Question

A 1.2-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 cm from the pivot. The block is spun at 40 rpm , 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?

Answer 1

Since angular momentum is always conserved, wecan set the initial angular momentum to the final angular momentum and can find the final angular velocity:



The moment of inertia (I) for the block (which may be considered as a point particle, in fact it must be considered as one because we are not given its dimensions) is mr². There are two values for I, one where the block is a distance of 20cm (0.20m) from the pivot, and one that is 1.2m from the pivot. Therefore:


------------------>mass divides out
= (0.20m)²(40rev/mim) / (1.2m)²
= 1.11 rev/min

If this should be in SI units, then:

1.1rpm = 1.1rev/min(2πrad/1.0rev)(1.0min/60s)
= 0.115rad/s