A ball of mass m is fastened to a string. The ball swings in a vertical circle of radius R with the other end of the string held fixed. Neglecting the air resistance, the difference between the string's tension at the bottom of the circle and at the quarter of the cirlce is

Options

A. mg
B. 3mg
C. 4 mg
D. 6 mg

Explanation

To solve the problem of finding the difference between the tension in the string at the bottom of the circle and at a quarter of the circle, we can follow these steps:

### Step 1: Identify the points of interest
- We have two points to consider:
- Point A (bottom of the circle)
- Point B (quarter of the circle)

### Step 2: Analyze forces at Point A
- At the bottom of the circle (Point A), the forces acting on the ball are:
- Tension (T_A) acting upwards
- Weight (mg) acting downwards
- The net force provides the centripetal force required for circular motion:
\[ T_A - mg = \frac{mv_A^2}{R} \]
- Rearranging gives:
\[ T_A = mg + \frac{mv_A^2}{R} \]

### Step 3: Analyze forces at Point B
- At the quarter of the circle (Point B), the forces acting on the ball are:
- Tension (T_B) acting towards the center of the circle
- Weight (mg) acting downwards
- The vertical component of the tension must balance the weight, and the net force provides the centripetal force:
\[ T_B \cos(90^\circ) - mg = \frac{mv_B^2}{R} \]
- Since the angle at Point B is 90 degrees, we can simplify this to:
\[ T_B = \frac{mv_B^2}{R} \]

### Step 4: Relate velocities at Points A and B
- Using conservation of mechanical energy between the two points:
\[ \text{Potential Energy at A} + \text{Kinetic Energy at A} = \text{Potential Energy at B} + \text{Kinetic Energy at B} \]
- At Point A, the height is 0, and at Point B, the height is R:
\[ 0 + \frac{1}{2} mv_A^2 = mgR + \frac{1}{2} mv_B^2 \]
- Rearranging gives:
\[ \frac{1}{2} mv_A^2 - \frac{1}{2} mv_B^2 = mgR \]
- This can be rewritten as:
\[ v_A^2 - v_B^2 = 2gR \]

### Step 5: Substitute back into the tension equations
- Substitute \(v_A^2 - v_B^2 = 2gR\) into the tension difference equation:
\[ T_A - T_B = \left(mg + \frac{m v_A^2}{R}\right) - \frac{m v_B^2}{R} \]
- This simplifies to:
\[ T_A - T_B = mg + \frac{m(v_A^2 - v_B^2)}{R} \]
- Substituting \(v_A^2 - v_B^2 = 2gR\):
\[ T_A - T_B = mg + \frac{m(2gR)}{R} \]
- Simplifying gives:
\[ T_A - T_B = mg + 2mg = 3mg \]

### Final Answer
The difference between the tension at the bottom of the circle and at the quarter of the circle is:
\[ T_A - T_B = 3mg \]

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