A particle mass `m` begins to slide down a fixed smooth sphere from the top of its vertical diameter. Calculate its tangential acceleration, radial acceleration and total acceleration when it breaks off.

Options

A. `(2g)/(3)`
B. `(sqrt(5)g)/(3)`
C. `g`
D. `(g)/(3)`

Answer

To solve the problem of a particle of mass \( m \) sliding down a fixed smooth sphere from the top of its vertical diameter, we need to calculate its tangential acceleration, radial acceleration, and total acceleration when it breaks off. Here’s a step-by-step solution:

### Step 1: Determine the position of the particle
- The particle starts at the top of the sphere and slides down to a point \( P \) at an angle \( \theta \) from the vertical.

### Step 2: Analyze forces acting on the particle
- At point \( P \), the forces acting on the particle are:
- The gravitational force \( mg \) acting downward.
- The normal force \( N \) exerted by the sphere acting perpendicular to the surface.

### Step 3: Resolve gravitational force into components
- The gravitational force can be resolved into two components:
- Radial component: \( mg \cos \theta \) (towards the center of the sphere)
- Tangential component: \( mg \sin \theta \) (along the direction of motion)

### Step 4: Apply Newton's second law in the radial direction
- The net radial force is given by:
\[ N - mg \cos \theta = -m \cdot a_r \]
where \( a_r \) is the radial acceleration.
- At the point of breaking off, the normal force \( N = 0 \):
\[ -mg \cos \theta = -m \cdot a_r \implies a_r = g \cos \theta \]

### Step 5: Apply conservation of energy
- The total mechanical energy is conserved. Initially, the particle has potential energy and no kinetic energy:
\[ E_1 = mgh = mgR \quad \text{(at the top)} \]
- At point \( P \):
\[ E_2 = \text{Potential Energy} + \text{Kinetic Energy} = mg(R - R \cos \theta) + \frac{1}{2} mv^2 \]
Setting \( E_1 = E_2 \):
\[ mgR = mgR(1 - \cos \theta) + \frac{1}{2} mv^2 \]
Simplifying gives:
\[ mgR \cos \theta = \frac{1}{2} mv^2 \implies v^2 = 2gR \cos \theta \]

### Step 6: Radial acceleration
- The radial acceleration can be expressed as:
\[ a_r = \frac{v^2}{R} = \frac{2gR \cos \theta}{R} = 2g \cos \theta \]

### Step 7: Tangential acceleration
- The tangential acceleration \( a_t \) is given by:
\[ a_t = g \sin \theta \]

### Step 8: Calculate values at the breaking point
- At the breaking point, we have found that \( \cos \theta = \frac{2}{3} \):
- Calculate \( a_r \):
\[ a_r = 2g \cos \theta = 2g \cdot \frac{2}{3} = \frac{4g}{3} \]
- Calculate \( a_t \):
\[ a_t = g \sin \theta = g \sqrt{1 - \cos^2 \theta} = g \sqrt{1 - \left(\frac{2}{3}\right)^2} = g \sqrt{\frac{5}{9}} = \frac{g \sqrt{5}}{3} \]

### Step 9: Total acceleration
- The total acceleration \( a \) is given by:
\[ a = \sqrt{a_r^2 + a_t^2} \]
Substituting the values:
\[ a = \sqrt{\left(\frac{4g}{3}\right)^2 + \left(\frac{g \sqrt{5}}{3}\right)^2} = \sqrt{\frac{16g^2}{9} + \frac{5g^2}{9}} = \sqrt{\frac{21g^2}{9}} = \frac{g \sqrt{21}}{3} \]

### Final Results
- **Radial Acceleration \( a_r \)**: \( \frac{4g}{3} \)
- **Tangential Acceleration \( a_t \)**: \( \frac{g \sqrt{5}}{3} \)
- **Total Acceleration \( a \)**: \( \frac{g \sqrt{21}}{3} \)