Solution to Question 6

Problem Analysis:

A ring of mass $m$ slides on a cord fixed at A and passing over a peg B to a load of mass $m$. The ring is released from rest at the level of the pegs. We need to find the ring’s acceleration when it passes its equilibrium position.

Step 1: Finding the Equilibrium Position

Let $y$ be the depth of the ring. The potential energy of the system $U(y)$ involves the ring dropping and the load rising. The distance from peg to ring is $x = \sqrt{b^2 + y^2}$. Total string length used in the ‘V’ is $2x$. The load rises by $(2x – 2b)$.

$$ U(y) = -mgy + mg(2\sqrt{b^2+y^2}) $$

For equilibrium, $dU/dy = 0$:

$$ -mg + mg \frac{2y}{\sqrt{b^2+y^2}} = 0 \implies \frac{y}{\sqrt{b^2+y^2}} = \frac{1}{2} $$

Since $\cos\theta = \frac{y}{\sqrt{b^2+y^2}}$ (where $\theta$ is the angle with the vertical), we have:

$$ \cos\theta = \frac{1}{2} \implies \theta = 60^\circ $$

At this position, $\sin\theta = \frac{\sqrt{3}}{2}$ and the length of the cord segment $x = \frac{b}{\sin\theta} = \frac{2b}{\sqrt{3}}$.

Step 2: Velocity at Equilibrium (Energy Conservation)

The ring starts at $y=0$ and falls to equilibrium.
Loss in PE of Ring = $mg y_{eq} = mg (x \cos 60^\circ) = mg (b/\sqrt{3})$.
Gain in PE of Load = $mg(2x – 2b) = 2mg b(\frac{2}{\sqrt{3}} – 1)$.

Using Conservation of Energy ($\Delta K + \Delta U = 0$):

$$ \frac{1}{2}mv^2 + \frac{1}{2}mv_{load}^2 = \Delta PE_{ring} – \Delta PE_{load} $$

Constraint: $v_{load} = 2v \cos\theta$. At $60^\circ$, $v_{load} = 2v(1/2) = v$. Thus:

$$ mv^2 = mg \frac{b}{\sqrt{3}} – 2mg b \left(\frac{2}{\sqrt{3}} – 1\right) $$ $$ v^2 = gb \left[ \frac{1}{\sqrt{3}} – \frac{4}{\sqrt{3}} + 2 \right] = gb \left( 2 – \sqrt{3} \right) $$

Step 3: Acceleration Analysis

We apply Newton’s Laws at the equilibrium position.
Ring: $mg – 2T\cos 60^\circ = ma \implies mg – T = ma$.
Load: $T – mg = m a_{load}$.

Combining these eliminates $T$:

$$ (mg – ma) – mg = m a_{load} \implies a_{load} = -a $$

Kinematic Constraint (Acceleration):
Differentiating $v_{load} = 2v \cos\theta$:

$$ a_{load} = 2 [ a \cos\theta + v (-\sin\theta)\dot{\theta} ] $$

Using $\dot{\theta} = -\frac{v \sin\theta}{x}$ (derived from geometry for horizontal motion relative to vertical): Wait, strictly $\dot{\theta} = -\frac{b v}{x^2}$.

$$ a_{load} = 2 \left( a(1/2) + v(-\sin\theta)\left(-\frac{bv}{x^2}\right) \right) = a + \frac{2b v^2 \sin\theta}{x^2} $$

Substituting values ($x = 2b/\sqrt{3}$, $\sin\theta = \sqrt{3}/2$):

$$ \frac{2b \sin\theta}{x^2} = \frac{2b (\sqrt{3}/2)}{4b^2/3} = \frac{b\sqrt{3}}{4b^2/3} = \frac{3\sqrt{3}}{4b} $$

So, $a_{load} = a + \frac{3\sqrt{3}}{4b} v^2$.

Step 4: Solving for Acceleration

Substitute $a_{load} = -a$:

$$ -a = a + \frac{3\sqrt{3}}{4b} v^2 $$ $$ -2a = \frac{3\sqrt{3}}{4b} [ gb(2-\sqrt{3}) ] $$ $$ -2a = \frac{3\sqrt{3}g}{4} (2 – \sqrt{3}) = \frac{g}{4} (6\sqrt{3} – 9) $$ $$ a = -\frac{3g}{8} (2\sqrt{3} – 3) $$