Question

A light inextensible cord tied at on one end to a peg A on a wall, passes over a frictionless peg B fixed in level with the peg A and supports a load at the other end. Between the pegs, the cord is wearing a small ring of mass equal to that of the load. Initially the ring is held in level with the pegs as shown in the figure and then released. Determine acceleration of the ring when it passes an equilibrium position. Acceleration of free fall is $g$.

Answer 1

$\frac{g(15-6\sqrt{3})}{8}$

Answer 2

To determine the acceleration of the ring when it passes an equilibrium position, we first need to identify the equilibrium position and then apply Newton's second law considering the dynamics of the system.

Let $m$ be the mass of the ring and also the mass of the load.
Let $2d$ be the horizontal distance between the pegs A and B. We can set peg A at $(-d, 0)$ and peg B at $(d, 0)$.
Let the ring be at position $(0, -y)$, where $y$ is the vertical distance below the line connecting pegs A and B.

1. Determine the tension in the cord:
The cord passes over a frictionless peg B and supports a load of mass $m$. Since the cord is light and inextensible and the peg is frictionless, the tension $T$ in the entire cord is uniform and equal to the weight of the load.
Therefore, $T = mg$.

2. Forces on the ring:
The ring is subjected to three forces:

• Its weight $mg$ acting vertically downwards.
• Tension $T$ from the cord segment connected to peg A.
• Tension $T$ from the cord segment connected to peg B.

Due to symmetry, the ring moves vertically downwards along the line $x = 0$. Let $\theta$ be the angle each segment of the cord makes with the horizontal.
The vertical component of tension from each side is $T \sin(\theta)$.
The horizontal components of tension $T \cos(\theta)$ from both sides are equal and opposite, thus cancelling each other out.

The net vertical force on the ring is $F_{net} = 2T \sin(\theta) - mg$.

3. Identify the equilibrium position:
An equilibrium position is where the net force on the ring is zero.
Setting $F_{net} = 0$:
$2T \sin(\theta) - mg = 0$
Substitute $T = mg$:
$2mg \sin(\theta) - mg = 0$
$2 \sin(\theta) = 1$
$\sin(\theta) = 1/2$
This implies $\theta = 30^\circ$.
This is the angle the cord segments make with the horizontal when the ring is at its equilibrium position.

4. Determine the acceleration at the equilibrium position:
The problem asks for the acceleration of the ring when it passes an equilibrium position. When an object passes through its equilibrium position in an oscillatory motion (like SHM), its net force is zero, but its velocity is maximum. Since $F_{net} = ma$, if $F_{net} = 0$ at this specific point, then the acceleration $a$ must also be zero.

However, in problems involving constrained motion, the acceleration might not be zero even if the net force in the direction of motion is zero, due to centripetal acceleration. Let's re-examine the forces and motion.

Let $y$ be the vertical displacement of the ring from the line AB. The ring moves along the y-axis.
The speed of the ring is $v_r = dy/dt$.
The length of the cord from A to the ring is $L_s = \sqrt{d^2 + y^2}$.
The relationship between the speed of the ring $v_r$ and the speed of the load $v_l$ can be found by differentiating the total cord length $L_{total} = 2L_s + L_{load}$ (where $L_{load}$ is the length of cord hanging vertically with the load).
$dL_{total}/dt = 0$
$2 * (1/2) * (2y / \sqrt{d^2 + y^2}) * (dy/dt) + dL_{load}/dt = 0$
$2 * (y / \sqrt{d^2 + y^2}) * v_r + v_l = 0$
Since $\sin(\theta) = y / \sqrt{d^2 + y^2}$, we have:
$2 \sin(\theta) * v_r + v_l = 0$
So, $v_l = -2 \sin(\theta) * v_r$. (The negative sign indicates that if the ring moves down, the load moves up).

