Please HELP! A uniform rod of mass 3.05×10 2 kg and length 0.380 m rotates in a
ID: 1464363 • Letter: P
Question
Please HELP! A uniform rod of mass 3.05×102 kg and length 0.380 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.250 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.40×102 m on each side from the center of the rod, and the system is rotating at an angular velocity 26.0 rev/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? B)What is the angular speed of the rod after the rings leave it?
Explanation / Answer
I(rod) = ML^2/12 = 3.05x 10^-2 x (0.38)^2/12 = 3.67 x 10^-4 kg-m^2
I(ring at initial position) = sigmaMR^2 = 2 x 0.25 x (5.40 x 10^-2)^2 = 1.46 x 10^-3
=>I(initial) = (0.367+1.46) x 10^-3 = 1.827 x 10^-3 kg-m^2
=>omega(initial) = 2 x pi x n = 2 x 3.14 x (26/60) = 2.72 rad/sec
Thus angular momentum L(initial) = I x omega = 1.827 x 10^-3 x 2.72 = 4.97 x 10^-3 N-m-s
I(ring at final position) = sigmaMR^2 = 2 x 0.25 x (19 x 10^-2)^2 = 0.002 x 10^-3
=>I(final) = (0.367+0.002) x 10^-3 = 0.369 x 10^-3 kg-m^2
Let the angular velocity now is omega(final), Thus by the law of angular momentum conservation:-
=>L(initial) = L(final)
=>4.97 x 10^-3 = 0.369 x 10^-3 x omega(final)
=>omega(final) = 13.47 rad/sec
By omega = 2 x pi x n
=>13.47 = 2 x 3.14 x n
=>n = 2.144 rev/sec = 128.65 rev/min; answer for (A)
As the ring leaves the rod, Again by the law of angular momentum conservation:-
=>L = I(rod) x omega(rod)
=>4.97 x 10^-3 = 3.67 x 10^-4 x omega(rod)
omega(rod) = 13.54 rad/sec; answer for (B)