Circular MotionHard
Question
Three identical particles are joined together by a thread as shown in figure. All the three particles are moving on a smooth horizontal plane about point O. If the speed of the outermost particle is v0, then the ratio of tensions in the three sections of the string is :
(Assume that the string remains straight)
(Assume that the string remains straight)
Options
A.3 : 5 : 7
B.3 : 4 : 5
C.7 : 11 : 6
D.3 : 5 : 6
Solution
ω = const., for all three particles
ω =
TC = mω2 3l
TB – TC = mω2 2l
TB = 5 mω2l
TA – TB = mω2l
TA = 6 mω2l
TC : TB : TA : : 3 : 5 : 6 Ans.
ω =
TC = mω2 3l
TB – TC = mω2 2l
TB = 5 mω2l
TA – TB = mω2l
TA = 6 mω2l
TC : TB : TA : : 3 : 5 : 6 Ans.
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