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?

1 answer

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 =
=1.02 (units?)