Question

A DVD (radius 6.0 cm) is spinning freely with an angular velocity of 1150 rpm when a bug drops onto and sticks to the DVD a distance 4.4 cm from the center. If the DVD slows to 900 rpm, what is the ratio of the bug's mass to the DVD's mass? (Ignore the effect of the hole in the center of the DVD.) mbug mOVD

Answer 1

let mass of dvd be m_D and that of bug be m_B

The problem can be solved using conservation of angular momentum

initial angular velocity = 1150*2*pi/60=120.4 rad/s

final angular velocity = 900*2*pi/60 = 94.25 rad/s

moment of inertia of disc = 0.5*m_D*0.06² = 0.0018m_D

moment of inertia of bug = m_B*0.044² = 0.0019m_B

initial angular momentum =0.0018m_D* 120.4 = 0.2167m_D

final angular momentum = (0.0018m_D+0.0019m_B)*94.25= = 0.1696m_D+0.1791m_B

From conservation of angular momentum,

0.2167m_D =  0.1696m_D+0.1791m_B

0.0471m_D = 0.1791m_B

m_B/m_D = 0.0471/0.1791=0.26