In the figure shown, PQ is a uniform sphere of mass M and radius R hinged at point P , with PQ as horizontal diameter. QT is a uniform horizontal rod of mass M and length 2R rigidly attached to the sphere at Q, supported by a vertical spring of spring constant k. If the system is in equilibrium, then the potential energy stored in the spring is

Options

A. `+MgR`
B. `(MgR)/(2)`
C. `(1)/(2) K ( R^(2))`
D. `(M^(2) g^(2))/( 2k)`

Answer

To solve the problem, we need to find the potential energy stored in the spring when the system is in equilibrium. Let's break down the solution step by step.

### Step 1: Understand the System
We have a uniform sphere (PQ) of mass \( M \) and radius \( R \) hinged at point \( P \). There is a horizontal rod (QT) of mass \( M \) and length \( 2R \) attached to the sphere at point \( Q \). The system is supported by a vertical spring with spring constant \( k \).

### Step 2: Determine the Forces Acting on the System
In equilibrium, the downward gravitational force acting on the system must be balanced by the upward force exerted by the spring. The total weight of the system can be calculated as:
\[ \text{Total Weight} = \text{Weight of Sphere} + \text{Weight of Rod} = Mg + Mg = 2Mg \]

### Step 3: Calculate the Moment about the Hinge Point
The center of mass of the sphere is at a distance \( R \) from the hinge point \( P \), and the center of mass of the rod is at a distance \( R \) from the hinge point (since it is uniformly distributed along its length). Therefore, the total moment about point \( P \) due to the weights is:
\[ \text{Moment due to Sphere} = Mg \cdot R \]
\[ \text{Moment due to Rod} = Mg \cdot (R + R) = 2Mg \cdot R \]
Thus, the total moment is:
\[ \text{Total Moment} = MgR + 2MgR = 3MgR \]

### Step 4: Relate the Spring Force to the Displacement
Let \( x \) be the displacement of the spring from its natural length. The force exerted by the spring is given by Hooke's Law:
\[ F_{\text{spring}} = kx \]
In equilibrium, this spring force must balance the total weight:
\[ kx = 3Mg \]

### Step 5: Solve for \( x \)
From the equation \( kx = 3Mg \), we can solve for \( x \):
\[ x = \frac{3Mg}{k} \]

### Step 6: Calculate the Potential Energy Stored in the Spring
The potential energy \( U \) stored in the spring is given by the formula:
\[ U = \frac{1}{2} k x^2 \]
Substituting the value of \( x \):
\[ U = \frac{1}{2} k \left(\frac{3Mg}{k}\right)^2 \]
\[ U = \frac{1}{2} k \cdot \frac{9M^2g^2}{k^2} \]
\[ U = \frac{9M^2g^2}{2k} \]

### Final Answer
Thus, the potential energy stored in the spring is:
\[ U = \frac{9M^2g^2}{2k} \]