Question

A bead A of mass $m$ can slide without friction on a fixed horizontal rod. Another bead B of the same mass is suspended from the bead A with help of a light inextensible cord of length $l$. Both the beads are held motionless in such a way that the thread is straight and inclined with the vertical at an angle $\theta$ as shown in the figure.

(a) Determine accelerations of the beads immediately after they are released simultaneously.

(b) Determine accelerations of the beads when the thread becomes vertical.

Answer 1

a) $\vec{a}_A = \frac{g \sin\theta \cos\theta}{1 + \sin^2\theta} \hat{i}, \vec{a}_B = -\frac{g \sin\theta \cos\theta}{1 + \sin^2\theta} \hat{i} - \frac{2g \sin^2\theta}{1 + \sin^2\theta} \hat{j}; b) \vec{a}_A = 0, \vec{a}_B = 2g(1 - \cos\theta) \hat{j}$

Answer 2

Explanation of the solution:

Part (a): Immediately after release

1. Free Body Diagrams and Newton's Laws:
   • For bead A (mass $m$), moving horizontally: The horizontal component of tension $T \sin\theta$ causes its acceleration $a_A$. So, $T \sin\theta = m a_A$.
   • For bead B (mass $m$):
     • Horizontal motion: The horizontal component of tension $-T \sin\theta$ (opposite to A's acceleration) causes its horizontal acceleration $a_{Bx}$. So, $-T \sin\theta = m a_{Bx}$.
     • Vertical motion: The vertical component of tension $T \cos\theta$ and gravity $mg$ cause its vertical acceleration $a_{By}$. So, $T \cos\theta - mg = m a_{By}$.
2. Conservation of Horizontal Momentum: Since there are no external horizontal forces, the horizontal momentum of the system (A+B) is conserved. As the system starts from rest, the total horizontal momentum remains zero. This implies $m a_A + m a_{Bx} = 0$, leading to $a_{Bx} = -a_A$.
3. Kinematic Constraint (Inextensible String): At the instant of release, velocities are zero. The acceleration of B relative to A, $\vec{a}_{B/A} = \vec{a}_B - \vec{a}_A$, must be perpendicular to the string (purely tangential, no centripetal component as $v=0$). The string is along the direction $(\sin\theta \hat{i} - \cos\theta \hat{j})$ relative to A. Therefore, the dot product of $\vec{a}_{B/A}$ with this direction must be zero: $(a_{Bx} - a_A)\sin\theta - a_{By}\cos\theta = 0$.
4. Solving the System: Substitute $a_{Bx} = -a_A$ into the kinematic constraint to get $a_{By} = -2a_A \tan\theta$. Then, substitute $T = m a_A / \sin\theta$ (from A's equation) and the expression for $a_{By}$ into B's vertical equation. This allows solving for $a_A$. Once $a_A$ is found, $a_{Bx}$ and $a_{By}$ are determined.

Part (b): When the thread becomes vertical

1. Velocities (Conservation of Energy and Momentum):
   • Horizontal momentum conservation still holds: $m v_A + m v_{Bx} = 0 \implies v_{Bx} = -v_A$.
   • When the string is vertical, bead B is directly below bead A. The velocity of B relative to A, $\vec{v}_{B/A}$, must be along the vertical string. This implies $v_{Bx} - v_A = 0$, so $v_{Bx} = v_A$.
   • Combining these, $v_A = v_{Bx} = 0$. This means the horizontal velocities of both beads are zero when the string is vertical.
   • Conservation of mechanical energy: The initial potential energy is $-mgl \cos\theta$ (taking the rod as $y=0$). The final potential energy is $-mgl$. The final kinetic energy is $\frac{1}{2}m v_A^2 + \frac{1}{2}m v_B^2$. Since $v_A=0$ and $v_B$ is purely vertical ($v_B = v_{By}$), the final kinetic energy is $\frac{1}{2}m v_{By}^2$. Equating initial and final energies gives $\frac{1}{2}m v_{By}^2 = mgl(1 - \cos\theta)$, so $v_{By}^2 = 2gl(1 - \cos\theta)$.
2. Accelerations:
   • For bead A: $T \sin(0) = m a_A \implies a_A = 0$.
   • For bead B:
     • Horizontal: $-T \sin(0) = m a_{Bx} \implies a_{Bx} = 0$.
     • Vertical: $T \cos(0) - mg = m a_{By} \implies T - mg = m a_{By}$.
   • Kinematic Constraint (Acceleration): When the string is vertical, the acceleration of B relative to A is purely centripetal, directed upwards along the string. So, $a_{B/A, y} = v_{By}^2/l$.
   • Since $a_A=0$, $a_{By} = v_{By}^2/l$. Substitute the value of $v_{By}^2$ to get $a_{By} = 2g(1 - \cos\theta)$.

Answer:

The accelerations of the beads are: (a) Immediately after they are released simultaneously:

• Acceleration of bead A: $\vec{a}_A = \frac{g \sin\theta \cos\theta}{1 + \sin^2\theta} \hat{i}$
• Acceleration of bead B: $\vec{a}_B = -\frac{g \sin\theta \cos\theta}{1 + \sin^2\theta} \hat{i} - \frac{2g \sin^2\theta}{1 + \sin^2\theta} \hat{j}$

(b) When the thread becomes vertical:

• Acceleration of bead A: $\vec{a}_A = 0$
• Acceleration of bead B: $\vec{a}_B = 2g(1 - \cos\theta) \hat{j}$