Now, let's consider the acceleration. Differentiate with respect to time:
$a_l = d(v_l)/dt = -2 * [ (d(\sin(\theta))/dt) * v_r + \sin(\theta) * a_r ]$
We need $d(\sin(\theta))/dt$.
$\sin(\theta) = y / \sqrt{d^2 + y^2}$.
$d(\sin(\theta))/dt = d/dt (y / \sqrt{d^2 + y^2})$
$d(\sin(\theta))/dt = [ (dy/dt \sqrt{d^2 + y^2}) - (y * (1/2) * (2y / \sqrt{d^2 + y^2}) * dy/dt) ] / (d^2 + y^2)$
$d(\sin(\theta))/dt = [ v_r (d^2 + y^2) - y^2 v_r ] / (d^2 + y^2)^{3/2}$
$d(\sin(\theta))/dt = [ v_r d^2 ] / (d^2 + y^2)^{3/2}$
Since $\cos(\theta) = d / \sqrt{d^2 + y^2}$, we have $d^2 = (d^2 + y^2) \cos^2(\theta)$.
So, $d(\sin(\theta))/dt = [ v_r (d^2 + y^2) \cos^2(\theta) ] / (d^2 + y^2)^{3/2} = v_r \cos^2(\theta) / \sqrt{d^2 + y^2}$.
Substituting $\sqrt{d^2 + y^2} = y / \sin(\theta)$:
$d(\sin(\theta))/dt = v_r \cos^2(\theta) / (y / \sin(\theta)) = v_r \sin(\theta) \cos^2(\theta) / y$.
This is $v_r \cos(\theta) * (d/L_s) / L_s = v_r \cos(\theta) * \cos(\theta) / L_s = v_r \cos^2(\theta) / L_s$.
So, $a_l = -2 * [ (v_r^2 \cos^2(\theta) / L_s) + \sin(\theta) * a_r ]$.

Now, apply Newton's second law for the load and the ring.
For the load: $T - mg = m a_l$
Since $T = mg$, this gives $mg - mg = m a_l$, so $m a_l = 0$, which means $a_l = 0$.
This is a crucial point. The load's acceleration is zero at the instant the ring passes through the equilibrium position. This means the load is momentarily at rest or moving at a constant velocity. Given the system's dynamics, the load must be momentarily at rest, or rather, its vertical component of velocity changes sign.

Substitute $a_l = 0$ into the acceleration relationship:
$0 = -2 * [ (v_r^2 \cos^2(\theta) / L_s) + \sin(\theta) * a_r ]$
This implies:
$(v_r^2 \cos^2(\theta) / L_s) + \sin(\theta) * a_r = 0$
$\sin(\theta) * a_r = - v_r^2 \cos^2(\theta) / L_s$
$a_r = - (v_r^2 \cos^2(\theta)) / (L_s \sin(\theta))$

At the equilibrium position, $\theta = 30^\circ$.
$\sin(30^\circ) = 1/2$
$\cos(30^\circ) = \sqrt{3}/2$
$L_s = y / \sin(\theta) = y / (1/2) = 2y$. Also, $y = d \tan(30^\circ) = d/\sqrt{3}$. So $L_s = 2d/\sqrt{3}$.
$a_r = - (v_r^2 * (\sqrt{3}/2)^2) / ((2d/\sqrt{3}) * (1/2))$
$a_r = - (v_r^2 * 3/4) / (d/\sqrt{3})$
$a_r = - (3\sqrt{3} v_r^2) / (4d)$

This acceleration $a_r$ is the vertical acceleration of the ring. It is not zero. The initial assumption that $F_{net} = 0$ means $a=0$ is only true if the object is not constrained to move along a path where its velocity changes direction. Here, the ring is moving along a curved path (or rather, a straight vertical line, but its speed is changing). The $F_{net}$ calculation $2T \sin(\theta) - mg$ gives the net force along the vertical direction.

The tension $T$ is not simply $mg$ if the load is accelerating.
Let $T$ be the tension in the cord.
For the load: $mg - T = m a_l$ (taking downward as positive for the load).
For the ring (vertical motion): $2T \sin(\theta) - mg = m a_r$ (taking downward as positive for the ring).

We have $a_l = -2 * [ (v_r^2 \cos^2(\theta) / L_s) + \sin(\theta) * a_r ]$. (This $a_l$ is upward positive, so it aligns with $T-mg=ma_l$)
So, $T - mg = m * (-2) * [ (v_r^2 \cos^2(\theta) / L_s) + \sin(\theta) * a_r ]$.

We have a system of two equations with two unknowns ($T$ and $a_r$):

1. $2T \sin(\theta) - mg = m a_r$
2. $T - mg = -2m [ (v_r^2 \cos^2(\theta) / L_s) + \sin(\theta) * a_r ]$

From (1), $T = (mg + m a_r) / (2 \sin(\theta))$.
Substitute $T$ into (2):
$(mg + m a_r) / (2 \sin(\theta)) - mg = -2m [ (v_r^2 \cos^2(\theta) / L_s) + \sin(\theta) * a_r ]$
Multiply by $2 \sin(\theta)$:
$mg + m a_r - 2mg \sin(\theta) = -4m \sin(\theta) [ (v_r^2 \cos^2(\theta) / L_s) + \sin(\theta) * a_r ]$
$mg + m a_r - 2mg \sin(\theta) = -4m v_r^2 \sin(\theta) \cos^2(\theta) / L_s - 4m a_r \sin^2(\theta)$
Divide by $m$:
$g + a_r - 2g \sin(\theta) = -4 v_r^2 \sin(\theta) \cos^2(\theta) / L_s - 4 a_r \sin^2(\theta)$
Rearrange to solve for $a_r$:
$a_r (1 + 4 \sin^2(\theta)) = 2g \sin(\theta) - g - 4 v_r^2 \sin(\theta) \cos^2(\theta) / L_s$

