Consider a slender rod of mass 3.45×102 kg and length 0.410 m (of uniform shape
ID: 1463033 • Letter: C
Question
Consider a slender rod of mass 3.45×102 kg and length 0.410 m (of uniform shape and density) that is rotating in a horizontal plane about a fixed axis perpendicular to the rod and passing through the center of the rod. The axis is frictionless. Two disks, each of mass 0.180 kg , and radius of 0.083 m are mounted on the rod so that they can slide freely along the rod. You are initially holding them at rest at a distance 5.10×102 m on each side from the center of the rod and the system is set to rotate at an angular velocity 35.0 rev/min . Without otherwise changing the system, you let them go, and the disks being to slide outward along the rod and fly off at the ends. You should know that the moment of inertia of a rod, about an axis passing perpendicular to the rod, and for one passing through its center and the other through one end are 112ML2 and 13ML2 respectively; and that for a disk with an axis passing through its center but for an axis in the plane of the disk and for an axis perpendicular to the plane of the disk are 14MR2 and 12MR2 respectively.
What is the angular speed of the system at the instant when the disks reach the ends of the rod?
Explanation / Answer
I1 = (1/12)(mrod)L2 + 2(mdisc)r2
where, L is the length of the rod; r is the distance where the disc is placed from the center of the rod.
I1 = [0.0833*3.45*10-2*0.412]+[2*0.18*(5.1*10-2)2] = 4.83*10-4+9.36*10-4
I1 = 1.42*10-3
I1 = 0.0014
I2= (1/12)(mrod)L2+ 2(mdisc)(L/2)2
I2 = [0.0833*3.45*10-2*0.412]+[2*0.18*0.2052] = 4.83*10-4+0.01513
I2 = 0.0156
Using conservation of angular momentum;
2 = 1(I1/I2) = 35.0*[0.0014 / 0.0156]
2 = 3.1410 rev/min