Question

Three balls of masses $1kg$, $2kg$ and $3kg$ are attached to three strings of lengths $2m$ each as shown in figure. They are rotated in a horizontal circle with angular speed $\omega =2rad/s$ about point 'O'. Then the value of $T_1+T_2-3T_3$ is

• A. $8N$
• B. $0N$
• C. $32N$
• D. $72N$

Answer 1

The correct option is B $0N$

Given, $m_A=1kg;m_B=2kg;m_C=3kg$
$r_A=\text{length of}{m}_A\text{from 'O'}=2m$
$r_B=\text{length of}{m}_B\text{from 'O'}=4m$
$r_C=\text{length of}{m}_C\text{from 'O'}=6m$

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:

$T_3=m_C{r}_C{\omega }^2=3\times 6\times (2{)}^2=18\times 4=72N-(1)$
$T_2-T_3=m_B{r}_B{\omega }^2=2\times 4\times (2{)}^2=32N-(2)$
$T_1-T_2=m_A{r}_A{\omega }^2=1\times 2\times (2{)}^2=8N-(3)$

Adding equations (1) and (2):
$T_2=104N$

Putting in equation (3):
$T_1=8+T_2=112N$

Therefore, $T_1+T_2-3T_3=112+104-3\left(72\right)=0N$