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    OB is a light rigid rod of length 5l. Particles of masses 3m and 4m are attached at points A and B of the rod, where OA is 2l. The rod is made to rotate about O in a horizontal plane with constant angular speed omega. Calculate the ratio of the tensions in OA and AB.
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    Is the answer Ta:Tb = 13:10 ?
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    (Original post by Fermat)
    Is the answer Ta:Tb = 13:10 ?
    yes it is.
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    Ok, working coming up.
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    I can't do any sketches at the moment, so I'll try to describe it.

    The tension in the rod along AB is Tb.
    At B, Tb acts in the dirn AB.
    At A, Tb acts in the dirn BA.

    In the section of the rod OA, OA has a tension Ta.
    At A, Ta acts in the dirn AO.
    So, at the point A, There are two tensions,
    #
    Ta acting in the dirn AO, and Tb acting in the dirm AB.

    The rod section AB
    Tb = (4m).(w²).(5l)
    Tb = 20mw²l = 20k
    ==============

    The rod section OA
    Ta - Tb = (3m).(w²).(2l)
    Ta - Tb = 6mw²l = 6k
    ===============

    Tb = 20k
    Ta - Tb = 6k

    Ta = 26k
    Ta:Tb = 26k: 20k = 13:10
    ===================

    Edit: To clarify, There are two tensions acting on the mass at A. The tension Tb acting away from A in the dirn AB, and the tension Ta acting in the dirn AO. So the net tension acting on the mass At A is Ta - Tb.
 
 
 
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