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 solve the problem of finding the tangential acceleration of a particle of mass `m` as it slides down from the top of a fixed smooth sphere and breaks off, we can follow these steps:

### Step 1: Understand the Situation
The particle starts from rest at the top of the sphere and slides down due to gravity. When it breaks off, the normal force acting on it becomes zero. We need to find the tangential acceleration at this point.

### Step 2: Identify Forces Acting on the Particle
At the point of breaking off, the forces acting on the particle are:
- The weight of the particle, `mg`, acting downwards.
- The normal force, `N`, which becomes zero when the particle leaves the surface of the sphere.

### Step 3: Set Up the Geometry
Let the radius of the sphere be `r`. When the particle has slid down to an angle `θ`, the height `h` it has fallen can be expressed as:
\[ h = r(1 - \cos \theta) \]

### Step 4: Apply Energy Conservation
Using the conservation of energy, the potential energy lost by the particle is converted into kinetic energy. The potential energy at the top is `mgR`, and at height `h`, it is `mg(R - h)`. Thus:
\[ mgR - mg(R - h) = \frac{1}{2} mv^2 \]
This simplifies to:
\[ mg h = \frac{1}{2} mv^2 \]
Substituting for `h`, we get:
\[ mg(r(1 - \cos \theta)) = \frac{1}{2} mv^2 \]
Cancelling `m` and rearranging gives:
\[ v^2 = 2g r(1 - \cos \theta) \]

### Step 5: Apply Centripetal Force Condition
At the point of breaking off, the centripetal force needed to keep the particle moving in a circular path is provided by the component of the weight acting towards the center of the sphere:
\[ mg \cos \theta = \frac{mv^2}{r} \]
Substituting for `v^2` from the previous step:
\[ mg \cos \theta = \frac{m(2g r(1 - \cos \theta))}{r} \]
Cancelling `m` and `r` gives:
\[ g \cos \theta = 2g(1 - \cos \theta) \]
This simplifies to:
\[ g \cos \theta + 2g \cos \theta = 2g \]
Thus:
\[ 3g \cos \theta = 2g \]
So:
\[ \cos \theta = \frac{2}{3} \]

### Step 6: Find Sin θ
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \sin^2 \theta = 1 - \left(\frac{2}{3}\right)^2 = 1 - \frac{4}{9} = \frac{5}{9} \]
Thus:
\[ \sin \theta = \frac{\sqrt{5}}{3} \]

### Step 7: Calculate Tangential Acceleration
The tangential acceleration \( a_t \) at the point of breaking off is given by:
\[ a_t = g \sin \theta \]
Substituting for \( \sin \theta \):
\[ a_t = g \left(\frac{\sqrt{5}}{3}\right) = \frac{\sqrt{5}g}{3} \]

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