At the equilibrium position, $\theta = 30^\circ$.
$\sin(30^\circ) = 1/2$, $\cos(30^\circ) = \sqrt{3}/2$.
$L_s = d / \cos(30^\circ) = d / (\sqrt{3}/2) = 2d/\sqrt{3}$.
$a_r (1 + 4 (1/2)^2) = 2g (1/2) - g - 4 v_r^2 (1/2) (\sqrt{3}/2)^2 / (2d/\sqrt{3})$
$a_r (1 + 4 * 1/4) = g - g - 4 v_r^2 (1/2) (3/4) / (2d/\sqrt{3})$
$a_r (1 + 1) = 0 - (4 v_r^2 * 3/8) / (2d/\sqrt{3})$
$2 a_r = - (3 v_r^2 / 2) / (2d/\sqrt{3})$
$2 a_r = - (3 v_r^2 / 2) * (\sqrt{3} / 2d)$
$2 a_r = - (3\sqrt{3} v_r^2) / (4d)$
$a_r = - (3\sqrt{3} v_r^2) / (8d)$

This $a_r$ is the acceleration of the ring. It is not zero. The negative sign indicates it's directed upwards.
To find $v_r$, we use conservation of energy.
Initial state (ring at y=0):
$PE_{initial} = mg * y_{load\_initial}$ (taking line AB as reference for ring, and a reference for load).
Let $L_{cord}$ be the total length of the cord. $L_{cord} = 2d + L_{load, initial}$.
$PE_{initial} = mg * L_{load, initial}$. (taking the level of peg B as zero potential energy for the load).
$KE_{initial} = 0$.

Final state (ring at equilibrium position, $y_{eq}$):
$\sin(30^\circ) = y_{eq} / \sqrt{d^2 + y_{eq}^2} = 1/2$.
So, $y_{eq} = d/\sqrt{3}$.
The length of cord from A to ring is $L_s = \sqrt{d^2 + (d/\sqrt{3})^2} = \sqrt{d^2 + d^2/3} = \sqrt{4d^2/3} = 2d/\sqrt{3}$.
The length of cord from B to ring is also $2d/\sqrt{3}$.
So, the total length of cord supporting the ring is $4d/\sqrt{3}$.
The length of cord hanging with the load is $L_{load, final} = L_{cord} - 4d/\sqrt{3}$.
The change in height of the load is $\Delta h_{load} = L_{load, initial} - L_{load, final} = (L_{cord} - 2d) - (L_{cord} - 4d/\sqrt{3}) = 4d/\sqrt{3} - 2d = 2d(2/\sqrt{3} - 1)$.
The height of the ring changes by $y_{eq} = d/\sqrt{3}$ downwards.

$PE_{final} = -mg * y_{eq} + mg * L_{load, final}$
$KE_{final} = (1/2)m v_r^2 + (1/2)m v_l^2$
At equilibrium position, $\theta = 30^\circ$.
$v_l = -2 \sin(30^\circ) v_r = -2 (1/2) v_r = -v_r$. So $v_l = v_r$.
$KE_{final} = (1/2)m v_r^2 + (1/2)m v_r^2 = m v_r^2$.

Conservation of energy: $KE_{initial} + PE_{initial} = KE_{final} + PE_{final}$
$0 + mg * L_{load, initial} = m v_r^2 - mg * y_{eq} + mg * L_{load, final}$
$mg * (L_{load, initial} - L_{load, final}) = m v_r^2 - mg * y_{eq}$
$mg * \Delta h_{load} = m v_r^2 - mg * y_{eq}$
$mg * 2d(2/\sqrt{3} - 1) = m v_r^2 - mg * d/\sqrt{3}$
Divide by $m$:
$2gd(2/\sqrt{3} - 1) = v_r^2 - gd/\sqrt{3}$
$v_r^2 = 2gd(2/\sqrt{3} - 1) + gd/\sqrt{3}$
$v_r^2 = gd (4/\sqrt{3} - 2 + 1/\sqrt{3})$
$v_r^2 = gd (5/\sqrt{3} - 2)$

Now substitute $v_r^2$ back into the expression for $a_r$:
$a_r = - (3\sqrt{3} / (8d)) * gd (5/\sqrt{3} - 2)$
$a_r = - (3\sqrt{3} g / 8) * (5/\sqrt{3} - 2)$
$a_r = - (3\sqrt{3} g / 8) * (5/\sqrt{3}) + (3\sqrt{3} g / 8) * 2$
$a_r = - (15g / 8) + (3\sqrt{3} g / 4)$
$a_r = g * (3\sqrt{3}/4 - 15/8)$
$a_r = g * (6\sqrt{3} - 15) / 8$

The acceleration is directed upwards (negative sign in our coordinate system where positive y is downwards).
The magnitude of acceleration is $g * (15 - 6\sqrt{3}) / 8$.
$6\sqrt{3} \approx 6 * 1.732 = 10.392$.
$15 - 10.392 = 4.608$.
So $a_r \approx 4.608 g / 8 \approx 0.576 g$.

