Question

FE4 Two identical uniform thin rigid rods are held together at one end with an electromagnet. One of the rods is attached to a fixed pivot point, around which the magnetic two rods are rotating at a constant rate (there is no gravity present here). Suddenly the connector electromagnet is tumed off, and the rod not attached to the pivot flies away. Find the percentage change in kinetic energy of the system before and after this separation occurs. Erroneous Solution: Let's call the mass and length of each rod m and I, respectively, and the initial rotational speed es. Then the initial kinetic energy of the system is: fixed KE - 5102 - {}(21)(21) ]co?=$ ml?co? The end of the pivoted rod that is not at the pivot is moving at a linear speed of: v=10 When the other rod is released, it is moving at this speed, so its final kinetic energy is: KE = m2 = 13 muc? Releasing the other rod doesny change the rotational speed of the pivoted rod, so its final kinetic energy is: The ratio of the final kinetic energy of the system to the initial kinetic energy is: KE, _KE,, + REgg 5mlv++6ml?cs?, KE KE 4mico? The system loses half its energy.

find the error in the calculation

Answer 1

Solution -

The error is that you did not took the conservation of angular momentum principle into the account after the electromagnet has been turned off and the other rod has been released so in this case the angular velocity of the pivoted end rod would change once you equate the conservation of angular momentum for initial and final conditions .

Here in solution you assumed the that releasing the other rod would not change the angular velocity and in the calculation you have kept it as it is and this made your answer wrong.

Apply the conservation of the momentum principle and then calculate the angular velocity of pivoted rod in final situation when other rod has separated off the use that value in final calculation you will get the desired results .

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