Question

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 $Q$. If the speed of the outermost particle is $V_o$, then the ratio of tensions, $T_1$:$T_2$:$T_3$, in the three sections of the string is (Assume that the string remains straight).

• A. $3:5:6$
• B. $3:4:5$
• C. $7:11:6$
• D. $6:5:3$

Answer 1

The correct option is A $3:5:6$

Let angular velocity of system is $w$.

For ball $c$,

$v_0=3wL$

$w=\frac{v_0}{3L}$

For ball $A$, force due to rotation,

$F_A=m{w}^{2}L=\frac{m{v}_0}{9L}$

Similarly,

$F_B=\frac{2m{v}_0}{9L}$

$F_C=\frac{m{v}_0}{3L}$

So, tension in the string joining $B$ and $C$ is $\frac{m{v}_0}{3L}$

$T_1=\frac{m{v}_0}{3L}$

$T_2=\frac{2m{v}_0}{9L}+\frac{m{v}_0}{3L}=\frac{5m{v}_0}{9L}$

So,

$T_1:T_2:T_3$ will be

$\frac{1}{3}:\frac{5}{9}:\frac{2}{3}$

$⟹3:5:6$