A stone hanging from a massless string of length 15 m is projected horizontally with speed `12ms^(-1)`. The speed of the particle at the point where the tension in the string is equal to the weight of the particle, is close to

Options

A. `10ms^(-1)`
B. `7ms^(-1)`
C. `12ms^(-1)`
D. `5ms^(-1)`

Answer

To solve the problem step by step, we will analyze the motion of the stone hanging from the string and apply the principles of physics, particularly the concepts of tension, centripetal force, and conservation of energy.

### Step 1: Understanding the Problem
We have a stone hanging from a massless string of length \( L = 15 \, \text{m} \) that is projected horizontally with an initial speed \( u = 12 \, \text{m/s} \). We need to find the speed \( v \) of the stone at the point where the tension in the string equals the weight of the stone.

### Step 2: Forces Acting on the Stone
At the point where the tension \( T \) in the string equals the weight of the stone \( mg \), we can write:
\[ T = mg \]

### Step 3: Analyzing the Motion
As the stone swings down, it moves in a circular path with radius equal to the length of the string \( L \). The centripetal force required for circular motion is provided by the net force acting towards the center of the circular path.

### Step 4: Applying Newton's Second Law
At an angle \( \theta \) from the vertical, the forces acting on the stone can be resolved into components. The net force towards the center of the circle is given by:
\[ T - mg \cos \theta = \frac{mv^2}{L} \]

Since \( T = mg \), we substitute this into the equation:
\[ mg - mg \cos \theta = \frac{mv^2}{L} \]

### Step 5: Simplifying the Equation
Factoring out \( mg \):
\[ mg(1 - \cos \theta) = \frac{mv^2}{L} \]

We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g(1 - \cos \theta) = \frac{v^2}{L} \]

### Step 6: Energy Conservation
We can also apply the principle of conservation of mechanical energy. The initial kinetic energy when the stone is projected is:
\[ KE_i = \frac{1}{2} m u^2 \]

As the stone swings down to height \( h \), its potential energy changes. The height \( h \) can be expressed as:
\[ h = L - L \cos \theta = L(1 - \cos \theta) \]

The potential energy at height \( h \) is:
\[ PE = mgh = mgL(1 - \cos \theta) \]

The total mechanical energy at the starting point is equal to the total mechanical energy at the point where the tension equals the weight:
\[ \frac{1}{2} m u^2 = \frac{1}{2} m v^2 + mgL(1 - \cos \theta) \]

### Step 7: Setting Up the Equation
Substituting \( h \) into the energy conservation equation:
\[ \frac{1}{2} m (12^2) = \frac{1}{2} m v^2 + mgL(1 - \cos \theta) \]

### Step 8: Solving for \( v \)
We can cancel \( m \) from the equation:
\[ \frac{1}{2} (12^2) = \frac{1}{2} v^2 + gL(1 - \cos \theta) \]

Substituting \( g = 9.8 \, \text{m/s}^2 \) and \( L = 15 \, \text{m} \):
\[ \frac{1}{2} (144) = \frac{1}{2} v^2 + 9.8 \times 15 (1 - \cos \theta) \]

### Step 9: Calculate Values
Calculating \( 9.8 \times 15 = 147 \):
\[ 72 = \frac{1}{2} v^2 + 147(1 - \cos \theta) \]

### Step 10: Final Calculation
To find \( v \), we can rearrange and solve for \( v^2 \):
\[ \frac{1}{2} v^2 = 72 - 147(1 - \cos \theta) \]

Assuming \( \theta \) is small and approximating \( \cos \theta \approx 1 \), we can simplify further. However, we can also directly calculate \( v \) using the earlier derived relationship:
\[ v^2 = 48 \]
\[ v = \sqrt{48} \approx 6.93 \, \text{m/s} \]

### Conclusion
The speed of the stone at the point where the tension in the string equals the weight of the particle is approximately \( 7 \, \text{m/s} \).