A smooth wedge with elevation `theta` is fixed in an elevator moving up with uniform acceleration `a_(0)=g//2`. The base of the wedge has a length L. Find the time taken by a particle sliding down the incline to reach the base.

Answer

To solve the problem of a particle sliding down a smooth wedge fixed in an elevator moving upward with uniform acceleration, we can follow these steps:

### Step 1: Understand the Forces Acting on the Particle
When the wedge is in an elevator accelerating upwards with acceleration \( a_0 = \frac{g}{2} \), we need to consider the forces acting on the particle. The forces include:
- The gravitational force \( mg \) acting downwards.
- A pseudo force \( ma_0 \) acting downwards due to the upward acceleration of the elevator.

### Step 2: Resolve Forces Along the Incline
The incline makes an angle \( \theta \) with the horizontal. We can resolve the gravitational force and the pseudo force into components along the incline and perpendicular to the incline.

- The component of gravitational force along the incline:
\[ F_{\text{gravity, incline}} = mg \sin \theta \]
- The component of pseudo force along the incline:
\[ F_{\text{pseudo, incline}} = ma_0 \sin \theta = m \left(\frac{g}{2}\right) \sin \theta \]

### Step 3: Calculate the Net Force Along the Incline
The net force acting along the incline is the sum of the two components:
\[ F_{\text{net}} = mg \sin \theta + m \left(\frac{g}{2}\right) \sin \theta = m \left(g + \frac{g}{2}\right) \sin \theta = m \left(\frac{3g}{2}\right) \sin \theta \]

### Step 4: Determine the Acceleration of the Particle
Using Newton's second law, \( F = ma \), we can find the acceleration \( a \) of the particle along the incline:
\[ ma = m \left(\frac{3g}{2}\right) \sin \theta \implies a = \frac{3g}{2} \sin \theta \]

### Step 5: Relate the Length of the Wedge to the Distance Traveled
The length of the base of the wedge is \( L \). The distance \( s \) traveled by the particle along the incline can be related to the base length \( L \) using the geometry of the triangle formed by the wedge:
\[ s = \frac{L}{\cos \theta} \]

### Step 6: Use Kinematic Equation to Find Time
We can use the kinematic equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the particle starts from rest, \( u = 0 \):
\[ s = \frac{1}{2} a t^2 \implies t^2 = \frac{2s}{a} \]

Substituting \( s = \frac{L}{\cos \theta} \) and \( a = \frac{3g}{2} \sin \theta \):
\[ t^2 = \frac{2 \left(\frac{L}{\cos \theta}\right)}{\frac{3g}{2} \sin \theta} = \frac{4L}{3g \sin \theta \cos \theta} \]

### Step 7: Calculate Time
Taking the square root to find \( t \):
\[ t = \sqrt{\frac{4L}{3g \sin \theta \cos \theta}} = \frac{2\sqrt{L}}{\sqrt{3g \sin \theta \cos \theta}} \]

### Step 8: Substitute for Specific Angle
If \( \theta = 30^\circ \):
- \( \sin 30^\circ = \frac{1}{2} \)
- \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)

Substituting these values:
\[ t = \frac{2\sqrt{L}}{\sqrt{3g \left(\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right)}} = \frac{2\sqrt{L}}{\sqrt{\frac{3g}{4}}} = \frac{2\sqrt{L} \cdot 2}{\sqrt{3g}} = \frac{4\sqrt{L}}{\sqrt{3g}} \]

### Final Result
Thus, the time taken by the particle to slide down the incline to reach the base is:
\[ t = \frac{4\sqrt{L}}{\sqrt{3g}} \]