)^2]*w2 Equate these and solve for w2. When the rings leave the rod, the only moment is that of the rod, and its angular momentum is Ir*w3, which must equal the initial momentum (Ir + 2m_2*r^2)*w1 Set these equal and solve for w3. Look up the angular momentum Ir of a rod of length L rotating about its centerpoint (I think it's m_1 * (L^2)/12, but you should check it.)

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