Question

A 0.16 meter long, 0.15 kg thin rigid rod has a small 0.22 kg mass stuck on one of its ends and a small 0.080 kg mass stuck on the other end. The rod rotates at 1.7 rad/s through its physical center without friction. What is the magnitude of the angular momentum of the system taking the center of the rod as the origin? Treat the masses on the ends as point masses.

Answer 1

Answer:

The angular momentum of the system is 0.00381 kg*m^2/s

Explanation:

The moment of inertia of rod is:

$I=I_{stick} +I_{small1} +I_{small2} \\I=\frac{m_{1}L_{1}^{2} }{12} +m_{2} r_{2}^{2} +m_{3} r_{3}^{2} \\I=\frac{0.15*0.16^{2} }{12} +(0.22*0.08^{2} )+(0.08*0.08^{2} )\\I=0.00224 kgm^{2}$

The angular momentum is:

$L=Iw\\L=0.00224*1.7\\L=0.00381 kgm^{2} /s$