A uniform rod of mass 3.40×102 kg and length 0.430 m rotates in a horizontal pla

Question

A uniform rod of mass 3.40×102 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 0.230 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 32.0 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.

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

Explanation / Answer

conservation of angular momentum

I1W1 = I1W2

I1 = IROD+2*IRING

I1 = 1/12ML^2+2*0.23*0.049^2 = 5.24*10^-4+212.6*10^-4 = 0.00163 Kg m^2

I2 = 0.000524 KG M^2

W2 = I1W1/I2 = 0.00163*32/0.000524 = 99.5 REV/MIN

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