A stone of mass 1kg is tied with a string and it is whirled in a vertical circle of radius 1 m. If tension at the highest point is 14 N, then velocity at lowest point will be

Options

A. `3m//s`
B. `4m//s`
C. `6m//s`
D. `8m//s`

Answer

To solve the problem, we will follow these steps:

### Step 1: Analyze the forces at the highest point
At the highest point of the vertical circle, the forces acting on the stone are:
- The gravitational force (weight) acting downwards, \( F_g = mg \)
- The tension in the string, \( T \), also acting downwards.

The net centripetal force required to keep the stone moving in a circle is provided by the sum of these forces:
\[ T + mg = \frac{mv^2}{r} \]
Where:
- \( m = 1 \, \text{kg} \) (mass of the stone)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( r = 1 \, \text{m} \) (radius of the circle)
- \( T = 14 \, \text{N} \) (tension at the highest point)

### Step 2: Substitute known values into the equation
Substituting the known values into the equation:
\[ 14 + (1 \times 10) = \frac{1 \times v^2}{1} \]
This simplifies to:
\[ 14 + 10 = v^2 \]
\[ 24 = v^2 \]

### Step 3: Solve for velocity at the highest point
Taking the square root to find the velocity \( v \) at the highest point:
\[ v = \sqrt{24} = 2\sqrt{6} \, \text{m/s} \]

### Step 4: Apply the work-energy theorem between the highest and lowest points
Using the work-energy theorem, the work done by gravity as the stone moves from the highest point (A) to the lowest point (B) is equal to the change in kinetic energy:
\[ W = K_f - K_i \]
Where:
- \( K_f = \frac{1}{2} m v_B^2 \) (kinetic energy at the lowest point)
- \( K_i = \frac{1}{2} m v_A^2 \) (kinetic energy at the highest point)

The work done by gravity as the stone moves down a height of \( 2r \) (since it moves from the top to the bottom of the circle) is:
\[ W = -mg(2r) = -10 \times 1 \times 2 = -20 \, \text{J} \]

### Step 5: Set up the equation for kinetic energy
Now, substituting the values into the work-energy equation:
\[ -20 = \frac{1}{2} m v_B^2 - \frac{1}{2} m v_A^2 \]
Substituting \( m = 1 \, \text{kg} \) and \( v_A^2 = 24 \):
\[ -20 = \frac{1}{2} (1) v_B^2 - \frac{1}{2} (1) (24) \]
This simplifies to:
\[ -20 = \frac{1}{2} v_B^2 - 12 \]

### Step 6: Solve for \( v_B^2 \)
Rearranging gives:
\[ \frac{1}{2} v_B^2 = -20 + 12 \]
\[ \frac{1}{2} v_B^2 = -8 \]
Multiplying through by 2:
\[ v_B^2 = 16 \]

### Step 7: Find the final velocity at the lowest point
Taking the square root:
\[ v_B = \sqrt{16} = 4 \, \text{m/s} \]

### Final Answer
The velocity at the lowest point is \( 4 \, \text{m/s} \).

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