
Two particles of equal mass are connected to a rope AB of negligible mass, such that one is at end A and other dividing the length of the rope in the ratio $1:2$ from A. The rope is rotated about end B in a horizontal plane. Ratio of the tensions in the smaller part to the other is (ignore effect of gravity)
A. $4:3$
B. $1:4$
C. $1:2$
D. $1:3$
Answer
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Hint: In order to solve this question we need to understand Newton’s laws of motion and pseudo forces. So according to Newton’s third law of motion, it states that for every action there is equal and opposite reaction, so if mass is rotated in a circle then it creates tension in rope so for this action there must be some reaction and that is why a centripetal force acts in opposite direction to tension, Centripetal force is a pseudo force which only acts to conserve Newton’s universal law for all inertial frames moving with non-relativistic velocity.
Complete step by step answer:
For outer mass $m$, let tension be ${T_1}$ and it rotates in radius (shorter length) $\dfrac{L}{3}$
Also let the speed of mass be $v$. So Upon rotation centripetal force acts which is given by, $F = \dfrac{{m{v^2}}}{r}$
$\Rightarrow F = \dfrac{{m{v^2}}}{{\dfrac{L}{3}}}$
$\Rightarrow F = \dfrac{{3m{v^2}}}{L}$
So in case of equilibrium, ${T_1} = \dfrac{{3m{v^2}}}{L}$
For the inner part of length $\dfrac{{2L}}{3}$ and mass $2m$ let the tension be, ${T_2}$.
So tension in inner part is, ${T_2} = {T_1} + {F_2}$
Where, ${F_2}$ is centripetal force due to mass $2m$ and length $\dfrac{{2L}}{3}$ is given by, ${F_2} = \dfrac{{2m{v^2}}}{{\dfrac{{2L}}{3}}}$
${F_2} = \dfrac{{3m{v^2}}}{L}$
Putting values we get,
${T_2} = \dfrac{{3m{v^2}}}{L} + \dfrac{{3m{v^2}}}{L}$
$\Rightarrow {T_2} = \dfrac{{6m{v^2}}}{L}$
$\Rightarrow \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{\dfrac{{3m{v^2}}}{L}}}{{\dfrac{{6m{v^2}}}{L}}}$
$\therefore \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{1}{2}$
So the correct option is C.
Note: It should be remembered that velocity with which masses are rotating is non relativistic so it does not involve relativistic properties like time dilation and length contraction. Also centripetal force is a pseudo force which acts in the opposite direction of real force and with the same acceleration, but it only appears in a rotating frame of reference.
Complete step by step answer:
For outer mass $m$, let tension be ${T_1}$ and it rotates in radius (shorter length) $\dfrac{L}{3}$
Also let the speed of mass be $v$. So Upon rotation centripetal force acts which is given by, $F = \dfrac{{m{v^2}}}{r}$
$\Rightarrow F = \dfrac{{m{v^2}}}{{\dfrac{L}{3}}}$
$\Rightarrow F = \dfrac{{3m{v^2}}}{L}$
So in case of equilibrium, ${T_1} = \dfrac{{3m{v^2}}}{L}$
For the inner part of length $\dfrac{{2L}}{3}$ and mass $2m$ let the tension be, ${T_2}$.
So tension in inner part is, ${T_2} = {T_1} + {F_2}$
Where, ${F_2}$ is centripetal force due to mass $2m$ and length $\dfrac{{2L}}{3}$ is given by, ${F_2} = \dfrac{{2m{v^2}}}{{\dfrac{{2L}}{3}}}$
${F_2} = \dfrac{{3m{v^2}}}{L}$
Putting values we get,
${T_2} = \dfrac{{3m{v^2}}}{L} + \dfrac{{3m{v^2}}}{L}$
$\Rightarrow {T_2} = \dfrac{{6m{v^2}}}{L}$
$\Rightarrow \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{\dfrac{{3m{v^2}}}{L}}}{{\dfrac{{6m{v^2}}}{L}}}$
$\therefore \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{1}{2}$
So the correct option is C.
Note: It should be remembered that velocity with which masses are rotating is non relativistic so it does not involve relativistic properties like time dilation and length contraction. Also centripetal force is a pseudo force which acts in the opposite direction of real force and with the same acceleration, but it only appears in a rotating frame of reference.
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