A rough horizontal plate rotates with angular velocity `omega` about a fixed vertical axis. A particle or mass `m` lies on the plate at a distance `(5a)/(4)` from this axis. The coefficient of friction between the plate and the particle is `(1)/(3)` . The largest value of `omega^(2)` for which the particle will continue to be at rest on the revolving plate is

Options

A. `(a)(g)/(3a)`
B. `(b)(4g)/(5a)`
C. `(c)(4g)/(9a)`
D. `(d)(4g)/(15a)`

Explanation

To solve the problem, we need to determine the largest value of \( \omega^2 \) for which the particle will remain at rest on the rotating plate. Here are the steps to arrive at the solution:

### Step 1: Identify the Forces Acting on the Particle
The forces acting on the particle are:
- The gravitational force \( mg \) acting downward.
- The normal force \( N \) acting upward.
- The centrifugal force \( F_c \) acting outward, which is given by \( F_c = m \omega^2 r \), where \( r \) is the distance from the axis of rotation.

### Step 2: Set Up the Normal Force
Since the particle is on a horizontal plate and not moving vertically, the normal force \( N \) is equal to the gravitational force:
\[ N = mg \]

### Step 3: Determine the Frictional Force
The maximum static frictional force \( F_f \) that can act on the particle is given by:
\[ F_f = \mu N \]
Substituting for \( N \):
\[ F_f = \mu mg \]
Given that the coefficient of friction \( \mu = \frac{1}{3} \), we have:
\[ F_f = \frac{1}{3} mg \]

### Step 4: Equate the Forces
For the particle to remain at rest on the plate, the frictional force must balance the centrifugal force:
\[ F_f = F_c \]
Substituting the expressions for the forces:
\[ \frac{1}{3} mg = m \omega^2 r \]

### Step 5: Simplify the Equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{3} g = \omega^2 r \]

### Step 6: Substitute the Value of \( r \)
We are given that \( r = \frac{5a}{4} \). Substituting this into the equation:
\[ \frac{1}{3} g = \omega^2 \left(\frac{5a}{4}\right) \]

### Step 7: Solve for \( \omega^2 \)
Rearranging the equation to solve for \( \omega^2 \):
\[ \omega^2 = \frac{g}{3} \cdot \frac{4}{5a} \]
\[ \omega^2 = \frac{4g}{15a} \]

### Final Result
Thus, the largest value of \( \omega^2 \) for which the particle will continue to be at rest on the revolving plate is:
\[ \omega^2 = \frac{4g}{15a} \]