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Centripetal Acceleration

Three balls o...

Question

Three balls of masses $1 k g$ , $2 k g$ and $3 k g$ are attached to three strings of lengths $2 m$ each as shown in figure. They are rotated in a horizontal circle with angular speed $ω = 2 r a d / s$ about point 'O'. Then the value of $T_{1} + T_{2} - 3 T_{3}$ is

8 N

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0 N

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32 N

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72 N

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Solution

The correct option is B $0 N$
Given, $m_{A} = 1 k g; m_{B} = 2 k g; m_{C} = 3 k g$
$r_{A} = length of m_{A} from 'O' = 2 m$
$r_{B} = length of m_{B} from 'O' = 4 m$
$r_{C} = length of m_{C} from 'O' = 6 m$

FBD: Only horizontal forces are shown because all the bodies are moving in horizontal plane, so $a_{y} = 0$ and vertical forces are balanced.

Applying equation of circular motion along radial direction, for all three bodies i.e $m_{C}, m_{B}, m_{A}$ respectively gives:

$T_{3} = m_{C} r_{C} ω^{2} = 3 \times 6 \times (2)^{2} = 18 \times 4 = 72 N - (1)$
$T_{2} - T_{3} = m_{B} r_{B} ω^{2} = 2 \times 4 \times (2)^{2} = 32 N - (2)$
$T_{1} - T_{2} = m_{A} r_{A} ω^{2} = 1 \times 2 \times (2)^{2} = 8 N - (3)$

Adding equations (1) and (2):
$T_{2} = 104 N$

Putting in equation (3):
$T_{1} = 8 + T_{2} = 112 N$

Therefore, $T_{1} + T_{2} - 3 T_{3} = 112 + 104 - 3 (72) = 0 N$

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