A uniform rod of mass 2.60×102 kg and length 0.430 m rotates in a horizontal pla

Question

A uniform rod of mass 2.60×102 kg and length 0.430 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.190 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 4.90×102 m on each side from the center of the rod, and the system is rotating at an angular velocity 35.0 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

We will treat the rings as point masses
Initially the moment of the system is
0.0260*0.43^2/12+2*0.190*0.049^2
I=0.0013129

When the rings are at the very end of the rod the moment is
0.0260*0.43^2/12+2*0.190*0.2^2
I=0.019206

Set up the momentum equation

0.0013129*35=0.019206*
solve for

=0.001617*28/0.016407

2.39 rev/min

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