A particle of mass `m` starts to slide down from the top of the fixed smooth sphere. What is the tangential acceleration when it break off the sphere ?

Options

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

Explanation

To find the tangential acceleration of a particle of mass `m` when it breaks off from the surface of a fixed smooth sphere, we can follow these steps:

### Step 1: Understand the Problem
The particle starts at the top of the sphere and slides down. We need to determine the tangential acceleration at the moment it leaves the surface of the sphere.

### Step 2: Identify Forces Acting on the Particle
At any point while sliding down, the forces acting on the particle are:
- The gravitational force acting downward, which is `mg`.
- The normal force exerted by the surface of the sphere, which acts perpendicular to the surface.

### Step 3: Resolve the Weight into Components
When the particle is at an angle θ from the vertical:
- The component of gravitational force acting tangential to the sphere is `mg sin(θ)`.
- The component of gravitational force acting normal to the sphere is `mg cos(θ)`.

### Step 4: Use Energy Conservation to Find Velocity
Using the conservation of energy, we can find the velocity of the particle at the point just before it leaves the surface:
- The height `h` from which the particle falls is given by \( h = r - r \cos(θ) = r(1 - \cos(θ)) \).
- Using the energy conservation equation:
\[ \frac{1}{2}mv^2 = mg h \]
Substituting for `h` gives:
\[ \frac{1}{2}mv^2 = mg(r(1 - \cos(θ))) \]
Simplifying this, we find:
\[ v^2 = 2g r(1 - \cos(θ)) \]

### Step 5: Find the Condition for Leaving the Surface
The particle will leave the surface when the normal force becomes zero. Therefore, we set up the equation:
\[ mg \cos(θ) = \frac{mv^2}{r} \]
Substituting for `v^2` from the previous step:
\[ mg \cos(θ) = \frac{m(2g r(1 - \cos(θ)))}{r} \]
This simplifies to:
\[ g \cos(θ) = 2g(1 - \cos(θ)) \]
Dividing by `g` and rearranging gives:
\[ \cos(θ) = \frac{2}{3} \]

### Step 6: Calculate Tangential Acceleration
The tangential acceleration `a_t` at the moment of leaving the surface is given by:
\[ a_t = g \sin(θ) \]
Using the identity \( \sin^2(θ) + \cos^2(θ) = 1 \):
\[ \sin(θ) = \sqrt{1 - \cos^2(θ)} = \sqrt{1 - \left(\frac{2}{3}\right)^2} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]
Thus,
\[ a_t = g \cdot \frac{\sqrt{5}}{3} \]

### Final Result
The tangential acceleration when the particle breaks off the sphere is:
\[ a_t = \frac{g \sqrt{5}}{3} \]