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Q. 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 $v_{0}$, then the ratio of tensions in the three sections of the string is
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Laws of Motion

Solution:

Let $\omega$ is the angular speed of revolution
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$T_{3}=m\omega^{3}3l$
$T_{2}-T_{3}=m \omega^{2}2l$
$\Rightarrow T_{2}=m\omega^{2}5l$
$T_{1}-T_{2}=m\omega^{2}2l$
$\Rightarrow T_{1}=m\,\omega^{2}6l$
$T_{3}: T_{2}: T_{1}=3:5:6$