A 1.4-m-long massless rod is pivoted at one end and swings around in a circle on

Question

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 60 cm from the pivot. The block is spun at 60 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?

Part A

Give your answer in rpm.

Express your answer using two significant figures.

Explanation / Answer

Since angular momentum is always conserved, you can set the initial angular momentum to the final angular momentum and solve for the final angular velocity:

L(i) = L(f)
I(i) = I(f)

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 60cm (0.60m) from the pivot, and one that is 1.4m from the pivot. Therefore:

mr²(i) = mr²(f)
(f) = r²(i) / r²------------------->mass divides out
= (0.60m)²(60rev/mim) / (1.4m)²
= 11.02 rev/min

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