Solution:
T c = mω²(3l)= 3mω²
T B = mω²(2l) + T c = 5mω²l
T A = mω²(l) + T B = 6mω²l
So,
T C : T B : T A = 3:5:6
Three identical particles are joined together by a thread as shown in figure. All the three particles are moving in a horizontal plane. If the velocity of the outermost particle is v0, then the ratio of tensions in the three sections of the string is :
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