A table with smooth horizontal surface is turning at an angular speed `omega` about its axis. A groove is made on the surface along a radius and a particle is gently placed inside the groove at a distance a from the centre. Find the speed of the speed of the particle as its distance from the centre becomes L.

Options

A. `omega sqrt(L^2-a^2)`
B. `omega sqrt(a^2-L^2)`
C. `omega sqrt(2L^2-a^2)`
D. `omega sqrt(L^2)`

Explanation

To solve the problem step by step, we will analyze the motion of the particle placed in the groove of the rotating table.

### Step 1: Understand the system
We have a table rotating with an angular speed \( \omega \). A particle is placed in a groove at a distance \( a \) from the center of the table. As the table rotates, the particle will move outward along the groove until it reaches a distance \( L \) from the center.

**Hint:** Visualize the setup. The particle experiences circular motion due to the rotation of the table.

### Step 2: Write the expression for centripetal acceleration
The centripetal acceleration \( a_c \) experienced by the particle at a distance \( x \) from the center is given by:
\[ a_c = \omega^2 x \]

**Hint:** Remember that centripetal acceleration depends on the distance from the center and the angular speed.

### Step 3: Relate acceleration to velocity
We know that centripetal acceleration can also be expressed as:
\[ a_c = \frac{dv}{dt} \]
where \( v \) is the tangential speed of the particle. Thus, we can write:
\[ \frac{dv}{dt} = \omega^2 x \]

**Hint:** You can think of \( \frac{dv}{dt} \) as how the speed changes with time as the particle moves outward.

### Step 4: Use the chain rule
To relate the change in velocity to the change in position, we can use the chain rule:
\[ \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} \]
Since \( \frac{dx}{dt} = v \) (the speed of the particle), we can rewrite the equation as:
\[ \frac{dv}{dt} = v \frac{dv}{dx} \]

**Hint:** This step involves using the relationship between velocity and position to express the acceleration in terms of position.

### Step 5: Set up the equation
Now we can equate the expressions for centripetal acceleration:
\[ v \frac{dv}{dx} = \omega^2 x \]
Rearranging gives:
\[ v dv = \omega^2 x dx \]

**Hint:** This equation relates the change in speed to the change in position along the groove.

### Step 6: Integrate both sides
We need to integrate both sides to find the relationship between speed and position. The limits for \( v \) will be from \( 0 \) to \( v \) (initial speed to final speed), and for \( x \) from \( a \) to \( L \):
\[ \int_0^v v \, dv = \int_a^L \omega^2 x \, dx \]
This results in:
\[ \frac{v^2}{2} = \omega^2 \left( \frac{L^2}{2} - \frac{a^2}{2} \right) \]

**Hint:** When integrating, remember to apply the limits correctly.

### Step 7: Solve for \( v \)
Now we can simplify and solve for \( v \):
\[ \frac{v^2}{2} = \frac{\omega^2}{2} (L^2 - a^2) \]
Multiplying both sides by 2 gives:
\[ v^2 = \omega^2 (L^2 - a^2) \]
Taking the square root:
\[ v = \omega \sqrt{L^2 - a^2} \]

**Hint:** The final expression gives the speed of the particle when it reaches distance \( L \) from the center.

### Final Answer
The speed of the particle as its distance from the center becomes \( L \) is:
\[ v = \omega \sqrt{L^2 - a^2} \]