A ball whirls around in a vertical circle at the end of a string . The other end of the string is fixed at the centre of the circle. Assuming the total energy of the ball-Earth system remains constant. What is the difference of tension in string at bottom and top during circular motion (`T_(b)` - `T_(t)` = ?)

Options

A. `5 mg`
B. `3 mg`
C. `6 mg`
D. `3.5 mg`

Explanation

To solve the problem of finding the difference in tension in the string at the bottom and top of the vertical circular motion of a ball, we can follow these steps:

### Step 1: Understand the Forces at the Bottom and Top
At the bottom of the circular path, the tension in the string (T_b) must counteract both the gravitational force acting on the ball (mg) and provide the necessary centripetal force to keep the ball moving in a circle. At the top of the circular path, the tension in the string (T_t) must only provide the centripetal force, as gravity also acts in the same direction.

### Step 2: Write the Equation for Tension at the Bottom
At the bottom of the circle, the forces acting on the ball can be expressed as:
\[ T_b - mg = \frac{mv_1^2}{r} \]
Where:
- \( T_b \) = Tension at the bottom
- \( m \) = Mass of the ball
- \( g \) = Acceleration due to gravity
- \( v_1 \) = Velocity of the ball at the bottom
- \( r \) = Radius of the circular path

Rearranging gives:
\[ T_b = \frac{mv_1^2}{r} + mg \]

### Step 3: Write the Equation for Tension at the Top
At the top of the circle, the forces can be expressed as:
\[ T_t + mg = \frac{mv_2^2}{r} \]
Where:
- \( T_t \) = Tension at the top
- \( v_2 \) = Velocity of the ball at the top

Rearranging gives:
\[ T_t = \frac{mv_2^2}{r} - mg \]

### Step 4: Find the Difference in Tension
Now we need to find the difference \( T_b - T_t \):
\[ T_b - T_t = \left(\frac{mv_1^2}{r} + mg\right) - \left(\frac{mv_2^2}{r} - mg\right) \]

This simplifies to:
\[ T_b - T_t = \frac{mv_1^2}{r} - \frac{mv_2^2}{r} + 2mg \]
\[ T_b - T_t = \frac{m}{r}(v_1^2 - v_2^2) + 2mg \]

### Step 5: Use Conservation of Energy
According to the conservation of mechanical energy, the total mechanical energy at the bottom and the top can be equated:
\[ \frac{1}{2}mv_1^2 = \frac{1}{2}mv_2^2 + mg(2r) \]

Rearranging gives:
\[ \frac{1}{2}mv_1^2 - \frac{1}{2}mv_2^2 = mg(2r) \]
Multiplying through by 2:
\[ mv_1^2 - mv_2^2 = 4mgr \]

### Step 6: Substitute Back into the Tension Difference Equation
Substituting this result back into the tension difference equation:
\[ T_b - T_t = \frac{m}{r}(4mgr) + 2mg \]
\[ T_b - T_t = 4mg + 2mg \]
\[ T_b - T_t = 6mg \]

### Final Answer
Thus, the difference in tension in the string at the bottom and top during circular motion is:
\[ T_b - T_t = 6mg \]

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