1. In the figure below, a small disk of radius r = 2.00 cm has been glued to the
ID: 2245040 • Letter: 1
Question
1.In the figure below, a small disk of radius r = 2.00 cm has been glued to the edge of a larger disk of radius R = 4.00 cm so that the disks lie in the same plane. The disks can be rotated around a perpendicular axis through point O at the center of the larger disk. The disks both have a uniform density (mass per unit volume) of 1.50 ? 103 kg/m3 and a uniform thickness of5.20 mm. What is the rotational inertia of the two-disk assembly about the rotation axis through O?
In the figure below, a small disk of radius r = 2.00 cm has been glued to the edge of a larger disk of radius R = 4.00 cm so that the disks lie in the same plane. The disks can be rotated around a perpendicular axis through point O at the center of the larger disk. The disks both have a uniform density (mass per unit volume) of 1.50 ? 103 kg/m3 and a uniform thickness of5.20 mm. What is the rotational inertia of the two-disk assembly about the rotation axis through O? The thin uniform rod in the figure below has length 1.0 m and can pivot about a horizontal, frictionless pin through one end. It is released from rest at angle ? = 40 degree above the horizontal. Use the principle of conservation of energy to determine the angular speed of the rod as it passes through the horizontal position.Explanation / Answer
mass of small disk = volume * density = pi*r^2 * 0.0052 * 1500 = pi * 0.02^2 *0.0052 * 1500 = 0.009801769 kg
inertia of small disk through it's center = 0.5 * m_small * r^2 = 0.5 * 0.009801769 * 0.02^2 = 0.0000019603538
so.. according to parallel axix theorem...
inertia of small disk through the center of large disk = inertia of small disk through it's ccenter + m_small * R_large^2 = 0.0000019603538 + ( 0.009801769*0.04^2 ) = 0.0000176431842
mass of large disk = volume * density = pi*r^2 * 0.0052 * 1500 = pi * 0.04^2 *0.0052 * 1500 = 0.039207076 kg
inertia of small disk through it's center = 0.5 * m_slarge * r^2 = 0.5 * 0.039207076 * 0.04^2 = 0.0000313656608
so.. total inertia of the ssytem through the center of large disk = 0.0000176431842 + 0.0000313656608
= 0.000049008845
2)
inertia of rod = (1/3)*m*l^2 = (1/3)* m * 1 = m/3
let the angular speed be w ...
change in potential energy = m*g*(1/2)*sin 40 = 3.149659 * m
final kinetic energy = 0.5 * m/3 * w^2 = 3.149659 * m
so.. w = 4.3471779 rad / sec
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