Name: Variable Mass System
Uploaded: 2019-07-19
Duration: 12 min 25 s
Description: In this video, we will look at how the laws of motion and their mathematical formulation apply to a variable mass system like a rocket. Don't miss it.

Question

A small ring $A$ of mass $m = 1 / 2 kg$ is attached at an end of a light string the other end of which is tied to a block $B$ of mass $2 m .$ The ring is free to move on a fixed smooth horizontal rod as shown. Find tension in the string when it becomes vertical.

Accepted Answer

Given,
Mass of ring is, $m = 1 / 2 kg$
Mass of block is, $B$

Let us assume,
Velocity of ring in vertical position is, $v_{r}$
Velocity of block in vertical position is, $v_{b}$
Tension in string at vertical position is, $T$

Since net force on the system in horizontal direction is zero, so linear momentum remains conserved in horizontal direction.
When string is in vertical position, it is represented as below figure,

Initial momentum of the system is, $P_{i} = 0$
Final momentum is, $P_{f} = m v_{r} - 2 m v_{b}$
From conservation of momentum,

$P_{i} = P_{f}$
$\Rightarrow 0 = m v_{r} - 2 m v_{b}$
$\Rightarrow 0 = \frac{1}{2} \times v_{r} - 2 \times \frac{1}{2} \times v_{b}$
$\Rightarrow v_{r} = 2 v_{b} \dots (1)$

Since there is not any external force and friction, we can apply conservation of mechanical energy, diagram showing potential and kinetic energy is shown in below figure,

Let us consider the datum line at level of wire as shown in figure, so initially system has zero potential energy and zero kinetic energy.
From mechanical energy conservation, we have
$U_{i} + K_{i} = U_{f} + K_{f}$
$\Rightarrow 0 = - 2 m g l + \frac{1}{2} m {v_{r}}^{2} + \frac{1}{2} 2 m {v_{b}}^{2}$
$\Rightarrow \frac{1}{2} m {v_{r}}^{2} + \frac{1}{2} 2 m {v_{b}}^{2} = 2 m g l$
$\Rightarrow {v_{r}}^{2} + 2 {v_{b}}^{2} = 4 g l \dots (2)$

Using $(1)$ in $(2)$ , we have,
$\Rightarrow {(2 v_{b})}^{2} + 2 {v_{b}}^{2} = 4 g l$
$\Rightarrow 4 {v_{b}}^{2} + 2 {v_{b}}^{2} = 4 g l$
$\Rightarrow v_{b} = \sqrt{\frac{2}{3} g l}$
Using $(1)$ , $v_{r} = 2 \times \sqrt{\frac{2}{3} g l} = \sqrt{\frac{8}{3} g l}$

To find the tension in string, we have to just find velocity of block with respect to ring to find the centripetal force.
The case is shown in below figure,
Velocity of block with respect to ring is, $\vec{v_{b, r}} = \vec{v_{b}} - \vec{v_{r}}$
$\Rightarrow v_{b, r} = - \sqrt{\frac{2}{3} g l} - \sqrt{\frac{8}{3} g l} = - \sqrt{g l} (\frac{(\sqrt{2} + \sqrt{8})}{\sqrt{3}}) \dots (3)$ shown in below figure,

As the motion of block performs vertical circular motion, so applying Newton's second law at the bottom most point, we have
$\Rightarrow T - 2 m g = \frac{2 m {v_{b, r}}^{2}}{l}$
$\Rightarrow T - 2 m g = \frac{2 m {(\sqrt{g l} (\frac{(\sqrt{2} + \sqrt{8})}{\sqrt{3}}))}^{2}}{l}$
$\Rightarrow T - 2 m g = \frac{2 m g {(\sqrt{2} + \sqrt{8})}^{2}}{3}$
$\Rightarrow T - 2 m g = \frac{2 m g \times 18}{3}$
$\Rightarrow T - 2 m g = 12 m g$
$\Rightarrow T = 14 m g$
$\Rightarrow T = 14 \times \frac{1}{2} \times 10 = 70 N$

A small ring $A$ of mass $m = 1 / 2 kg$ is attached at an end of a light string the other end of which is tied to a block $B$ of mass $2 m .$ The ring is free to move on a fixed smooth horizontal rod as shown. Find tension in the string when it becomes vertical.

Important Questions on Centre of Mass, Momentum and Collisions

Learn and Prepare for any exam you want !

A small ring A of mass m=1/2 kg is attached at an end of a light string the other end of which is tied to a block B of mass 2m. The ring is free to move on a fixed smooth horizontal rod as shown. Find tension in the string when it becomes vertical.

Important Questions on Centre of Mass, Momentum and Collisions

Learn and Prepare for any exam you want !

A small ring $A$ of mass $m = 1 / 2 kg$ is attached at an end of a light string the other end of which is tied to a block $B$ of mass $2 m .$ The ring is free to move on a fixed smooth horizontal rod as shown. Find tension in the string when it becomes vertical.