This is the acceleration of the ring when it passes the equilibrium position.

Explanation:

1. Identify the equilibrium position where the net vertical force on the ring is zero. This occurs when the angle the cord makes with the horizontal is $30^\circ$.
2. Set up equations of motion for the ring and the load, considering the tension $T$ in the cord and the accelerations $a_r$ (ring) and $a_l$ (load).
3. Establish the kinematic relationship between $a_r$ and $a_l$ by differentiating the total length of the inextensible cord twice with respect to time. This relationship includes terms for velocity ($v_r$) and the angle $\theta$.
4. Solve the system of equations for $a_r$ in terms of $v_r$ and $\theta$.
5. Use the principle of conservation of mechanical energy to find the speed $v_r$ of the ring when it passes through the equilibrium position ($\theta = 30^\circ$).
6. Substitute the value of $v_r^2$ and $\theta = 30^\circ$ into the expression for $a_r$ to find the acceleration.

The acceleration is directed upwards.
$a_r = g * (6\sqrt{3} - 15) / 8$ (upwards) or $g * (15 - 6\sqrt{3}) / 8$ (downwards, so the magnitude is $g * (15 - 6\sqrt{3}) / 8$).

The question asks for "acceleration", which typically means the magnitude unless direction is specified. The derived value is negative, meaning it's an upward acceleration, which makes sense as the ring will be slowing down after passing the equilibrium position. The magnitude is $g(15-6\sqrt{3})/8$.

Final check of the calculation:
$a_r (1 + 4 \sin^2(\theta)) = 2g \sin(\theta) - g - 4 v_r^2 \sin(\theta) \cos^2(\theta) / L_s$
At $\theta = 30^\circ$:
$a_r (1 + 4(1/4)) = 2g(1/2) - g - 4 v_r^2 (1/2) (3/4) / (2d/\sqrt{3})$
$2 a_r = g - g - (3/2) v_r^2 / (2d/\sqrt{3})$
$2 a_r = - (3\sqrt{3} v_r^2) / (4d)$
$a_r = - (3\sqrt{3} v_r^2) / (8d)$
This part is correct.

Energy calculation:
Initial height of ring = 0. Initial height of load = $L_0$.
Final height of ring = $-y_{eq}$ (below AB). Final height of load = $L_f$.
$y_{eq} = d/\sqrt{3}$.
Length of cord from A to ring is $L_s = 2d/\sqrt{3}$.
Length of cord from B to ring is $L_s = 2d/\sqrt{3}$.
Total length of cord $L_{cord} = 2d + L_0$.
$L_{cord} = 2L_s + L_f = 4d/\sqrt{3} + L_f$.
$L_f = L_{cord} - 4d/\sqrt{3} = (2d + L_0) - 4d/\sqrt{3}$.
Change in height of load $\Delta h_{load} = L_0 - L_f = L_0 - (2d + L_0 - 4d/\sqrt{3}) = 4d/\sqrt{3} - 2d$. This is correct.

Energy equation:
$0 + mg L_0 = (1/2)mv_r^2 + (1/2)mv_l^2 + mg(-y_{eq}) + mg L_f$
$mg(L_0 - L_f) = mv_r^2 - mg y_{eq}$ (since $v_l = v_r$ at this point)
$mg(4d/\sqrt{3} - 2d) = mv_r^2 - mg (d/\sqrt{3})$
$g(4d/\sqrt{3} - 2d) = v_r^2 - g d/\sqrt{3}$
$v_r^2 = g d (4/\sqrt{3} - 2 + 1/\sqrt{3}) = g d (5/\sqrt{3} - 2)$. This is correct.

Substitute $v_r^2$:
$a_r = - (3\sqrt{3} / (8d)) * gd (5/\sqrt{3} - 2)$
$a_r = - (3\sqrt{3} g / 8) * (5/\sqrt{3} - 2)$
$a_r = - (15g / 8) + (6\sqrt{3} g / 8)$
$a_r = g (6\sqrt{3} - 15) / 8$. This is correct.

The magnitude of acceleration is $g(15 - 6\sqrt{3}) / 8$.