Question

A table with a smooth horizonal surface is placed in a cabin which moves in a horizontal circle of a large radius $R$. A smooth pulley of small radius is fixed to the table. Two masses $m$ and $2m$ placed on the table are connected through a light string going over the pulley. System is released from rest with respect to the cabin. The initial tension in the string will be:

• A. $\frac{1}{3}m{\omega }^{2}R$
• B. $\frac{4}{3}m{\omega }^{2}R$
• C. $\frac{2}{3}m{\omega }^{2}R$
• D. $m{\omega }^{2}R$

Answer 1

The correct option is B $\frac{4}{3}m{\omega }^{2}R$

Let mass of block $A=m$, mass of block $B=2m$.
Direction of centrifugal force ($F\&F^{\prime }$) will be radially outward on both the blocks, since acceleration of rotating frame $(a={\omega }^{2}R)$ is radially inwards towards centre of circle.
$\Rightarrow F=m{\omega }^{2}R$ on block $A$
and $F^{\prime }=2m{\omega }^{2}R$ on block $B$

$R=$ radius of horizontal circle in which cabin is rotating.
Blocks will be moving with same magnitude of linear acceleration $\left(a\right)$ relative to rotating frame, due to string constraint.
FBD of blocks:


$\because$ Radius of circle is very large, hence length of string and radius of pulley can be neglected. Radius of each mass can be taken to be equal to $R$, from centre of horizontal circle.

FBD of block $A$:


$T-m{\omega }^{2}R=ma...\left(i\right)$
FBD of block $B$:


$2m{\omega }^{2}R-T=2ma...\left(ii\right)$
On adding Eq. $\left(i\right)\&\left(ii\right)$
$m{\omega }^{2}R=3ma$
$\Rightarrow a=\frac{{\omega }^{2}R}{3}....\left(iii\right)$

Substituting in Eq $\left(i\right)$,
$T-m{\omega }^{2}R=\frac{m{\omega }^{2}R}{3}$
$\Rightarrow T=\frac{m{\omega }^{2}R}{3}+m{\omega }^{2}R$
$\therefore T=\frac{4}{3}m{\omega }^{2}R$
So, initial tension in the string is $\frac{4}{3}m{\omega }^{2}R$