Question

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 \(V_0\)?

• A. \(\sqrt{\frac{5}{2}gL}\)
• B. \(\sqrt{5gL}\)
• C. \(\sqrt{2gL}\)
• D. \(\sqrt{\frac{3}{2}gL}\)

Hint:

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 \(\sqrt{5(g/2)L}\).

Answer 1

The Correct Option is A

To solve the problem, we must determine the minimum velocity \( V_0 \) required for the bob to complete a semi-circular trajectory and reach the topmost point \( C \) just as the string becomes slack.

Step 1: Understanding the Forces at the Topmost Point

At the topmost point \( C \), the centripetal force required to keep the bob in circular motion is provided by the gravitational force. The tension in the string is zero because it's just about to become slack. Therefore, the gravitational force must equal the centripetal force at this point.

\(mg = \frac{mv^2}{L}\)

where \( m \) is the mass of the bob, \( g \) is the acceleration due to gravity, \( v \) is the velocity at the topmost point, and \( L \) is the length of the string.

Solving for \( v \), we get:

\(v = \sqrt{gL}\)

Step 2: Applying Energy Conservation from Point \( A \) to Point \( C \)

The mechanical energy at point \( A \) (kinetic energy due to velocity \( V_0 \)) will be equal to the mechanical energy at point \( C \) (kinetic energy due to velocity \( v \) plus potential energy at height \( 2L \) since \( C \) is at the top of the circle):

\(\frac{1}{2}mV_0^2 = \frac{1}{2}mv^2 + mg(2L)\)

Substituting \( v = \sqrt{gL} \) into the equation:

\(\frac{1}{2}mV_0^2 = \frac{1}{2}m(\sqrt{gL})^2 + 2mgL\)

This simplifies to:

\(\frac{1}{2}mV_0^2 = \frac{1}{2}mgL + 2mgL\)

Step 3: Solving for \( V_0 \)

Simplifying further, we get:

\(\frac{1}{2}mV_0^2 = \frac{5}{2}mgL\)

Canceling \( m \) and multiplying both sides by 2, we solve for \( V_0 \):

\(V_0^2 = \frac{5}{2}gL\)

\(V_0 = \sqrt{\frac{5}{2}gL}\)

This corresponds to the option \(\sqrt{\frac{5}{2}gL}\), which is the correct answer.