Question

In a very large container with a fluid, a simple pendulum in just-slacking position is kept at rest in the orientation shown. Density of bob B is one-third that of the fluid. If B is currently vertically separated from A by half the total length (L) of the string, then find the maximum speed of bob B on releasing it.

• A. $\sqrt{\frac{11gL}{4}}$
• B. $\sqrt{\frac{11gL}{2}}$
• C. $\sqrt{5gL}$
• D. $\sqrt{6gL}$

Answer 1

Answer: (D)

$\sqrt{6gL}$

Answer 2

The net upward force on the bob due to buoyancy and gravity is $F_{net} = (\rho_f - \rho_B)Vg$. Given $\rho_B = \frac{1}{3}\rho_f$, let $m$ be the mass of the bob. Then $m = \rho_B V$. The buoyant force is $F_B = \rho_f Vg = 3 \rho_B Vg = 3mg$. The net upward force is $F_{net} = F_B - mg = 3mg - mg = 2mg$.

The potential energy of the bob is given by $U(\theta) = (mg - F_B)y$, where $y$ is the vertical displacement from the pivot A. Let $y = -L\cos\theta$. $U(\theta) = (mg - 3mg)(-L\cos\theta) = (-2mg)(-L\cos\theta) = 2mgL\cos\theta$.

The initial condition "B is currently vertically separated from A by half the total length (L) of the string" means that the vertical distance from A to the bob is $L/2$. If A is at the origin, and the bob is below A, then $y_0 = -L/2$. Since $y_0 = -L\cos\theta_0$, we have $-L/2 = -L\cos\theta_0$, which implies $\cos\theta_0 = 1/2$. Thus, the initial angle is $\theta_0 = 60^\circ$.

The initial total energy of the bob is $E_0 = K_0 + U(\theta_0)$. Since the bob is released from rest, $K_0 = 0$. $E_0 = 0 + 2mgL\cos(60^\circ) = 2mgL(1/2) = mgL$.

The maximum speed occurs at the equilibrium position. The equilibrium position is where the net force along the string is zero. The forces along the string are tension $T$ and the component of the net force $2mg$ along the string, which is $2mg\cos\theta$. So, $T + 2mg\cos\theta = 0$. For tension $T$ to be non-negative, $2mg\cos\theta \le 0$, which means $\cos\theta \le 0$. This occurs for $\theta \in [90^\circ, 270^\circ]$. The equilibrium position corresponds to the minimum of the potential energy function $U(\theta) = 2mgL\cos\theta$. The minimum occurs at $\theta = 180^\circ$ (where $\cos\theta = -1$). At the equilibrium position $\theta_{eq} = 180^\circ$, the potential energy is $U_{eq} = 2mgL\cos(180^\circ) = -2mgL$.

By conservation of energy, the total energy at the equilibrium position is equal to the initial energy: $E_{eq} = K_{max} + U_{eq} = E_0$ $\frac{1}{2}mv_{max}^2 + (-2mgL) = mgL$ $\frac{1}{2}mv_{max}^2 = mgL + 2mgL = 3mgL$ $v_{max}^2 = 6gL$ $v_{max} = \sqrt{6gL}$