Physics Solution: Coupled Bead Motion

Coupled Bead Motion Analysis

Let the mass of each bead be $m$. We will analyze the motion in three stages: motion on the rod alone, motion on the helix alone, and finally the combined motion of the system.

1. Method I: Force Analysis

Analysis of Individual Motions

Case I: Bead on the Rod
The bead falls freely under gravity $g$ along the vertical rod. $$h = \frac{1}{2} g T_1^2 \implies T_1^2 = \frac{2h}{g} \quad \dots(1)$$

Case II: Bead on the Helix
Let $\theta$ be the constant angle the tangent to the helix makes with the vertical. The effective acceleration along the helical path is the component of gravity along the tangent, $g \cos \theta$. The length of the path corresponding to vertical height $h$ is $L = h \sec \theta$.

$$L = \frac{1}{2} (g \cos \theta) T_2^2 \implies h \sec \theta = \frac{1}{2} g \cos \theta T_2^2$$ $$T_2^2 = \frac{2h}{g \cos^2 \theta} \quad \dots(2)$$

Fig 1: Free Body Diagrams of the individual beads

Coupled Motion Analysis

In the combined setup, the bead on the helix (Bead A) and the bead on the rod (Bead B) are connected by an inextensible vertical thread. This constraint implies that their vertical acceleration $a_z$ must be identical.

Since the helix path is longer than the vertical path, Bead A would naturally descend slower than Bead B if unconnected. Therefore, the thread will become taut. Bead B (on the rod) will pull Bead A down, and conversely, Bead A will pull Bead B up.

Forces Involved:

Bead B (Rod): Gravity $mg$ acts downwards. Tension $T$ acts upwards.
Bead A (Helix): Gravity $mg$ and Tension $T$ both act vertically downwards.

Equation for Bead B (Rod):
Applying Newton’s Second Law in the vertical direction: $$mg – T = m a_z \quad \dots(3)$$

Equation for Bead A (Helix):
The total vertical force is $mg + T$. The component of this force driving motion along the tangent of the helix is $(mg + T) \cos \theta$. Let $a_{tangent}$ be the acceleration along the helical path. $$(mg + T) \cos \theta = m a_{tangent}$$

From kinematics, if the vertical acceleration is $a_z$, the acceleration along the tangent (at angle $\theta$ to vertical) is $a_{tangent} = a_z / \cos \theta$. Substituting this into the force equation:

$$(mg + T) \cos \theta = m \left( \frac{a_z}{\cos \theta} \right)$$ Multiply by $\cos \theta$: $$(mg + T) \cos^2 \theta = m a_z \quad \dots(4)$$

Solving for Time $T_3$

We eliminate Tension $T$ between equations (3) and (4). From (3), $T = m(g – a_z)$. Substitute this into (4):

$$ (mg + m(g – a_z)) \cos^2 \theta = m a_z $$ $$ (2g – a_z) \cos^2 \theta = a_z $$ $$ 2g \cos^2 \theta = a_z (1 + \cos^2 \theta) $$ $$ a_z = \frac{2g \cos^2 \theta}{1 + \cos^2 \theta} $$

Now, applying the kinematic equation for the combined system falling height $h$ in time $T_3$:

$$ h = \frac{1}{2} a_z T_3^2 \implies T_3^2 = \frac{2h}{a_z} $$ Substituting $a_z$: $$ T_3^2 = 2h \cdot \frac{1 + \cos^2 \theta}{2g \cos^2 \theta} = \frac{h}{g \cos^2 \theta} (1 + \cos^2 \theta) $$ $$ T_3^2 = \frac{h}{g \cos^2 \theta} + \frac{h}{g} $$

Recall our results from the individual motions (Equations 1 and 2): $$ \frac{h}{g} = \frac{T_1^2}{2} \quad \text{and} \quad \frac{h}{g \cos^2 \theta} = \frac{T_2^2}{2} $$

Substituting these values into the expression for $T_3^2$: $$ T_3^2 = \frac{T_2^2}{2} + \frac{T_1^2}{2} $$

Final Answer The assembly will take $$T_3 = \sqrt{\frac{T_1^2 + T_2^2}{2}}$$ to slide down the height $h$.

2. Work-Energy Theorem Aliter

Fig 2: Combined system with inextensible thread

Individual Motions Recalled

Case 1: Bead on vertical rod. Falling under gravity $g$ through height $h$: $$T_1^2 = \frac{2h}{g} \quad \dots \text{(i)}$$

Case 2: Bead on helix. Let $\alpha$ be the angle the helical path makes with the horizontal (Note: $\alpha = 90^\circ – \theta$). The effective acceleration is $g \sin \alpha$. The path length is $L = h / \sin \alpha$. $$L = \frac{1}{2}(g \sin \alpha) T_2^2 \implies \frac{h}{\sin \alpha} = \frac{1}{2} g \sin \alpha T_2^2$$ $$T_2^2 = \frac{2h}{g \sin^2 \alpha} \quad \dots \text{(ii)}$$

Combined System: Energy Method

The two beads are tied by a light inextensible thread. The thread is “almost vertical,” implying that both beads share the same vertical displacement $dy$, vertical velocity $v$, and vertical acceleration $a_{sys}$.

We apply the Work-Energy Theorem for a small vertical displacement $dy$:

Work Done: Gravity acts on both beads moving down $dy$.
$$dW = mg(dy) + mg(dy) = 2mg\, dy$$
Kinetic Energy ($K$):
- Bead on rod has velocity $v$. $K_1 = \frac{1}{2}mv^2$.
- Bead on helix has vertical velocity $v$, but actual velocity $v_{path} = v / \sin \alpha$.
  $K_2 = \frac{1}{2}m \left(\frac{v}{\sin \alpha}\right)^2$.

Total Kinetic Energy is: $$K = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 \frac{1}{\sin^2 \alpha} = \frac{1}{2}mv^2 \left( 1 + \frac{1}{\sin^2 \alpha} \right)$$

Differentiating $K$ with respect to time to equate to Power ($P = dW/dt = 2mgv$):

$$ \frac{dK}{dt} = m v \frac{dv}{dt} \left( 1 + \frac{1}{\sin^2 \alpha} \right) = m v a_{sys} \left( 1 + \frac{1}{\sin^2 \alpha} \right) $$

Equating Power terms ($\frac{dK}{dt} = 2mgv$): $$ m v a_{sys} \left( 1 + \frac{1}{\sin^2 \alpha} \right) = 2mgv $$ $$ a_{sys} = \frac{2g}{1 + \frac{1}{\sin^2 \alpha}} $$

Calculation of Time $T$

The time $T$ to cover height $h$ with constant acceleration $a_{sys}$ is $T = \sqrt{2h/a_{sys}}$.

$$ T = \sqrt{\frac{2h \left( 1 + \frac{1}{\sin^2 \alpha} \right)}{2g}} = \sqrt{\frac{h}{g} + \frac{h}{g \sin^2 \alpha}} $$

Substituting $T_1$ and $T_2$ from equations (i) and (ii) into this expression:

$$ T = \sqrt{\frac{T_1^2}{2} + \frac{T_2^2}{2}} $$