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 find the velocity of the stone at the lowest point of the vertical circle, we can 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 downward: \( mg \)
- The tension in the string acting downward: \( T \)

The centripetal force required to keep the stone moving in a circle is provided by the sum of the tension and the weight of the stone. Therefore, we can write the equation for centripetal force as:

\[ \frac{mv^2}{r} = T + mg \]

### Step 2: Substitute known values
Given:
- Mass \( m = 1 \, \text{kg} \)
- Radius \( r = 1 \, \text{m} \)
- Tension at the highest point \( T = 14 \, \text{N} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

Substituting these values into the equation:

\[ \frac{1 \cdot v^2}{1} = 14 + 1 \cdot 10 \]

This simplifies to:

\[ v^2 = 14 + 10 = 24 \]

### Step 3: Calculate the velocity at the highest point
Now we can find the velocity \( v \) at the highest point:

\[ v = \sqrt{24} = 2\sqrt{6} \approx 4.89 \, \text{m/s} \]

### Step 4: Apply conservation of energy to find velocity at the lowest point
Using the principle of conservation of mechanical energy, the total mechanical energy at the highest point will be equal to the total mechanical energy at the lowest point.

At the highest point, the total energy \( E_a \) is given by:

\[ E_a = \frac{1}{2}mv_a^2 + mgh_a \]

At the lowest point, the total energy \( E_b \) is given by:

\[ E_b = \frac{1}{2}mv_b^2 + mgh_b \]

Since the height at the highest point \( h_a = 2 \, \text{m} \) (1 m above the lowest point) and at the lowest point \( h_b = 0 \, \text{m} \), we can write:

\[ E_a = \frac{1}{2}m(4.89)^2 + mg(2) \]
\[ E_b = \frac{1}{2}mv_b^2 + mg(0) \]

Setting \( E_a = E_b \):

\[ \frac{1}{2}m(4.89)^2 + mg(2) = \frac{1}{2}mv_b^2 \]

### Step 5: Solve for \( v_b \)
Substituting \( m = 1 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \):

\[ \frac{1}{2}(4.89)^2 + 10(2) = \frac{1}{2}v_b^2 \]

Calculating the left side:

\[ \frac{1}{2}(24) + 20 = \frac{1}{2}v_b^2 \]
\[ 12 + 20 = \frac{1}{2}v_b^2 \]
\[ 32 = \frac{1}{2}v_b^2 \]
\[ v_b^2 = 64 \]

### Step 6: Calculate \( v_b \)
Taking the square root:

\[ v_b = \sqrt{64} = 8 \, \text{m/s} \]

### Final Answer
Thus, the velocity at the lowest point is \( 8 \, \text{m/s} \).

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