Posted by **Anonymous** on Saturday, April 28, 2012 at 4:00am.

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?

- physics -
**Elena**, Saturday, April 28, 2012 at 4:25am
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?)

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