posted by Anonymous on .
A 1.4-m-long massless rod is pivoted at one end and swings around in a circle on a frictionless table. A block with a hole through the center can slide in and out along the rod. Initially, a small piece of wax holds the block 20 from the pivot. The block is spun at 50 , then the temperature of the rod is slowly increased. When the wax melts, the block slides out to the end of the rod. What is the final angular velocity?
The moment of inertia (I) for the block (the point mass, as we are not given its dimensions) is mr^2:
I1 =m•r1^2 =m •0.2^2 =0.04•m,
I2 = m•r1^2 =m •1.4^2=1.96•m,
Acording to the law of conservation of angular momentum
L1 = L2
I1•ω1 = I2•ω2,
ω2 = I1•ω1/I2 =0.04•m•50/1.96•m =