Question

Two beads A and B of equal mass $m$ are connected by a light rod, which acts as a chord for the ring. They are constrained to move on the frictionless ring in a vertical plane. The beads are released from rest as shown in figure. The tension in the chord just after release will be:

• A. $\frac{mg}{4}$
  
• B. $\sqrt{2}mg$
• C. $\frac{mg}{2}$
  
• D. $\frac{mg}{\sqrt{2}}$

Answer 1

The correct option is D $\frac{mg}{\sqrt{2}}$

Just after the release, bead B moves vertically downwards and bead A moves leftwards horizontally with the same acceleration i.e $a$


From FBD of A, applying equation of circular motion along tangential direction (i.e direction of acceleration)
$T\cos {45}^{\circ }=ma-\left(1\right)$
i.e $\frac{T}{\sqrt{2}}=ma$

Similarly for bead B:


$mg-T\cos {45}^{\circ }=ma-\left(2\right)$

Substituting from equation (1) in equation (2),
$mg-ma=ma$
or $a=\frac{g}{2}-\left(3\right)$
Substituting Eq.(3) in Eq. (1), we get:

$T=ma\sqrt{2}=\frac{mg}{\sqrt{2}}$