a 1.5 m long massless rod l Verizon LTE 11:48 PM * 27% . s.com Problem 9.72 Mast

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a 1.5 m long massless rod l Verizon LTE 11:48 PM * 27% . s.com Problem 9.72 MasteringPhysics 4 of 15 A 1.5-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 I0 cm from the pivot. The block is spun at 50 rpm, then the temperature of the rod is slowly increased. When the wax melts, the block slides out to the end of the rod. Part A What is the final angular velocity? Give your answer in rpm Express your answer using two significant figures. 0.267 m 1 2 3 4 56 7 8 9 0 Submit Request Answer

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 10cm (0.10m) from the pivot, and one that is 1.5m from the pivot. Therefore:

mr?²?(i) = mr?²?(f)

?(f) = r?²?(i) / r?² ----------------->mass cancel out both sides

= (0.10m)² (50rev/mim) / (1.5m)²

?(f) = 0.222 rev/min

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