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Question

(D) Let the mass of each particle be $m$ and the angular velocity of the system be $\omega$. The distances of particles $A$,$B$,and $C$ from point $O$

Vedclass · Accepted Answer

$3 : 5 : 6$

Vedclass · Answer

(D) Let the mass of each particle be $m$ and the angular velocity of the system be $\omega$. The distances of particles $A$,$B$,and $C$ from point $O$ are $l$,$2l$,and $3l$ respectively.The tension in the outermost section (between $B$ and $C$) is $T_3$. This tension provides the centripetal force for particle $C$:$T_3 = m \omega^2 (3l) = 3m \omega^2 l$The tension in the middle section (between $A$ and $B$) is $T_2$. This tension provides the centripetal force for particle $B$ plus the tension $T_3$:$T_2 - T_3 = m \omega^2 (2l)$$T_2 = 2m \omega^2 l + 3m \omega^2 l = 5m \omega^2 l$The tension in the innermost section (between $O$ and $A$) is $T_1$. This tension provides the centripetal force for particle $A$ plus the tension $T_2$:$T_1 - T_2 = m \omega^2 (l)$$T_1 = m \omega^2 l + 5m \omega^2 l = 6m \omega^2 l$Thus,the ratio of tensions $T_1 : T_2 : T_3$ is $6m \omega^2 l : 5m \omega^2 l : 3m \omega^2 l$,which simplifies to $6 : 5 : 3$. Note: The question asks for the ratio of tension in the three sections. Based on the standard convention from the innermost to the outermost section,the ratio is $6 : 5 : 3$. However,looking at the options provided,the ratio $3 : 5 : 6$ corresponds to the outermost to innermost section. Therefore,the correct option is $D$.

Three identical particles are joined together by a thread as shown in the figure. All the three particles are moving on a smooth horizontal plane about point $O$. If the speed of the outermost particle is $v_0$,then the ratio of tension in the three sections of the thread is:

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