A solid bob of a material having density twice that of water is suspended with a massless and inextensible string of length \(L\). The whole set-up is placed inside a water-filled tank. The bob is imparted a horizontal velocity \(V_0\) at the lowest point \(A\), while the other end of the string is fixed, such that the bob completes a semi-circular trajectory in the vertical plane. The string becomes slack only when the bob reaches the topmost point \(C\). Assume that the effects of viscosity and water currents are negligible. The acceleration due to gravity is \(g\). What is the expression for \(V_0\)?
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Whenever a system is immersed in a fluid, replace gravity \(g\) with effective gravity \(g_{\text{eff}} = g\left(1 - \frac{\rho_{\text{fluid}}}{\rho_{\text{solid}}}\right)\).
Since \(\rho_{\text{solid}} = 2 \rho_{\text{fluid}}\), we get \(g_{\text{eff}} = g/2\).
The standard critical velocity formula \(\sqrt{5gL}\) then directly becomes \(\sqrt{5(g/2)L}\).
Step 1: Understanding the Question:
The problem concerns vertical circular motion of a suspended bob inside a fluid medium (water).
Because the bob is submerged, it experiences a constant upward buoyant force in addition to gravity.
This modifies the effective acceleration due to gravity acting on the system. Step 2: Key Formula or Approach:
• Net downward force (effective weight):
\[ F_{\text{eff}} = W - F_B = m g_{\text{eff}} \]
• For a bob to just complete vertical circular motion, the minimum velocity at the lowest point \(A\) is given by:
\[ V_0 = \sqrt{5 g_{\text{eff}} L} \] Step 3: Detailed Explanation:
• Let the density of the bob be \(\rho_b\) and the density of water be \(\rho_w\). We are given \(\rho_b = 2\rho_w\).
• The mass of the bob is \(m = \rho_b V\), where \(V\) is its volume.
• The buoyant force exerted by water on the bob is:
\[ F_B = \rho_w V g \]
• The gravitational force on the bob is:
\[ W = \rho_b V g = 2 \rho_w V g \]
• The effective downward force is:
\[ F_{\text{eff}} = W - F_B = 2 \rho_w V g - \rho_w V g = \rho_w V g \]
• Thus, the effective gravitational acceleration is:
\[ g_{\text{eff}} = \frac{F_{\text{eff}}}{m} = \frac{\rho_w V g}{2 \rho_w V} = \frac{g}{2} \]
• For vertical circular motion, the condition for the string to just go slack at the highest point \(C\) is:
\[ V_0 = \sqrt{5 g_{\text{eff}} L} \]
• Substituting the value of \(g_{\text{eff}}\).:
\[ V_0 = \sqrt{5 \left(\frac{g}{2}\right) L} = \sqrt{\frac{5}{2} g L} \] Step 4: Final Answer:
The expression for \(V_0\) is \(\sqrt{\frac{5}{2}gL}\).
