A solid disk of 5kg and radius 0.5m turns at the rate of 10rpm. You then carefully place a 2kg book at the edge of table as it spins. What is the new rotation rate of the table(in rpm)?
I1 =MR²/2 =2•0.5²/2 =0.25 kg•m².
Assume the book is the3 point mass
I2=mR² =2•0.5²=0.5 kg•m².
The law of conservation of angular momentum
L1 = L2,
I1• ω1 =(I1+I2) •ω2,
I1• 2•π•f 1 =(I1+I2) • 2•π•f 2
f2= I1• f1/(I1+I2)=
=0.25•10/(0.25+0.5)=3.33 rpm